Reform Calculus: Part IIfaculty.atu.edu/mfinan/calreform2.pdf · Example 40.3 Find R x3 x4 +5dx....

Reform Calculus: Part II

Marcel B. FinanArkansas Tech University

c©All Rights Reserved

1

PREFACE

This supplement consists of my lectures of a sophomore-level mathematicsclass offered at Arkansas Tech University. The lectures are designed to ac-company the textbook ”Calculus: Single and Multivariable” written by theHarvard Consortium.This book has been written in a way that can be read by students. That is,the text represents a serious effort to produce exposition that is accessible toa student at freshmen or sophomore levels.This supplement is a continuation of the previous calculus book. The lec-tures cover Chapters 7, 8, 9, 10, and 11. These chapters are well suited for a4-hour one semester course of a second course Calculus.

Marcel B. FinanMay 2003

2

Contents

40 Integration by Substitution 5

41 The Method of Integration by Parts 10

42 Tables of Integrals 16

43 Integration by Partial Fractions. Trigonometric Substitu-tions 20

44 Numerical Approximation of Definite Integrals 29

45 Simpson’s Rule and Error Estimates 38

46 Improper Integrals 45

47 Comparison Tests for Improper Integrals 54

48 Areas and Volumes 60

49 Solids of Revolution- Arc Length 69

50 Density and Center of Mass 81

51 Applications to Physics 92

52 Applications to Economics 99

53 Continuous Random Variables: Distribution Function andDensity Function 108

54 The Median and Mean 117

55 Geometric Series 124

56 Convergence of Sequences and Series 130

57 Tests for Convergence 140

58 Power Series 150

3

59 Approximations by Taylor’s Polynomials 156

60 The Error in Taylor Polynomial Approximations: Taylor’sTheorem 162

61 Taylor Series 166

62 Constructing New Taylor Series from Known Ones 176

63 Introduction to Fourier Series 182

64 The Definition of a Differential Equation 189

65 Slope Fields 193

66 Euler’s Method 200

67 Seperation of Variables 204

68 Differential Equations: Growth and Decay Models 208

69 First Order Differential Equations Models 215

70 The Logistic Model 222

71 Second-Order Differential Equations Models 227

72 Second-Order Homogeneous Linear Differential Equations 233

4

40 Integration by Substitution

The purpose of this section is to evaluate the integral∫f ′(g(x))g′(x)dx. (1)

This is done by letting u = g(x). Then du = g′(x)dx. (See Section 23). Hence,we have the following∫

f ′(g(x))g′(x)dx =∫

f ′(u)du = f(u) + C = f(g(x)) + C (2)

The above procedure is referred to as the method of integration by sub-stitution. Thus, when the integrand looks like a compound function timesthe derivative of the inside then try using substitution to integrate. Notealso that this method of antidifferentiation reverses the chain rule of differ-entiation.The following examples illustrate the use of this method.

Example 40.1Find

∫3x2 cos x3dx.

Solution.Let u(x) = x3. Then du = 3x2dx and therefore∫

3x2 cos x3dx =∫

cos udu = sin u + C = sin x3 + C.

The method of substitution works even when the derivative of the inside ismissing a constant factor as shown in the next couple of examples.

Example 40.2Find

∫xex2+1dx.

Solution.Letting u(x) = x2 + 1 then du = 2xdx. Thus,∫

xex2+1dx = 12

∫2xex2+1dx

= 12

∫eudu = eu

2+ C

= 12ex2+1 + C.

You may wonder why 12

∫eudu = 1

2eu + C and not 1

2

∫eudu = 1

2(eu + C) =

eu

2+ C

2. The convention is always to add C to whatever antiderivative we

have calculated.

5

Example 40.3Find

∫x3√

x4 + 5dx.

Solution.Let u = x4 + 5. Then du = 4x3dx. Thus,∫

x3√

x4 + 5dx = 14

∫4x3

√x4 + 5dx

= 14

∫ √udu = 1

4· u

3232

+ C

= 16(x4 + 5)

32 + C

In some situations, the integrand consists of a fraction having a function inthe denominator and its derivative in the numerator. This leads to a naturallogarithm as seen in the next two examples.

Example 40.4Find

∫ ex

ex+1dx.

Solution.Let u = ex + 1. Then du = exdx. Thus,∫ ex

ex+1dx =

∫ duu

= ln |u|+ C

= ln |ex + 1|+ C.

Example 40.5Find

∫tan xdx.

Solution.Since tan x = sin x

cos xthen letting u = cos x we find that du = − sin xdx and

therefore ∫tan xdx = −

∫ du

u= − ln |u|+ C = − ln | cos x|+ C.

Next, we discuss the evaluation of a definite integral using the techniqueof substitution. From ( 2) we have that f(g(x)) is an antiderivative of thefunction f ′(g(x))g′(x). Applying the Fundamental Theorem of Calculus wecan write ∫ b

af ′(g(x))g′(x)dx = f(g(x))|ba = f(g(b))− f(g(a)).

6

If we let u = g(x) then the previous formula reduces to

∫ b

af ′(g(x))g′(x)dx = f(g(b))− f(g(a)) =

∫ g(b)

g(a)f ′(u)du.

Warning: When evaluating definite integrals, there is no constant of inte-gration in the final answer.

Example 40.6Compute

∫ 20 xex2

dx.

Solution.Let u(x) = x2. Then du = 2xdx, u(0) = 0, and u(2) = 4. Thus,

∫ 2

0xex2

dx =1

2

∫ 4

0eudu =

eu

2

∣∣∣∣40

=1

2(e4 − 1).

Example 40.7

Compute∫ π

40

tan3 xcos2 x

dx.

Solution.Let u = tan x. Then du = dx

cos2 x, u(0) = 0, and u(π

4) = 1. Thus,

∫ π4

0

tan3 x

cos2 xdx =

∫ 1

0u3du =

u4

4

∣∣∣∣∣1

0

=1

4.

Example 40.8

Find∫ √

1 +√

xdx.

Solution.Let u = 1 +

√x. Then du = dx

2√

x= dx

2(u−1)or dx = 2(u− 1)du. Thus,

∫ √1 +

√xdx =

∫2√

u(u− 1)du =∫(2u

√u− 2

√u)du

=∫(2u

32 − 2u

12 )du

= 2u5252

− 2u3232

+ C

= 45(1 +

√x)

52 − 4

3(1 +

√x)

32 + C.

7

Practice Problems

Exercise 40.1Find

∫x2(1 + 2x3)2dx.

Exercise 40.2Find

∫x(x2 − 4)

72 dx.

Exercise 40.3Find

∫ 1√4−x

dx.

Exercise 40.4Find

∫sin x(cos x + 5)7dx.

Exercise 40.5Find

∫x2ex3+1dx.

Exercise 40.6Find

∫ (ln x)2

xdx.

Exercise 40.7Find

∫ ex+1ex+x

dx.

Exercise 40.8Find

∫ 1+ex√

x+ex dx.

Exercise 40.9Find

∫ x+1x2+2x+19

dx.

Exercise 40.10Find

∫ x cos (x2)√sin (x2)

dx.

Exercise 40.11Compute

∫ π2

0 e− cos xdx.


∫ 81

e3√x

3√x2

dx.


∫ 41

cos√

x√x

dx.

8


∫ 1−1

11+x2 dx.


∫ 31

dx(x+7)2

.


∫ 21

sin xx

dx.

Exercise 40.17Find the exact area under the graph of f(x) = xex2

between x = 0 and x = 2.

Exercise 40.18Find the exact area enclosed by the graph of y = 4

x, the x-axis and the lines

x = 2 and x = 4.

Exercise 40.19Suppose

∫ 10 f(x)dx = 3. Calculate the following:

(a)∫ 0.50 f(2x)dx (b)

∫ 10 f(x− 1)dx (c)

∫ 1.51 f(3− 2x)dx.

Exercise 40.20If t is in years since 1990, the population P, of the world in billions can bemodeled by P (t) = 5.3e0.014t.(a) What does this model give the world population in 1990? in 2000?(b) Use the Fundamental Theorem of Calculus to find the average populationof the world during the 1990s.

Exercise 40.21Decide whether the following statements are true or false. Give an explana-tion for your answer.(a)

∫f ′(x) cos (f(x))dx = sin(f(x)) + C.

(b)∫ 1

f(x)dx = ln |f(x)|+ C.

(c)∫

x sin (5− x2)dx can be evaluated using substitution.(d)

∫sin7 θ cos6 θdθ can be written as a polynomial with cos θ as the variable.

9

41 The Method of Integration by Parts

The integration by parts formula is an antidifferentiation method which re-verses the product rule of differentiation. To see this, let u and v be twodifferentiable functions. Then the product rule asserts that the product func-tion uv is also differentiable and its derivative is given by

d

dx(uv) = u

dv

dx+

du

dxv.

This says that an antiderivative of u dvdx

+ v dudx

is the function uv. In terms ofindefinite integrals we have∫

(udv

dx+

du

dxv)dx = uv + C. (3)

But the differential of u is du = dudx

dx and that of v is dv = dvdx

dx. Hence, (3)becomes ∫

udv +∫

vdu = uv

or ∫udv = uv −

∫vdu (4)

Formula (4) is known as the integration by parts formula.

Example 41.1Find

∫xexdx.

Solution.Let u = x and dv = exdx. Then du = dx and v =

∫exdx = ex. Note that in

finding v we did not include the constant of integration. We will write theconstant C in the answer of

∫udv. Now, substituting in formula (4) to obtain∫

xexdx = xex −∫

exdx= xex − ex + C.

Remark 41.1If we chose u = ex and dv = xdx then we would have du = exdx and v = x2

2.

In this case, formula (4) yields∫xexdx =

x2

2ex −

∫ x2

2exdx

10

and the second integral is definitely worse than the one to the left of theformula we had in the previous example, i.e.

∫xexdx. It follows that for the

method to be useful it is important to choose u and dv in such a way to makethe integral on the right easier to find then the integral on the left.

Example 41.2Find

∫x sin xdx.

Solution.Let u = x and dv = sin x. Then du = dx and v = − cos x. Substituting informula (4) to obtain∫

x sin xdx = −x cos x−∫

(− cos x)dx= −x cos x +

∫cos xdx

= −x cos x + sin x + C.

There are examples which don’t look like good candidates for integrationby parts because they don’t appear to involve products, but for which themethod works well. Such examples often involves ln x or the inverse trigono-metric functions.

Example 41.3Calculate

∫ 51 ln xdx.

Solution.Let u = ln x and dv = dx. Then du = dx

xand v = x. Substituting in formula

(4) we obtain ∫ 51 ln xdx = x ln x|51 −

∫ 51 x 1

xdx

= x ln x|51 −∫ 51 dx

= x ln x|51 − x|51= 5 ln 5− ln 1− (5− 1)= 5 ln 5− 4.

Example 41.4Find

∫x3 ln xdx.

Solution.Let u = ln x and dv = x3dx. Then du = dx

xand v = x4

4. Substituting in

11

formula (4) to obtain∫x3 ln xdx = x4

4ln x−

∫ x4

41xdx

= x4

4ln x− 1

4

∫x3dx

= x4

4ln x− 1

4x4

4+ C

= x4

4ln x− 1

16x4 + C.

Sometimes evaluating an integral might require integration by parts morethan once as the following example shows.

Example 41.5Find

∫x2 cos xdx.

Solution.Let u = x2 and dv = cos xdx. Then du = 2xdx and v = sin x. Substitutingin formula (4) to obtain∫

x2 cos xdx = x2 sin x−∫

2x sin xdx= x2 sin x− 2

∫x sin xdx

= x2 sin x− 2(−x cos x + sin x) + C= x2 sin x + 2x cos x− 2 sin x + C

where we have used Example 41.2 to evaluate∫

x sin xdx.

The following example illustrates a very useful technique: Use integration byparts to transform the integral into an expression containing another copyof the same integral, possibly multiplied by a coefficient, then solve for theoriginal integral.

Example 41.6Find

∫sin2 xdx.

Solution.Method I:Using a half-angle formula to write sin2 x = 1−cos 2x

2. In this case the problem

reduces to integrating 1−cos 2x2

. That is,∫sin2 xdx = 1

2

∫(1− cos 2x)dx

= 12

∫dx− 1

2

∫cos 2xdx

= 12x− 1

4

∫cos udu

= 12x− 1

4sin u + C

= 12x− 1

4sin 2x + C

12

where the substitution u = 2x is used to evaluate the integral∫

cos 2xdx.

Method II:We now use integration by parts formula to evaluate the given integral. Letu = sin x and dv = sin xdx. Then du = cos x and v = − cos x. Substitutingin formula (4) to obtain∫

sin2 xdx = − sin x cos x−∫− cos2 xdx

= − sin x cos x +∫

cos2 xdx.

Using the trigonometric identities sin 2x = 2 sin x cos x and cos2 x+sin2 x = 1we can rewrite the right-hand side of the above integral as∫

sin2 xdx = −12sin 2x +

∫(1− sin2 x)dx

= −12sin 2x +

∫dx−

∫sin2 xdx

= −12sin 2x + x−

∫sin2 xdx

Moving the right integral to the left side to obtain

2∫

sin2 xdx = −1

2sin 2x + x

and finally dividing both sides by 2 to obtain∫sin2 xdx = −1

4sin 2x +

x

2+ C.

13

Practice Problems

Exercise 41.1Find

∫ 10 arctan xdx.

Exercise 41.2Find

∫x2e5xdx.

Exercise 41.3Find

∫x3 ln xdx.

Exercise 41.4Find

∫x2 sin xdx.

Exercise 41.5Find

∫cos2 xdx.

Exercise 41.6Find

∫ ln xx2 dx.

Exercise 41.7Find

∫x5 ln 5xdx.

Exercise 41.8Find

∫ x√5−x

dx.

Exercise 41.9Find

∫(ln x)2dx.

Exercise 41.10Find

∫x arctan x2dx.

Exercise 41.11Find

∫x5 cos x3dx.

Exercise 41.12Evaluate

∫ 100 xe−xdx.


∫ 50 ln (1 + x)dx.

14


∫ 10 arcsin xdx.

Exercise 41.15Find

∫ex sin xdx.


∫cos2 θdθ in two different ways.

Exercise 41.17Find the exact value of the area under the first arch of f(x) = x sin x.

Exercise 41.18Derive the following formula∫

cosn xdx =1

ncosn−1 x sin x +

n− 1

n

∫cosn−2 xdx.

Exercise 41.19Let f be twice differentiable function such that f(0) = 6, f(1) = 5, andf ′(1) = 2. Evaluate the integral

∫ 10 xf ′′(x)dx.

Exercise 41.20Suppose that F (a) represents the area under the graph of y = x2e−x betweenx = 0 and x = a > 0.(a) Find a formula for F (a).(b) Is F increasing or decreasing function?(c) Is F concave up or down for 0 < a < 2?

Exercise 41.21In describing the behavior of an electron, we use wave functions Ψ1, Ψ2, Ψ3, · · ·of the form

Ψn(x) = Cn sin (nπx), n = 1, 2, 3, · · ·where x is the distance from a fixed point and Cn is a positive constant.(a) Find C1 so that Ψ1 satisfies∫ 1

0(Ψ1(x))2dx = 1.

(b) For any integer n, find Cn so that∫ 1

0(Ψn(x))2dx = 1.

15

42 Tables of Integrals

If the standard integration techniques presented previously fail to yield anantiderivative, the last measure of despair are integral tables. These tablesbasically consist of collections of functions together with their antiderivatives.In order to use them, you may have to re-write the integrand function firstin a standard form listed in the table. A short table of common formulas isgiven in the back cover of your textbook. The first example shows how toapply an integral with no manipulations of any kind.

Example 42.1Find

∫cos (2x) cos (7x)dx.

Solution.Using formula II-11 with a = 2 and b = 7 we have∫

cos (2x) cos (7x)dx =1

45(7 cos (2x) sin (7x)− 2 sin (2x) cos (7x)) + C.

The following two examples use reduction formulas.

Example 42.2rm Find

∫(x2 − 3x + 2)e−4xdx.

Solution.Using formula III-14 with p(x) = x2 − 3x + 2 and a = −4 to obtain∫

(x2−3x+2)e−4xdx = −1

4(x2−3x+2)e−4x− 1

16(2x−3)e−4x− 1

32e−4x +C.

Example 42.3Find

∫sin4 xdx.

Solution.Using the reduction formula IV-17 we find∫

sin4 xdx = −14sin3 x cos x + 3

4

∫sin2 xdx

= −14sin3 x cos x− 3

8sin x cos x + 3

8x + C.

Remember that some integrals can be evaluated without the use of an integraltable as shown in the next example.

16

Example 42.4Find

∫cos3 x sin4 xdx.

Solution.This integral can be evaluated without the use of a table of integral. Indeed,let u = sin x then du = cos xdx and∫

cos3 x sin4 xdx =∫(1− sin2 x) cos x sin4 xdx

=∫(1− u2)u4du =

∫(u4 − u6)du

= u5

5− u7

7+ C

= sin5 x5− sin7 x

7+ C.

Example 42.5Find

∫sin2 x cos2 xdx.

Solution.Using the trigonometric identity cos2 x + sin2 x = 1 and Formula IV-17 wehave∫

sin2 x cos2 xdx =∫

sin2 x(1− sin2 x)dx=

∫sin2 xdx−

∫sin4 xdx

= (−12sin x cos x + 1

2x)− (−1

4sin3 x cos x + 3

4

∫sin2 xdx)

= −12sin x cos x + 1

2x + 1

4sin3 x cos x− 3

4(−1

2sin x cos x + 1

2x) + C


2x + 1

4sin3 x cos x + 3

8sin x cos x− 3

8x + C


4sin3 x cos x + 1

8x + C.

The integrands in the following require some manipulations to fit the entriesin the table.

Example 42.6 (Factoring)Find

∫ 1x2+4x+3

dx.

Solution.Factoring the denominator and then using Formula V-26 we obtain∫ 1

x2 + 4x + 3dx =

∫ 1

(x + 1)(x + 3)dx =

1

2(ln |x + 1| − ln |x + 3|) + C.

Example 42.7 (Long division)Find

∫ x2+1x2−1

dx.

17

Solution.Using the long division of polynomials we find

x2 + 1

x2 − 1= 1 +

2

x2 − 1= 1 +

2

(x− 1)(x + 1).

This with Formula V-26 yield

∫ x2 + 1

x2 − 1dx = x + 2[

1

2(ln |x− 1| − ln |x + 1|) + C = x + ln

∣∣∣∣x− 1

x + 1

∣∣∣∣+ C.

Example 42.8 (Completing the square)Find

∫ 1x2+4x+5

dx.

Solution.Completing the square we find x2 + 4x + 5 = (x + 2)2 + 1. Now, lettingu = x + 2 we can write∫ 1

x2 + 4x + 5dx =

∫ 1

u2 + 1du.

Finally, using Formula V-24 we find∫ 1

x2 + 4x + 5dx =

∫ 1

u2 + 1du = arctan u + C = arctan (x + 2) + C.

Example 42.9 (Using substitution)Find

∫xe2x2

cos (2x2)dx.

Solution.Let u = 2x2. Then du = 4xdx so that∫

xe2x2

cos (2x2)dx =1

4

∫eu cos udu.

Now, applying Formula II-9 to the last integral we find∫xe2x2

cos (2x2)dx =1

8e2x2

(cos (2x2) + sin (2x2)) + C.

18

Practice Problems.

Exercise 42.1Find

∫x5 ln xdx.

Exercise 42.2Find

∫sin x cos4 xdx.

Exercise 42.3Find

∫sin (3x) sin (5x)dx.

Exercise 42.4Find

∫sin3 (3x) cos3 (3x)dx.

Exercise 42.5Find

∫x4e3xdx.

Exercise 42.6Find

∫ x2

x2+4dx.

Exercise 42.7Find

∫sin3 xdx.

Exercise 42.8Find

∫ dx4−x2 .

Exercise 42.9Find

∫ 1x2+4x+4

dx.

Exercise 42.10Suppose n is a positive integer and Ψn = Cn sin (nπx) is the wave functionused in describing the behavior of an electron. If n and m are differentintegers, find ∫ 1

0Ψn(x) ·Ψm(x)dx.

Exercise 42.11Decide whether the following statements are true or false. Give an explana-tion for your answer.(a)

∫ 1x2+4x+5

dx involves a natural logarithm.

(b)∫ 1

x2+4x−5dx involves an arctangent.

19

43 Integration by Partial Fractions. Trigono-

metric Substitutions

Method of Integration by Partial FractionsThe method of integration by partial fractions is a technique for integratingrational functions, i.e. functions of the form

R(x) =P (x)

Q(x)

where P (x) and Q(x) are polynomials.The idea consists of writing the rational function as a sum of simpler frac-tions called partial fractions. This can be done in the following way:

Step 1. Use long division to find two polynomials r(x) and q(x) such that

P (x)

Q(x)= q(x) +

r(x)

Q(x).

Note that if the degree of P (x) is smaller than that of Q(x) then q(x) = 0and r(x) = P (x).

Step 2. Write Q(x) as a product of factors of the form (ax + b)n or (ax2 +bx+c)n where ax2+bx+c is irreducible, i.e. ax2+bx+c = 0 has no real zeros.

Step 3. Decompose r(x)Q(x)

into a sum of partial fractions in the followingway:(1) For each factor of the form (x− α)k write

A1

x− α+

A2

(x− α)2+ · · ·+ Ak

(x− α)k,

where the numbers A1, A2, · · · , Ak are to be determined.(2) For each factor of the form (ax2 + bx + c)k write

B1x + C1

ax2 + bx + c+

B2x + C2

(ax2 + bx + c)2+ · · ·+ Bkx + Ck

(ax2 + bx + c)k,

where the numbers B1, B2, · · · , Bk and C1, C2, · · · , Ck are to be determined.

20

Step 4. Multiply both sides by Q(x) and simplify. This leads to an ex-pression of the form

r(x) = a polynomial whose coefficients are combinations of Ai, Bi, and Ci.

Finally, we find the constants, Ai, Bi, and Ci by equating the coefficients oflike powers of x on both sides of the last equation.

Example 43.1Decompose into partial fractions R(x) = x3+x2+2

x2−1.

Solution.Step 1. x3+x2+2

x2−1= x + 1 + x+3

x2−1.

Step 2. x2 − 1 = (x− 1)(x + 1).Step 3. x+3

(x+1)(x−1)= A

x+1+ B

x−1.

Step 4. Multiply both sides of the last equation by (x− 1)(x + 1) to obtain

x + 3 = A(x− 1) + B(x + 1).

Expand the right hand side, collect terms with the same power of x, andindentify coefficients of the polynomials obtained on both sides:

x + 3 = (A + B)x + (B − A).

Hence, A + B = 1 and B −A = 3. Adding these two equations gives B = 2.Thus, A = −1 and so

x3 + x2 + 2

x2 − 1= x + 1− 1

x + 1+

2

x− 1.

Now, after decomposing the rational function into a sum of partial fractionsall we need to do is to integrate expressions of the form A

(x−α)n or Bx+C(ax2+bx+c)n .

Example 43.2Find

∫ 1x(x−3)

dx.

Solution.We write

1

x(x− 3)=

A

x+

B

x− 3.

21

Multiply both sides by x(x− 3) and simplify to obtain

1 = A(x− 3) + Bx

or1 = (A + B)x− 3A.

Now equating the coefficients of like powers of x to obtain −3A = 1 andA + B = 0. Solving for A and B we find A = −1

3and B = 1

3. Thus,∫ 1

x(x−3)dx = −1

3

∫ dxx

+ 13

∫ dxx−3

= −13ln |x|+ 1

3ln |x− 3|+ C

= 13ln∣∣∣x−3

x

∣∣∣+ C.

Example 43.3Find

∫ 3x+6x2+3x

dx.

Solution.We factor the denominator and split the integrand into partial fractions:

3x + 6

x(x + 3)=

A

x+

B

x + 3.

Multiplying both sides by x(x + 3) to obtain

3x + 6 = A(x + 3) + Bx= (A + B)x + 3A

Equating the coefficients of like powers of x to obtain 3A = 6 and A+B = 3.Thus, A = 2 and B = 1. Finally,∫ 3x + 6

x2 + 3xdx = 2

∫ dx

x+∫ dx

x + 3= 2 ln |x|+ ln |x + 3|+ C.

Example 43.4Find

∫ x2+1x(x+1)2

dx.

Solution.We factor the denominator and split the integrand into partial fractions:

x2 + 1

x(x + 1)2=

A

x+

B

x + 1+

C

(x + 1)2.

22

Multiplying both sides by x(x + 1)2 and simplifying to obtain

x2 + 1 = A(x + 1)2 + Bx(x + 1) + Cx= (A + B)x2 + (2A + B + C)x + A.

Equating coefficients of like powers of x we find A = 1, 2A + B + C = 0 andA + B = 1. Thus, B = 0 and C = −2. Now integrating to obtain

∫ x2 + 1

x(x + 1)2dx =

∫ dx

x− 2

∫ dx

(x + 1)2= ln |x|+ 2

x + 1+ C.

Example 43.5Find

∫ 2x2−x−1(x2+1)(x−2)

dx.

Solution.We first write

2x2 − x− 1

(x2 + 1)(x− 2)=

Ax + B

x2 + 1+

C

x− 2.

Multiply both sides by (x2 + 1)(x− 2) and simplify

2x2 − x− 1 = (Ax + B)(x− 2) + C(x2 + 1)= (A + C)x2 + (−2A + B)x− 2B + C.

Equating coefficients of like powers of x

A + C = 2, − 2A + B = −1, − 2B + C = −1.

Solving for A, B, and C we find A = B = C = 1. Thus∫ 2x2−x−1(x2+1)(x−2)

dx =∫ x+1

x2+1dx +

∫ dxx−2

=∫ x

x2+1dx +

∫ dxx2+1

+∫ dx

x−2

= 12ln (x2 + 1) + arctan x + ln |x− 2|+ C.

Trigonometric Substitutions

This section deals with integrands involving terms like√

x2 − a2,√

x2 + a2,and

√a2 − x2.

• Integrands involving√

a2 − x2,−a ≤ x ≤ a.

23

For each x in the interval [−a, a] there is a θ in the interval [−π2, π

2] such that

x = a sin θ. Thus, using the substitution x = a sin θ,−π2≤ θ ≤ π

2to obtain

√a2 − x2 =

√a2(1− sin2 θ)

=√

a2 cos2 θ = a cos θ

where we have used the Pythagorean identity cos2 θ + sin2 θ = 1. Note thatcos θ ≥ 0 since −π

2≤ θ ≤ π

2. It is important to point out here that by con-

structing a right triangle with one of the angle being θ then the hypotenuseof the triangle has length a, the opposite side has length x and the adjacent

side has length√

a2 − x2. It follows that cos θ =√

a2−x2

a. See Figure 106.

Figure 106

Example 43.6Find

∫ 1√4−x2 dx.

Solution.Let x = 2 sin θ,−π

2≤ θ < π

2. Then

√4− x2 =

√4− 4 sin2 θ =

√4 cos2 θ = 2 cos θ.

Moreover, dx = 2 cos θdθ. It follows that∫ 1√4− x2

dx =∫ 2 cos θ

2 cos θdθ = θ + C = arcsin

(x

2

)+ C.

• Integrands involving√

a2 + x2

In this case, we let x = a tan θ with −π2

< θ < π2. Such a substitution leads

to √a2 + x2 =

√a2 + a2 tan2 θ =

√a2(1 + tan2 θ) = a sec θ

since 1 + tan2 θ = sec2 θ and sec θ > 0 for −π2

< θ < π2.

24

Remark 43.1Letting θ be the angle of a right triangle with opposite side x, adjacent side

a, and hypotenuse√

a2 + x2 we find sec θ =√

a2+x2

a. See Figure 107.

Figure 107

Example 43.7Find

∫ 1√x2+9

dx.

Solution.Let x = 3 tan θ with −π

2θ < π

2. Then

√x2 + 9 =

√9 + 9 tan2 θ =

√9(1 + tan2 θ) = 3 sec θ.

Moreover, dx = 3 sec2 θdθ. Thus,

∫ 1√x2 + 9

dx =∫ 3 sec2 θ

3 sec θdθ =

∫sec θdθ = ln | sec θ + tan θ|+ C.

Now, considering a triangle with acute angle θ, opposite side x, and adjacent

side 3 we see that sec θ =√

9+x2

3and tan θ = x

3. Thus,

∫ 1√x2 + 9

dx = ln

∣∣∣∣∣√

9 + x2

3+

x

3

∣∣∣∣∣+ C.

• Integrands Involving√

x2 − a2, x ≥ a or x ≤ −a.Here, we let x = a sec θ with 0 ≤ θ < π, θ 6= π

2so that

√x2 − a2 =

√a2(sec2 θ − 1) =

√a2 tan2 θ = a |tan θ| .

25

Remark 43.2Letting θ be the angle of a right triangle with opposite side

√x2 − a2, adja-

cent side a, and hypotenuse x we find tan θ =√

x2−a2

a. See Figure 108.

Figure 108

Example 43.8Find

∫ 1x√

x2−1dx.

Solution.Let x = sec θ, 0 ≤ θ ≤ π, θ 6= π

2. Then dx = sec θ tan θdθ and

x2 − 1 = sec2 θ − 1 = tan2 θ.

Thus,∫ 1

x√

x2 − 1dx =

∫ sec θ tan θ

sec θ| tan θ|dθ = ±

∫dθ = ±θ + C =

∣∣∣sec−1x∣∣∣+ C.

Completing the Square to Use Trigonometric Substitution

Example 43.9Find

∫ 1√x2+6x+25

dx.

Solution.Completing the square we find x2 + 6x + 25 = (x + 3)2 + 16. So let x + 3 =4 tan θ,−π

2θ < π

2. Then dx = 4 sec2 θdθ and

√x2 + 6x + 25 =

√16 sec2 θ =

4 sec θ. Thus, ∫ 1√x2+6x+25

dx =∫ 4 sec2 θ

4 sec θdθ =

∫sec θdθ

= ln | sec θ + tan θ|+ C

= ln |√

x2+6x+254

+ x+34|+ C.

26

We can summarize the above substitutions in the following table

expression substitution identity√a2 − u2 u = a sin θ 1− sin2 θ = cos2 θ√a2 + u2 u = a tan θ 1 + tan2 θ = sec2 θ√u2 − a2 u = a sec θ sec2 θ − 1 = tan2 θ

27

Practice Problems

Exercise 43.1Find

∫ 1x2+4x+3

dx.

Exercise 43.2Find

∫ 14−x2 dx.

Exercise 43.3Find

∫ 11+(x+2)2

dx.

Exercise 43.4Find

∫ 1x2+4x+5

dx.

Exercise 43.5Find

∫ 1x2+4x+4

dx.

Exercise 43.6Find

∫ 1x2−1

dx.

Exercise 43.7Find

∫ 1x2−x

dx.

Exercise 43.8Find

∫ 3x+1x2−3x+2

dx.


∫ 13x−x2 dx.


∫ x2+1x2−3x+2

dx.


∫ x3

x2+3x+2dx.

Exercise 43.12Find

∫ 1x2+5x+4

dx.

Exercise 43.13Find numbers A, B and C such that

1

(x2 + 6x + 14)(x− 1)=

Ax + B

x2 + 6x + 14+

C

x− 1.

Exercise 43.14Write the fraction 1

e2x−4ex+3as the sum of two fractions.

28

44 Numerical Approximation of Definite In-

tegrals

Sometimes the integral of a function cannot be expressed with elementaryfunctions, i.e. polynomial, trigonometric, exponential, logarithmic, or asuitable combination of these. For example, the functions sin (x2) and sin x

x

don’t have simple antiderivatives, i.e. it is not possible to write down a sim-ple analytic formula in terms of elementary fucntions. In cases like these westill can find an approximate value for the integral of such functions on aninterval.

Left- and Right-hand Riemann SumsWe already know that we can approximate a definite integral

∫ ba f(x)dx by a

left- or right-hand Riemann sum for some finite n:

LEFT (f, n) =n−1∑i=0

f(xi)∆x

and

RIGHT (f, n) =n∑

i=1

f(xi)∆x

where x0, x1, · · · , xn are n+1 equally spaced points in [a, b] with x0 = a, xn =b, and ∆x = b−a

n.

Example 44.1Approximate

∫ 51

1xdx using a left- and right-hand Riemann sum with 4 inter-

vals, i.e. n = 4.

Solution.We first construct the following chart.

xi 1 2 3 4 5f(xi) 1 1

213

14

15

Thus,LEFT ( 1

x, 4) = (f(1) + f(2) + f(3) + f(4))∆x

= (1 + 12

+ 13

+ 14) · 1

= 2512≈ 2.0833

29

andRIGHT ( 1

x, 4) = (f(2) + f(3) + f(4) + f(5))∆x

= (12

+ 13

+ 14

+ 15) · 1

= 7760≈ 1.2833.

For comparison, we know that the exact value of the integral we are seekingto approximate is ∫ 5

1

1

xdx = ln 5 ≈ 1.6094.

Thus, for n = 4 the left- and right-hand Riemann sums give poor approx-imations. Figures 109(a) and 109(b) illustrate why the left and right rulesare so inaccurate.

Figure 109

By drawing pictures of the geometric regions involved one can see easily that

LEFT (f, n) ≤∫ b

af(x)dx ≤ RIGHT (f, n), when f(x) is increasing

and

RIGHT (f, n) ≤∫ b

af(x)dx ≤ LEFT (f, n), when f(x) is decreasing.

So if f(x) is increasing then LEFT (f, n) is an underestimate while RIGHT (f, n)is an overestimate. Similarly, when f is decreasing RIGHT (f, n) is an un-derestimate and LEFT (f, n) is an overestimate.

30

Midpoint ApproximationInstead of using the left- or right-endpoints we use the midpoint. Recall thatthe midpoint of an interval [x, y] is given by the midpoint formula x+y

2. The

midpoint approximation is given by

MID(f, n) =n∑

i=1

f(

xi−1 + xi

2

)∆x.


∫ 51

1xdx with the midpoint rule using 4 intervals.

Solution.Let mi denote the midpoint of the interval [xi−1, xi]. Then

mi 1.5 2.5 3.5 4.5f(mi)

23

25

27

29

Thus,

MID( 1x, 4) = (f(1.5) + f(2.5) + f(3.5) + f(4.5))∆x

= 23

+ 25

+ 27

+ 29

≈ 1.5746.

This answer is fairly close to the exact answer. The midpoint rule approxi-mates with rectangles on each subdivision that are partly above and partlybelow the graph, so the errors tend to balance out. See Figure 110.

31

Figure 110

Note that if the graph of f(x) is concave down on [a, b] then MID(f, n) isan overestimate of

∫ ba f(x)dx since the midpoint rectangle and the trapezoid

constracted by drawing the tangent line to the graph at the midpoint havethe same area since the triangles ABC and CEF are equal. (See Figure 111)

Figure 111

On the other hand, if f(x) is concave up on [a, b] then MID(f, n) is anunderestimate of

∫ ba f(x)dx. See Figure 112.

Figure 112

32

Trapezoid RuleThere is no reason why we should necessarily use rectangles to approximate∫ ba f(x)dx. We could use the function values of both endpoints of the interval

and approximate the interval by trapezoids instead. Recall that the area ofa trapezoid with base [xi−1, xi] and sides f(xi−1) and f(xi) is given by

f(xi−1) + f(xi)

2·∆x.

Thus, the trapezoid approximation is given by

TRAP (f, n) =n∑

i=1

f(xi−1) + f(xi)

2·∆x =

LEFT (f, n) + RIGHT (f, n)

2.


∫ 51

1xdx with the trapezoid rule using 4 intervals.

Solution.Using the following chart

xi 1 2 3 4 5f(xi) 1 1

213

14

15

we can write

TRAP ( 1x, 4) = (f(1)+f(2)

2+ f(2)+f(3)

2+ f(3)+f(4)

2+ f(4)+f(5)

2)∆x

= 34

+ 512

+ 724

+ 940

≈ 1.6833

This answer is fairly close to the exact answer.

Note that if the graph of f(x) is concave up then the area of each trapezoid islarger than the area under the graph so that TRAP (f, n) is an overestimateof∫ ba f(x)dx. See Figure 113(a). On the other hand, if the graph of f(x) is

concave down then the area fo each trapezoid is smaller than the area underthe graph so that TRAP (f, n) is an underestimate of the definite integral.

33

See Figure 113(b).

Figure 113

It follows from the above discussions that:

(i) If f(x) is concave up and increasing on [a,b] then

LEFT (f, n) ≤ MID(f, n) ≤∫ b

af(x)dx ≤ TRAP (f, n) ≤ RIGHT (f, n).

(ii) If f(x) is concave up and decreasing on [a,b] then

RIGHT (f, n) ≤ MID(f, n) ≤∫ b

af(x)dx ≤ TRAP (f, n) ≤ LEFT (f, n).

(iii) If f(x) is concave down and increasing on [a,b] then

LEFT (f, n) ≤ TRAP (f, n) ≤∫ b

af(x)dx ≤ MID(f, n) ≤ RIGHT (f, n).

(iv) If f(x) is concave down and decreasing on [a,b] then

RIGHT (f, n) ≤ TRAP (f, n) ≤∫ b

af(x)dx ≤ MID(f, n) ≤ LEFT (f, n).

34

Practice Problems

In Exercises 1 - 4 sketch the area given by the following approximationsto∫ ba f(x)dx. Identify each approximation as an overestimate or an underes-

timate.(a) LEFT (f, 2) (b) RIGHT (f, 2) (c) TRAP (f, 2) (d) MID(f, 2)

Exercise 44.1

Exercise 44.2

Exercise 44.3

35

Exercise 44.4

Exercise 44.5(a) Estimate

∫ 10

11+x2 dx by subdividing the interval [0, 1] into eight parts using:

(i) The left Riemann sum.(ii) The right Riemann sum.(iii) the trapeziod rule.(b) Since the exact value of the integral is π

4, you can estimate the value of π

using part (a). Explain why your first estimate is too large and your secondestimate is too small.

Exercise 44.6(a) Find LEFT (x2 + 1, 2) and RIGHT (x2 + 1, 2) for

∫ 40 (x2 + 1)dx.

(b) Illustrate your answers to part (a) graphically. Is each approximation anunderestimate or an overestimate?

Exercise 44.7(a) Find MID(x2 + 1, 2) and TRAP (x2 + 1, 2) for

∫ 40 (x2 + 1)dx.

(b) Illustrate your answers to part (a) graphically. Is each approximation anunderestimate or an overestimate?

Exercise 44.8Using the table, estimate the total distance traveled from time t = 0 to timet = 6 using LEFT, RIGHT, and TRAP.

t 0 1 2 3 4 5 6v 3 4 5 4 7 8 11

Exercise 44.9Using the figure below, order the following approximations to the integral

36

∫ 30 f(x)dx and its exact value from smallest to largest: LEFT (f, n), RIGHT (f, n), MID(f, n)

and TRAP (f, n).

Exercise 44.10(a) Values for f(x) are in the table. Which of the four approximation methodsin this section is most likely to give the best estimate of

∫ 120 f(x)dx? Estimate

the integral using this method.

x 0 3 6 9 12f(x) 100 97 90 78 55

(b) Assume f(x) si continuous with no critical points or points of inflectionon the interval 0 ≤ x ≤ 12. Is the estimate found in part (a) an overestimateor an underestimate?Explain.

Exercise 44.11Decide whether the following statements are true or false. Give an explana-tion for your answer. f is assumed to be continuous on [2, 6].(a) If n = 10, then the subdivision size is ∆x = 1

10.

(b) If we double the value of n, we make ∆x half as large.(c) LEFT(f, 10) ≤ RIGHT(f, 10).(d) As n approaches infinity, LEFT(f,n) approaches zero.

(e) LEFT(f, n)− RIGHT(f, n) = f(6)−f(2)n

.(f) Doubling n decreases the difference LEFT (f, n) − RIGHT (f, n) by ex-actly the factor 1

2.

(g) If LEFT (f, n) = RIGHT (f, n) for all n then f is a constant function.

37

45 Simpson’s Rule and Error Estimates

The trapezoid rule discussed in the previous section uses line segments toapproximate the graph of the integrand, i.e. we approximate the area underthe graph by using trapezoids. In this section, we will introduce an approx-imation technique, known as simpson’s rule, that approximates the graphof the integrand by using parabolas instead (i.e. functions with equationsy = Ax2 + Bx + C.).The process starts by dividing the interval [a, b] into n equal subintervalseach of length ∆x = b−a

nusing the partition points a = x0 < x1 < x2 <

· · · < xn = b. On the interval [xi−1, xi] we want to approximate f(x) by aquadratic function,i.e.

f(x) ≈ Ax2 + Bx + C

such thatf(xi−1) = Ax2

i−1 + Bxi−1 + Cf(xi) = Ax2

i + Bxi + Cf(mi) = Am2

i + Bmi + C

where mi is the midpoint of [xi−1, xi]. Thus,∫ xixi−1

f(x)dx ≈∫ xixi−1

(Ax2 + Bx + C)dx

=[

A3x3 + B

2x2 + Cx

]xi

xi−1

= A3(x3

i − x3i−1) + B

2(x2

i − x2i−1) + C(xi − xi−1)

= A3(xi − xi−1)(x

2i + xixi−1 + x2

i−1) + B2(xi − xi−1)(xi + xi−1) + C(xi − xi−1)

= ∆x3

[A(x2

i + xixi−1 + x2i−1) + 3

2B(xi + xi−1) + 3C

].

But

f(xi−1) + 4f(mi) + f(xi) = Ax2i−1 + Bxi−1 + C + 4A

(xi−1+xi

2

)2+ 4B

(xi−1+xi

2

)+ 4C + Ax2

i + Bxi + C

= 2[A(x2

i + xixi−1 + x2i−1) + 3

2B(xi−1 + xi) + 3C

].

It follows that∫ xi

xi−1

f(x)dx ≈ ∆x

3

(f(xi−1)

2+ 2f(mi) +

f(xi)

2

).

Hence, ∫ ba f(x)dx ≈ 1

3

∑ni=1

(f(xi−1)+f(xi)

2∆x + 2f(mi)∆x

)= 2·MID(f,n)+TRAP (f,n)

3.

38

We denote the expression on the right-hand side by SIMP (f, n).

Example 45.1Use Simpson’s rule to approximate the value of π.

Solution.First recall that

∫ 10

11+x2 dx = arctan x|10 = π

4. Let n = 2 in Simpson’s rule so

that ∆x = 12, x0 = 0, x1 = 1

2, x2 = 1, m1 = 1

4, and m2 = 3

4. Thus,∫ 1

01

1+x2 dx ≈ SIMP (f, 2)

= ∆x3

(f(x0)2

+ 2f(m1) + f(x1)2

+ f(x1)2

+ 2f(m2) + f(x2)2

)= 1

6(1

2+ 2 · 16

17+ 2

5+ 2

5+ 2 · 16

25+ 1

4)

≈ 0.7854.

This produces the approximation π = 4∫ 10

11+x2 dx ≈ 3.141568.

Error EstimatesWe next discuss error estimates of the five numerical methods discussed sofar. We define

Error = Exact V alue− Approximate V alue.

If the error is negative then the method produces an overestimate while ifthe error is positive the error produces an underestimate.In general, one does not know the exact error, for otherwise one can find theexact value. Often, we work on finding an upper bound on the error and getan idea of how much work is involved in making the error smaller and thusobtaining better estimation.

• Error in Left and Right RulesWith a bit of work, it can be shown that∣∣∣∣∣

∫ b

af(x)dx− LEFT (f, n)

∣∣∣∣∣ ≤ M(b− a)2

2n

and ∣∣∣∣∣∫ b

af(x)dx−RIGHT (f, n)

∣∣∣∣∣ ≤ M(b− a)2

2n

where M is the largest value of |f ′(x)| on the interval [a, b]. In words, theabsolute value of the error in either the left-hand rule or the right-hand rule

39

is bounded by a constant multiplied by 1n. Thus, doubling the number of in-

tervals will decrease the error by a factor of 12. Also, since M = max{|f ′(x)| :

a ≤ x ≤ b} then the error depends on how steeply the graph of f rises orfalls. See Figure 114.

Figure 114

Example 45.2Let A =

∫ 101

1xdx ≈ 2.302585. Then by the definition of LEFT(f,n)and

RIGHT(f,n) we have

LEFT (f, n) =∑n−1

i=0 f(xi)∆x=

∑n−1i=0

1xi

9n

=∑n−1

i=01

1+ 9n·i ·

9n

=∑n−1

i=09

n+9i.

Using a TI-83, one can find LEFT(f,n) for any n. For example, LEFT (f, 160)can be found by typing

sum(seq(9/(160 + 9N), N, 0, 159, 1))

and then hit enter.Similarly,

RIGHT (f, n) =n∑

i=1

9

n + 9i.

40

Thus, we have the following table

n RIGHT ( 1x, n) LEFT ( 1

x, n) A−RIGHT ( 1

x, n) A− LEFT ( 1

x, n)

10 1.960214 2.770214 0.342371 −0.46762920 2.116477 2.521477 0.186108 −0.21889240 2.205491 2.407991 0.097094 −0.10540680 2.253003 2.354253 0.049582 −0.051668160 2.277534 2.328159 0.025052 −0.025574320 2.289994 2.315307 0.012591 −0.012722

In agreement with the result above, the errors in the right-hand rule or left-hand rule approximations decrease by a factor of, roughly, 1

2when we double

the number of intervals. Also, note that the left errors are all negative sincethe function 1

xis decreasing so that LEFT is an overestimate. Similar remark

for the right errors.

• Error in the Trapezoid RuleIt can be shown that the absolute value of the error in the trapezoid rule isbounded by ∣∣∣∣∣

∫ b

af(x)dx− TRAP (f, n)

∣∣∣∣∣ ≤ K(b− a)3

n2

where K is the largest value of |f ′′(x)| in the interval [a, b]. Thus, if wedouble the number of intervals then we should expect the error to decreaseby a factor of 1

4.

Example 45.3Let A =

∫ 101

1xdx. Then by the definition of TRAP(f,n) we have

TRAP (f, n) =∑n

i=1f(xi−1)+f(xi)

2·∆x

=∑n

i=112

(1

xi−1+ 1

xi

)· 9

n

=∑n

i=112

(n

n+9(i−1)+ n

n+9i

)· 9

n.

Using a TI-83 we have the table

n TRAP ( 1x, n) A− TRAP ( 1

x, n)

10 2.365214 −0.06262920 2.318977 −0.01639240 2.306741 −0.00415680 2.303628 −0.001043160 2.302846 −0.000265320 2.302650 −0.000065

41

We see that the errors in the trapezoid rule approximations are significallysmaller than the corresponding errors for the left-hand and right-hand ruleapproximations. Moreover, in agreement with the result above, the errors inthe trapezoid rule approximations decrease by a factor of, roughly, 1

4when

we double the number of intervals. Note that the errors are all negativedue to the fact that the function 1

xis concave up so the trapezoid rule is an

overestimate. Note also, that the trapezoid rule converges to the value ofthe definite integral at a significantly faster rate than do the left-hand ruleor the right-hand rule.

• Error in Midpoint RuleAn anlaysis of error similar to ones discussed before shows that∣∣∣∣∣

∫ b

af(x)dx−MID(f, n)

∣∣∣∣∣ ≤ C · 1

n2,

where C depends on |f ′′(x)|. Hence, doubling the number of intervals willdecrease the error by, roughly, a factor of 1

4. Moreover, a more careful exam-

ination of the error would show that there is a sense in which it is typicallyon the order of 1

2the size of the error of the trapezoid rule.

Example 45.4Let A =

∫ 101

1xdx. Then by the definition of MID(f,n) we have

MID(f, n) =∑n

i=1 f(mi)∆x=

∑ni=1

1mi· 9

n

=∑n

i=12

xi−1+xi· 9

n

=∑n

i=118

2n+18i−9.

Using a TI-83 we have the following table

n MID( 1x, n) A−MID( 1

x, n)

10 2.272740 0.02984520 2.294504 0.00808140 2.300515 0.00207080 2.302064 0.000521160 2.302455 0.000130320 2.302552 0.000033

Notice that errors in the midpoint rule approximations decrease by a factorof, roughly, 1

4when we double the number of intervals. Note that the errors

42

are all positive since the function 1x

is concave up so that the midpoint ruleis an underestimate. Note also, that the error in each approximation isapproximately 1

2of the corresponding error for the trapezoid rule.

Remark 45.1The errors in either the midpoint rule of the trapezoid rule not only dependon n but also on the size of f ′′ and hence on the concavity of f. See Figure 115.

Figure 115

• Error of Simpson’s RuleIt may be shown that the absolute value of the error using Simpson’s rule isbounded by a constant multiple of 1

n4 , resulting in a dramatic improvementover both the trapezoid and midpoint rules. For Simpson’s rule, doublingthe number of intervals decreases the error by, roughly 1

16, a general fact that

we can see some evidence in the following example. The constant dependson the size of the fourth derivative of f.

Example 45.5Let A =

∫ 101

1xdx. Then by the definition of the Simpson’s rule we have

SIMP (f, n) =2MID(f, n) + TRAP (f, n)

3.

Using Examples 45.3 and 45.4 we have the following table

n SIMP ( 1x, n) A− SIMP ( 1

x, n)

10 2.303565 −0.00098020 2.302662 −0.00007740 2.302590 −0.00000580 2.302585 0.000000160 2.302455 0.000130320 2.302552 0.000033

43

Practice Problems

Exercise 45.1(a) Compute SIMP(2) for

∫ 40 (x2 + 1)dx.

(b) Use the Fundamental Theorem of Calculus to find∫ 40 (x2 + 1)dx exactly.

(c) What is the error in SIMP (x2 + 1, 2) for this integral?

Exercise 45.2In this problem you will investigate the behavior of the errors in the approx-imation of the integral ∫ 2

1

1

xdx ≈ 0.6931471806 · · ·

(a) For n = 2, 4, 8, 16, 32, 64, 128 intervals, find the left and right approxima-tions and the absolute values of the errors in each. How do the errors changeif n is doubled?(b) For the values of n in part (a), compute the midpoint and trapezoid ap-proximations and the absolute value of the errors in each. How do the errorschange if n is doubles?(c) For n = 2, 4, 8, 16, 32 intervals, compute Simpson’s error approximationsand the absolute values of the errors in each. How do the errors change if nis doubled?

Exercise 45.3(a) What is the exact value of

∫ 20 (x3 + 3x2)dx?

(b) Find SIMP (x3 + 3x2, n) for n = 2, 4, 100. What do you notice?

Exercise 45.4(a) What is the exact value of

∫ 40 exdx?

(b) Find LEFT (ex, 2), RIGHT (ex, 2), TRAP (ex, 2), MID(ex, 2), and SIMP (ex, 2).Compute the error for each.(c) Repeat part (b) with n = 4.(d) For each rule in part (b), as n goes from n = 2 to n = 4, does the errorgo down approximately as you would expect?Explain.

44

46 Improper Integrals

A very common mistake among students is when evaluating the integral∫ 1−1

1xdx. A non careful student will just argue as follows∫ 1

−1

1

xdx = [ln |x|]1−1 = 0.

Unfortunately, that’s not the right answer as we will see below. The impor-tant fact ignored here is that the integrand is not continuous at x = 0.Recall that the definite integral

∫ ba f(x)dx was defined as the limit of a left-

or right Riemann sum. We noted that the definite integral is always well-defined if:(a) f(x) is continuous on [a, b],(b) and if the domain of integration [a, b] is finite.Improper integrals are integrals in which one or both of these conditionsare not met, i.e.,

(1) The interval of integration is infinite:

[a,∞), (−∞, b], (−∞,∞),

e.g.: ∫ ∞

1

1

xdx.

(2) The integrand has an infinite discontinuity at some point c in the interval[a, b], i.e. the integrand is unbounded near c :

limx→c

f(x) = ±∞.

e.g.: ∫ 1

0

1

xdx.

An improper integral may not be well defined or may have infinite value. Inthis case we say that the integral is divergent. In case an improper integralhas a finite value then we say that it is convergent.We will consider only improper integrals with positive integrands since theyare the most common.

• Unbounded Intervals of Integration

45

The first type of improper integrals arises when the domain of integration isinfinite. In case one of the limits of integration is infinite, we define∫ ∞

af(x)dx = lim

b→∞

∫ b

af(x)dx

or ∫ b

−∞f(x)dx = lim

a→−∞

∫ b

af(x)dx.

If both limits are infinite, then we choose any number c in the domain of fand define ∫ ∞

−∞f(x)dx =

∫ c

−∞f(x)dx +

∫ ∞

cf(x)dx.

In this case, the integral is convergent if and only if both integrals on theright converge.

Example 46.1Does the integral

∫∞1

1x2 dx converge or diverge?

Solution.We have∫ ∞

1

1

x2dx = lim

b→∞

∫ b

1

1

x2dx = lim

b→∞[−1

x]b1 = lim

b→∞(−1

b+ 1) = 1.

In terms of area, the given integral represents the area under the graph off(x) = 1

x2 from x = 1 and extending infinitely to the right. The above im-proper integral says the following. Let b > 1 and obtain the area shown inFigure 116.

Figure 116

46

Then∫ b1

1x2 dx is the area under the graph of f(x) from x = 1 to x = b. As b

gets larger and larger this area is close to 1.

Example 46.2Does the improper integral

∫∞1

1√xdx converge or diverge?

Solution.We have∫ ∞

1

1√xdx = lim

b→∞

∫ b

1

1√xdx = lim

b→∞[2√

x]b1 = limb→∞

(2√

b− 2) = ∞.

So the improper integral is divergent.

Remark 46.1In general, some unbounded regions have finite areas while others have infi-nite areas. This is true whether a region extends to infinity along the x-axisor along the y-axis or both, and whether it extends to infinity on one or bothsides of an axis. For example the area under the graph of 1

x2 is finite whereasthat under the graph of 1√

xis infinite. This has to do with how fast each

function approaches 0 as x →∞. The function 1x2 approaches 0 more rapidly

than that of 1√x.

The following example generalizes the results of the previous two examples.

Example 46.3Determine for which values of p the improper integral

∫∞1

1xp dx diverges.

Solution.Suppose first that p = 1. Then∫∞

11xdx = limb→∞

∫ b1

1xdx

= limb→∞[ln |x|]b1 = limb→∞ ln b = ∞

so the improper integral is divergent.Now, suppose that p 6= 1. Then∫∞

11xp dx = limb→∞

∫ b1 x−pdx

= limb→∞[x−p+1

−p+1]b1

= limb→∞( b−p+1

−p+1− 1

−p+1).

47

If p < 1 then −p + 1 > 0 so that limb→∞ b−p+1 = ∞ and therefore the im-proper integral is divergent. If p > 1 then−p+1 < 0 so that limb→∞ b−p+1 = 0and hence the improper integral converges:∫ ∞

1

1

xpdx =

−1

−p + 1.

Example 46.4For what values of c is the improper integral

∫∞0 ecxdx convergent?

Solution.We have ∫∞

0 ecxdx = limb→∞∫ b0 ecxdx = limb→∞

1cecx|b0

= limb→∞1c(ecb − 1) = −1

provided that c < 0. Otherwise, i.e. if c ≥ 0, then the improper integral isdivergent.

Remark 46.2The previous two results are very useful and you may want to memorizethem.

Example 46.5Show that the improper integral

∫∞−∞

11+x2 dx converges.

Solution.Splitting the integral into two as follows:∫ ∞

−∞

1

1 + x2dx =

∫ 0

−∞

1

1 + x2dx +

∫ ∞

0

1

1 + x2dx.

Now, ∫ 0−∞

11+x2 dx = lima→−∞

∫ 0a

11+x2 dx = lima→−∞ arctan x|0a

= lima→−∞(arctan 0− arctan a) = −(−π2) = π

2.

Similarly, we find that∫∞0

11+x2 dx = π

2so that

∫∞−∞

11+x2 dx = π

2+ π

2= π.

• Unbounded IntegrandsSuppose f(x) is unbounded at x = a, that is limx→a+ f(x)dx = ∞. Then wedefine ∫ b

af(x)dx = lim

t→a+

∫ b

tf(x)dx.

48

Similarly, if f(x) is unbounded at x = b, that is limx→b− f(x)dx = ∞. Thenwe define ∫ b

af(x)dx = lim

t→b−

∫ t

af(x)dx.

Now, if f(x) is unbounded at an interior point a < c < b then we define∫ b

af(x)dx = lim

t→c−

∫ t

af(x)dx + lim

t→c+

∫ b

tf(x)dx.

If both limits exist then the integral on the right-hand side converges. Ifone of the limits does not exist then we say that the improper integral isdivergent.


∫ 10

1√xdx converges.

Solution.Indeed, ∫ 1

01√xdx = limt→0+

∫ 1t

1√xdx = limt→0+ 2

√x|1t

= limt→0+(2− 2√

t) = 2.

In terms of area, we pick an a > 0 as shown in Figure 117:

Figure 117

Then the shaded area is∫ 1a

1√xdx. As a approaches 0 from the right, the area

approaches the value 2.

Example 46.7Investigate the convergence of

∫ 20

1(x−2)2

dx.

49

Solution.We deal with this improper integral as follows∫ 2

01

(x−2)2dx = limt→2−

∫ t0

1(x−2)2

dx = limt→2− − 1(x−2)

|t0= limt→2−(− 1

t−2− 1

2) = ∞.

So that the given improper integral is divergent.

Example 46.8Investigate the improper integral

∫ 1−1

1xdx.

Solution.We first write ∫ 1

−1

1

xdx =

∫ 0

−1

1

xdx +

∫ 1

0

1

xdx.

On one hand we have,∫ 0−1

1xdx = limt→0−

∫ t−1

1xdx = limt→0− ln |x||t−1

= limt→0− ln |t| = ∞.

This shows that the improper integral∫ 0−1

1xdx is divergent and therefore the

improper integral∫ 1−1

1xdx is divergent.

• Improper Integrals of Mixed TypesNow, if the interval of integration is unbounded and the integrand is un-bounded at one or more points inside the interval of integration we can eval-uate the improper integral by decomposing it into a sum of improper integralwith finite interval but where the integrand is unbounded and an improperintegral with an infinite interval. If each component integrals converges, thenwe say that the original integral converges to the sum of the values of thecomponent integrals. If one of the component integrals diverges, we say thatthe entire integral diverges.


∫∞0

1x2 dx.

Solution.Note that the interval of integration is infinite and the function is undefinedat x = 0. So we write∫ ∞

0

1

x2dx =

∫ 1

0

1

x2dx +

∫ ∞

1

1

x2dx.

50

But ∫ 1

0

1

x2dx = lim

t→0−

∫ 1

t

1

x2dx = lim

t→0−−1

x|1t = lim

t→0−(1

t− 1) = ∞.

Thus,∫ 10

1x2 dx diverges and consequently the improper integral

∫∞0

1x2 dx diverges.

51

Practice Problems

Exercise 46.1Investigate the convergence of

∫∞0

xex dx.


∫ 0−∞

ex

ex+1dx.


∫∞−∞

1x2+25

dx.


∫ 40

1√16−x2 dx.


∫ 10

x4+1x

dx.


∫ 63

dx(4−x)2

.

Exercise 46.7Find the area under the curve y = xe−x for x ≥ 0.

Exercise 46.8Given that

∫∞−∞ e−x2

dx =√

π, calculate the exact value of∫ ∞

−∞e−(x−a)2

b dx.

Exercise 46.9Consider the improper integral ∫ ∞

exp ln xdx.

For what values of p does the integral converge or diverge? What is the valueof the integral when it converges?

Exercise 46.10Consider the improper integral ∫ e

0xp ln xdx.

For what values of p does the integral converge or diverge? What is the valueof the integral when it converges?

52

Exercise 46.11The gamma function is defined for all x > 0 by the rule

Γ(x) =∫ ∞

0tx−1e−tdt.

(a) Find Γ(1) and Γ(2).(b) Integrate by parts with respect to t to show that, for positive n,

Γ(n + 1) = nΓ(n).

(c) Find a simple expression for Γ(n) for positive integers.

53

47 Comparison Tests for Improper Integrals

Sometimes it is difficult to find the exact value of an improper integral byantidifferentiation, for instance the integral

∫∞0 e−x2

dx. However, it is stillpossible to determine whether an improper integral converges or diverges.The idea is to compare the integral to one whose behavior we already know,such us

• the p-integral∫∞1

1xp dx which converges for p > 1 and diverges otherwise;

• the integral∫∞0 ecxdx which converges for c < 0;

• the integral∫ 10

1xp dx which converges for p < 1 and diverges otherwise.

The comparison method consists of the following:

Theorem 47.1Suppose that f and g are continuous and 0 ≤ g(x) ≤ f(x) for all x ≥ a. Then

(a) if∫∞a f(x)dx is convergent, so is

∫∞a g(x)dx

(b) if∫∞a g(x)dx is divergent, so is

∫∞a f(x)dx.

This is only common sense: if the curve y = g(x) lies below the curve y =f(x), and the area of the region under the graph of f(x) is finite, then ofcourse so is the area of the region under the graph of g(x). For a proof ofthis theorem see Exercise 47.11.Similar results hold for the other types of improper integrals.

Example 47.1Determine whether

∫∞1

1√x3+5

dx converges.

Solution.For x ≥ 1 we have that x3 + 5 ≥ x3 so that

√x3 + 5 ≥

√x3. Thus,

1√x3+5

≤ 1√x3

. Letting f(x) = 1√x3

and g(x) = 1√x3+5

then we have that

0 ≤ g(x) ≤ f(x). From the previous section we know that∫∞1

1

x32dx is con-

vergent, a p-integral with p = 32

> 1. By the comparison test,∫∞1

1√x3+5

dx isconvergent.

The next question is to estimate such a convergent improper integral.

54

Example 47.2Estimate the value of

∫∞1

1√x3+5

dx with an error of less than 0.01.

Solution.We want to find b such that∣∣∣∣∣

∫ ∞

1

1√x3 + 5

dx−∫ b

1

1√x3 + 5

dx

∣∣∣∣∣ < 0.01.

But ∫ ∞

1

1√x3 + 5

dx =∫ b

1

1√x3 + 5

dx +∫ ∞

b

1√x3 + 5

dx.

Thus, the problem is to find b such that∣∣∣∣∣∫ ∞

b

1√x3 + 5

dx

∣∣∣∣∣ < 0.01.

From the example above, we have∫ ∞

b

1√x3 + 5

dx <∫ ∞

b

1√x3

dx =2√b.

So it suffices to choose b such that 2√b

< 0.01 or b > 40, 000, say for exampleb = 45000. In this case,∫ ∞

1

1√x3 + 5

dx ≈∫ 45,000

1

1√x3 + 5

dx = 1.69824.


∫∞4

dxln x−1

.

Solution.For x ≥ 4 we know that ln x− 1 < ln x < x. Thus, 1

ln x−1> 1

x. Let g(x) = 1

x

and f(x) = 1ln x−1

. Thus, 0 ≤ g(x) ≤ f(x). Since∫∞4

1xdx =

∫∞1

1xdx−

∫ 41

1xdx

which is divergent since∫∞1

1xdx is divergent being a p-integral with p = 1.

By the comparison test∫∞4

dxln x−1

is divergent.

Example 47.4Investigate the convergence of the improper integral

∫∞1

sin x+3√x

dx.

55

Solution.We know that −1 ≤ sin x ≤ 1. Thus 2 ≤ sin x + 3 ≤ 4. Since x ≥ 1 then2√x≤ sin x+3√

x≤ 4√

x. Note that the two integrals

∫∞1

2√xdx and

∫∞1

4√xdx are

both divergent being a multiple of a p-integral with p = 12

< 1. If we letg(x) = sin x+3√

xand f(x) = 4√

xthen we have no conclusion since

∫∞1 g(x)dx may

or may not converge and still∫∞1 g(x)dx ≤

∫∞1 f(x)dx. Now if we let g(x) =

2√x

and f(x) = sin x+3√x

then by the comparison test∫∞1

sin x+3√x

is divergent

since∫∞1 f(x)dx ≥

∫∞1 g(x)dx and

∫∞1 g(x)dx is divergent.


∫∞1 e−

12x2

dx.

Solution.If x ≥ 2 then x

2≥ 1. Multiply both sides of this inequality by x ≥ 2 to obtain

12x2 ≥ x. Now, multiply both sides of this last inequality by −1 to obtain

−12x2 ≤ −x and therefore e−

12x2 ≤ e−x since the function ex is an increasing

function. Thus, ∫ ∞

1e−

12x2

dx =∫ 2

1e−

12x2

dx +∫ ∞

2e−

12x2

dx.

But ∫ 2

1e−

12x2

dx ≈ 0.34

and ∫ ∞

2e−

12x2

dx ≤∫ ∞

2e−xdx ≤

∫ ∞

0e−xdx

so since∫∞0 e−xdx is convergent then

∫∞2 e−

12x2

dx is convergent. In conclu-

sion,∫∞1 e−

12x2

dx is convergent.

Sometimes it is laborious to find a convenient function f(x) with g(x) ≤ f(x),but we may know that g(x) is no larger than a constant multiple of f(x) forlarge enough x, and this is good enough. The most powerful test of this formin the course is this version of the limit comparison test:

Theorem 47.2Let f(x) and g(x) be two positive and continuous functions on [a,∞).

56

(a)limx→∞f(x)g(x)

= 0, or

(b)limx→∞f(x)g(x)

= L, where L is a finite positive number, or

(c)limx→∞f(x)g(x)

= ∞, then

(a) If∫∞a g(x)dx converges, then so does

∫∞a f(x)dx.

(b)∫∞a g(x)dx converges if and only if

∫∞a f(x)dx does.

(c) If∫∞a g(x)dx diverges, then so does

∫∞a f(x)dx.

Proof.(a) Suppose that limx→∞

f(x)g(x)

= 0. Let ε > 0 be given. Then there is a b > a

such that f(x)g(x)

< ε for all x ≥ b. Thus, f(x) < εg(x) for all x ≥ b. By the

comparison test, if∫∞a g(x)dx is convergent so does

∫∞a f(x)dx.

(b) Now, suppose that limx→∞f(x)g(x)

= L, where L is a finite positive constant.Let ε < L. Then there is a constant b > a such that for all x ≥ b we have∣∣∣∣∣f(x)

g(x)− L

∣∣∣∣∣ < ε.

That is,

L− ε <f(x)

g(x)< L + ε.

Thus, for x ≥ b we have (L − ε)g(x) < f(x) < (L + ε)g(x). Now the resultfollows from the comparison test.(c) Finally, suppose that limx→∞

f(x)g(x)

= ∞. Then given there is b > a such

that f(x)g(x)

≥ 1 for all x ≥ b. That is, g(x) ≤ f(x) for all x ≥ b. Therefore, if∫∞a f(x)dx converges, then

∫∞a g(x)dx converges.

Remark 47.1The Comparison Test and Limit Comparison Test also apply, modified asappropriate, to other types of improper integrals.


∫∞1

11+x2 dx is convergent.

Solution.Since the integral

∫∞1

dxx2 is convergent (p-integral with p > 1) and since

limx→∞

11+x2

1x2

= limx→∞x2

x2+1= 1 then by the limit comparison test we have

that∫∞1

dxx2+1

is also convergent.

57

Practice Problems.


∫∞1

x2

x4+1dx.


∫∞1

x2+1x3+3x+2

dx.


∫∞1

1e5x+2

dx.


∫∞1

dxx2+x

.


∫ 10

dx√x3+x

.


∫∞1

2+cos xx2 dx.

Exercise 47.7For what values of p does the integral

∫∞2

dxx(ln x)p converge or diverge?

Exercise 47.8For what values of p does the integral

∫ 21

dxx(ln x)p converge or diverge?

Exercise 47.9Find the value of a (to three decimal places) that makes∫ ∞

−∞ae−

x2

2 dx = 1.

Exercise 47.10In Planck’s Radiation Law, we encounter the integral∫ ∞

1

dx

x5(e1x − 1)

.

(a) Graph the functions y = x + 1 and y = ex. Conclude from the graph that1 + x ≤ ex for all x.(b) Replacing x by 1

xin (a), show that for all x

e1x − 1 >

1

x.

(c) Use the comparison test to show that the original integral converges.

58

Exercise 47.11Let f(x) ≥ 0 for all x ≥ a.(a) Show that the sequence an =

∫ na f(x)dx is increasing.

(b) Suppose that∫ ba f(x)dx ≤ M for all b ≥ a, where M > 0. Show that∫∞

a f(x)dx is convergent. Hint: You need to use the following result fromSection 56: If an is an increasing sequence of positive numbers such thatan ≤ M for all n ≥ 1 then liman exists.(c) Proof the comparison test of improper integrals.

59

48 Areas and Volumes

In this section we illustrate how the definite integral can be used to computethe the area of a region and the volume of certain solids. For example, theapproach to finding the area of a region will be to think of the region as ap-proximated by small elements, each of which is so geometrically simple thatits area can be calculated directly, for example the slices might be rectangles,circles or triangles. Next, the areas of each of these elements are added toobtain a Riemann sum. The limit of such Riemann sum gives the desiredarea. The same idea applies in finding the volume of a solid. We refer to thisprocess as the method of slicing.

Finding Areas by The Method of SlicingWhen calculating the area of a region using Riemann sums, slice the regioninto thin pieces in which the geometry is so simple that the area can beestimated.

Example 48.1Use horizontal slices to find the area of an isosceless triangle with vertices at(0, 0), (5, 5) and (10, 0).

Solution.We slice the triangle into n horizontal slices, each slice being approximatelya rectangle of width ∆y = yi − yi−1 and length wi. See Figure 118.

Figure 118

To find wi, note that the triangles ABC and AB’C’ are similar triangles sothat

wi

10=

5− yi

5.

60

Solving for wi, we find wi = 10− 2yi. Thus, the area of each piece is approx-imately (10− 2yi)∆y so that

Total Area ≈n∑

i=1

(10− 2yi)∆y.

Letting n →∞ we get

Total Area =∫ 5

0(10− 2y)dy = 10y − y2

∣∣∣50

= 25 square units.

Example 48.2Use horizontal slices to set up a definite integral representing the area of asemicircle of radius 7cm.

Solution.For simplicity, we assume that the circle is centered at (0, 0). We slice the

semicircle into n thin slices of width ∆y = yi − yi−1 and length 2√

49− y2i .

See Figure 119.

Figure 119

Thus, the area of a slice is approximately 2√

49− y2i ∆y so that

Total Area ≈n∑

i=1

2√

49− y2i ∆y.

61

Taking n →∞ we obtain

Total Area =∫ 70 2√

49− y2dy

= 2 · 12

[y√

49− y2 + 40 arcsin (y7)]70

= 492π cm2

• Finding Volumes by SlicingWhen calculating the volume of a solid using Riemann sums, slice the solidinto thin pieces in which the geometry is so simple that the volume can beestimated.

Example 48.3Use vertical slicing to find the volume of a cone with height 5 cm and radiusof base 5 cm.

Solution.We consider a cone centered at the origin and with axis the x-axis.We divide the cone into n thin disks each of thickness ∆x = xi − xi−1 andradius yi = xi. See Figure 120.

Figure 120

Thus, the volume of a slice is approximately πx2i ∆x so that

Total V olume ≈n∑

i=1

πx2i ∆x.

Letting n →∞ to obtain

Total V olume =∫ 5

0πx2dx = π

x3

3

∣∣∣∣∣5

0

=125

3π cm3.

62

Example 48.4The Great Pyramid of Egypt has a square base with side 755 feet long andheight 410 feet. Find the volume of the Great Pyramid in cubic feet.

Solution.The pyramid may be thought of as being made up of layers parallel to thebase. Each layer is a thin rectangular box with square base and with thick-ness ∆z. Figure 121 illustrates a slice of the pyramid.

Figure 121

The following figure illustrates a triangular cross-section of a typical layerwith a plane perpendicular at the center of the layer.

Figure 122

Let si denote the length of the base of a typical layer, then the similartriangles shown above imply that

si

755=

410− zi

410,

where zi denotes the height above the horizontal that the center of the layerlies. Solving for si one sees that the length of the rectangular box is given

63

by the formula

si = 755− 755

410zi.

The total volume is approximated by adding the volumes of the n layers

V ≈n∑

i=1

[755− 755

410zi]

2∆z.

Letting n →∞ we obtain

V =∫ 4100 [755− 755

410z]2dz = (755

410)2∫ 4100 (410− z)2dz

= (755410

)2[− (410−z)3

3]4100

= 4103

(755410

)2 ≈ 78 million ft3

64

Practice problems

In Exercises 1 - 5, write a Riemann sum and then use a definite integralrepresenting the area of the region, using the strip shown. Evaluate theintegral exactly.

Exercise 48.1

Exercise 48.2

Exercise 48.3

Exercise 48.4

65

Exercise 48.5

In Exercises 6 - 9, write a Riemann sum and then a definite integral repre-senting the volume of the region, using the slice shown. Evaluate the integralexactly.

Exercise 48.6

Exercise 48.7

66

Exercise 48.8

Exercise 48.9

Exercise 48.10Find the volume of a sphere of radius r by slicing.

67

Exercise 48.11A rectangular lake is 150 km long and 3 km wide. The vertical cross-sectionthrough the lake in Figure 123 shows that the lake is 0.2 km deep at the cen-ter. Set up and evaluate a definite integral giving the total volume of the lake.

Figure 123

68

49 Solids of Revolution- Arc Length

In this section we will discuss the use of definite integrals in solving problemsin geometry such as finding the

• volume of a solid of known cross-section,• volume of a solid of revolution,• arc length of a curve in the plane.

Recall the process of slicing which consists of the following steps:

◦ Divide the solid, region, or curve into small pieces whose volume, area,or length can be easily approximated.

◦ Add the volumes, areas, or lengths of all the pieces. (Thus obtaininga Riemann sum) that approximates the total volume, total area, or totallength.

◦ Take the limit of the Riemann sum from the previous step as n → ∞.This gives us a definite integral that gives the total volume, total area, ortotal length.

• Volume of a Solid with Known Cross SectionWhen we take a plane perpendicualr to a given solid then the common regionbetween the plane and the solid is knwon as a cross section. By knowncross sections we mean cross sections such as circles, squares, rectangles, orringsLet R be a solid lying alongside some interval [a, b] of the x-axis. See Figure124.

69

Figure 124

Divide the interval into n equal subintervals with mesh points x0 = a < x1 <x2 < · · · < xn−1 < xn = b. The planes that are perpendicular to the x-axisat the points x0, x1, x2 · · · , xn divide the solid into n slices. Since the crosssection of R changes little along a subinterval [xi−1, xi], the slab positionedalongside that subinterval can be considered a cylinder of height ∆x andwhose base has area A(xi). So the volume of the slice is

∆Vi ≈ A(xi)∆x.

The approximate total volume of the solid is

V =n∑

i=1

∆Vi ≈n∑

i=1

A(xi)∆x.

Once again we recognize a Riemann sum at the right. Letting n → ∞ weobtain the so-called Cavalieri’s principle:

V =∫ b

aA(x)dx.

Of course, the formula can be applied to any axis. For instance if a solid liesalongside some interval [a, b] on the y-axis, the formula becomes

V =∫ b

aA(y)dy.

70

Example 49.1Find the volume of a cone of radius r and height h.

Solution.Assume that the cone is placed with its vertex in the origin and its axis onthe x-axis as shown in Figure 125.

Figure 125

The cross section of the cone at each point x is a circular disk of radius y.Using similar triangles, we find y = xr

h. Hence its area is A(x) = π(xr

h)2. The

volume of the cone can now be computed by Cavalieri’s formula:

V =∫ h

0

πr2

h2x2dx =

πr2

h2[x3

3]h0 =

1

3πr2h.

Example 49.2There is a solid whose bottom face is the disk x2 + y2 ≤ 1 and every cross-section of the solid perpendicular to x-axis is a square. Find the volume ofthe solid.

Solution.We view the solid a cardboard model shown in Figure 126.

71

Figure 126

A typical cross-section is a square of length side s as shown in Figure 127.

Figure 127

The length s is given by the expression s = 2√

1− y2. Thus, the area of across section is A(y) = s2 = 4(1− y)2. By Cavalieri’s formula the volume is

V =∫ 1

−14(1− y2)dy = 4 y − 1

3y3

∣∣∣∣1−1

=4

3.

• Volume of Solids of RevolutionBy a solid of revolution we mean a solid obtained by revolving a regionaround a line. Consider the solid of revolution obtained by revolving a planeregion under the graph of f(x) around the x-axis. See Figure 128.

72

Figure 128

Each cross section is a circular disk of radius y, so its area is A(x) = πy2 =π[f(x)]2. Hence, by Cavalieri’s principle, the volume of the solid is

V =∫ b

aπ[f(x)]2dx.

Example 49.3The region bounded by the curve y =

√x + 1 and the x-axis between x = 0

and x = 9 is revolved around the x-axis. Find the volume of this solid ofrevolution.

Solution.

The solid of revolution is given in Figure 129.

73

Figure 129

A cross-section is a disk of area A(x) = π(√

x + 1)2. Thus, the total volumeis given by

V =∫ 90 π(

√x + 1)2dx =

∫ 90 π(x + 2

√x + 1)dx

= π[x2

2+ 4

3x

32 + x]90

= π 1712≈ 268.61 cubic units

If the revolution is performed around the y-axis, the roles of x and y areinterchanged so in that case the formula is∫ b

aπx2dy,

where x must be written as a function of y, i.e. x = f−1(y).

Example 49.4The curve y = x2, 0 ≤ x ≤ 1 is rotated about the y-axis. Find the volume ofthe resulting solid of revolution.

74

Solution.The solid of revolution is shown in Figure 130.

Figure 130

A cross-section is a disk of area A(y) = πy. Thus, by Vavalieri’s principle thevolume is

V =∫ 1

0πydy = π

y2

2

∣∣∣∣∣1

0

≈ 1.571. cubic units

If the region being revolved is the area between two curves y = f(x) andy = g(x), then each cross section is an annular ring (or washer) with outerradius f(x) and inner radius g(x) (assuming f(x) ≥ g(x) ≥ 0.) See Figure131.

Figure 131

75

The area of the annular ring is A(x) = π[(f(x))2−(g(x))2], hence the volumeof the solid will be

V =∫ b

aπ[(ytop)

2 − (ybottom)2]dx =∫ b

aπ[f(x)2 − g(x)2]dx.

If the revolution is performed around the y-axis, then

V =∫ b

aπ[(xright)

2 − (xleft)2]dy.

Example 49.5Find the volume of the solid obtained by revolving the area between y = x2

and y =√

x around the x-axis.

Solution.First we need to find the intersection points of these curves in order to findthe interval of integration. Setting x2 =

√x and solving for x we find (0, 0)

and (1, 1). Hence, we must integrate from x = 0 to x = 1.

V =∫ 10 π[(

√x)2 − (x2)2]dx =

∫ 10 π(x− x4)dx

= π[

12x2 − 1

5x5]10

= 3π10

.

• Arc LengthThe definite integral can also be used to compute the length of a smoothcurve (i.e. a curve with no corner points). Recall that when using the inte-gral to find the area of a region one approximates the region by rectanglesthe sum of whose areas approximates the area of the region. In finding thelength of an arc one approximates the arc by a finite set of straight line seg-ments. An approximation of the length of the arc is made by using the wellknown formula for the length of a line segment and taking a sum . A limitingprocess then yields the definite integral which is equal to the length of thearc.To elaborate the above statement, if an arc is just a line segment with end-points (x1, y1) and (x2, y2) then its length can be found by the Pythagoreantheorem:

s =√

(x2 − x1)2 + (y2 − y1)2 =√

∆x + ∆y.

Now, if the arc is the graph of a function f(x) defined on an interval [a, b],then we divide the interval into n equal subintervals. See Figure 132. The

76

corresponding points in the arc have coordinates (xi, f(xi)), so two consecu-tive points are seperated by a distance equal to

si =√

(xi − xi−1)2 + (f(xi)− f(xi−1))2.

But by the Mean Value Theorem there is a point x∗i in the interval [xi−1, xi]such that

f(xi)− f(xi−1) = f ′(x∗i )(xi − xi−1) = f ′(x∗i )∆x.

Hence,

si =√

(∆x)2 + [f ′(x∗i )∆x]2 =√

1 + [f ′(x∗i )]2∆x.

The total length of the arc is

s ≈n∑

i=1

si =n∑

i=1

√1 + [f ′(x∗i )]

2∆x.

Again, we recognize the sum on the right-hand side as a Riemann sum whichconverges to the following integral

s =∫ b

a

√1 + [f ′(x)]2dx =

∫ b

a

√1 + (

dy

dx)2dx.

Figure 132

Example 49.6Find the length of the arc defined by the curve y = x

32 between the points

(0, 0) and (1, 1).

77

Solution.Using the arc length formula we have

s =∫ 10

√1 + ( dy

dx)2dx =

∫ 10

√1 + [(x

32 )′]2dx

=∫ 10

√1 + (3

2x

12 )2dx =

∫ 10

√1 + 9x

4dx

= [ 127

(4 + 9x)32 ]10 = 1

27(13

32 − 8) unit length

78

Large Practice Problems

In Exercises 1 - 3, the region is rotated around the x-axis. Find the vol-ume.

Exercise 49.1Bounded by y = ex, y = 0, x = −1, x = 1.

Exercise 49.2Bounded by y = 4− x2, y = 0, x = −2, x = 1.

Exercise 49.3Bounded by y = cos x, y = 0, x = 0, x = π

2.

In Exercises 4 - 5, find the arc length of the given function from x = 0 tox = 2.

Exercise 49.4f(x) =

√4− x2.

Exercise 49.5f(x) =

√x3.

Exercise 49.6Find the length of the parametric curve x = cos (et), y = sin (et) for 0 ≤ t ≤ 1.

Exercise 49.7Consider the hyperbola x2 − y2 = 1 in Figure 133.

(a) The shaded region 2 ≤ x ≤ 3 is rotated around the x-axis. What isthe volume generated?(b) What is the arc length with y ≥ 0 from x = 2 to x = 3?

79

Figure 133

In Exercises 8 - 9, sketch the solid obtained by rotating each region aroundthe indicated axis. Using the sketch, show how to approximate the volumeof the solid by a Riemann sum, and hence find the volume.

Exercise 49.8Bounded by y = x3, x = 1, y = −1. Axis: y = −1.

Exercise 49.9Bounded by the first arch of y = sin x, y = 0. Axis: x-axis.

In Exercises 10 - 13 consider the region bounded by y = ex, teh x-axis, andthe lines x = 0 and x = 1. Find the volume of the following solids.

Exercise 49.10The solid obtained by rotating the region about the x-axis.

Exercise 49.11The solid obtained by rotating the region about the horizontal line y = −3.

Exercise 49.12The solid whose base is the given region and whose cross-sections perpendic-ular to the x-axis are squares.

Exercise 49.13(a) Write an integral which represents the circumference of a circle of radiusr.(b) Evaluate the integral, and show that you get the answer you expect.

80

50 Density and Center of Mass

In this section we discuss two of the applications of definite integrals, namely, the concepts of density and the center of mass.

DensityDensity is used in different situations to describe similar concepts. We presentsome of the situations.

• There is the density of a substance, which indicates how much mass pervolume unit (i.e. grams per cm3)the substance has.• There’s population density (i.e. people per mile).• Density of typed words on a page.• There’s density of fog, referring to the amount of water vapor in a volumeof unit of air ( i.e. kg/m3).

In all these cases we can use density to compute total mass, people, words,amount of water vapor, etc. If density is uniform, simply multiply the entirearea/volume/etc by the density. If density is not uniform, then we dividethe region /solid/etc. into small pieces so that the density is approximatelyuniform on each piece, then add all the pieces together to obtain a Riemannsum. Making the pieces smaller and smaller, i.e. letting n →∞ we obtain adefinite integral.

Example 50.1Find the total mass of a rod of length l and (line-)density δ(x) where x isthe distance a length element from the left end.

Solution.If the density δ was uniform then the mass ∆m of a length element ∆x would(by definition of δ) be simply

∆m = δ∆x.

However, the density is not uniform. So we can approximate the total massof the rod by slicing it into thin segments xi ≤ x ≤ xi+1 each of length ∆xwhere the density is constant there, say δ(x) ≈ δ(xi) for all x in the interval[xi, xi+1]. See Figure 134.

81

Figure 134

Then the mass of the ith piece is

∆mi ≈ δ(xi)∆x

and

M =n−1∑i=0

∆mi ≈n−1∑i=0

δ(xi)∆x.

Taking the limit n →∞ (infinitely many thin segments) we obtain

M =∫ l

0δ(x)dx.

Example 50.2Find the total mass of a circular object of radius R and (area-)density δ(r)where r is the radius of an area element from the center.

Solution.If the density δ was uniform then the mass ∆m of an area element ∆A would(by definition of δ) be simply

∆m = δ∆A.

Since the density is not uniform, we approximate the total mass of the circleby slicing it into thin concentric, circular rings, ri ≤ r ≤ ri+1 each of area∆A(ri) where the density is constant there, say δ(r) ≈ δ(ri) for all r in theinterval [ri, ri+1]. See Figure 135.

82

Figure 135

Then the mass of the ith ring is

∆mi ≈ δ(ri)∆A(ri).

But∆A(ri) = π(ri + ∆r)2 − πr2

i

= 2πri∆r + π(∆r)2.

Since each slice is assumed to be very thin then we can ignore (∆r)2. Thus,obtaining

∆mi2πriδ(ri)∆r.

It follows that the total mass is approximated by the sum

M =n−1∑i=0

∆mi ≈n−1∑i=0

2πriδ(ri)∆r.

Taking the limit n →∞ (infinitely many thin rings) we obtain

M =∫ R

02πrδ(r)dr.

Example 50.3Suppose we know the (volume-)density δ(r), i.e. the mass per volume ele-ment, of a spherical object of radius R as a function of the radius r from thecenter. Estimate the total mass of the object.

83

Solution.If the density δ was uniform then the mass ∆m of a volume element ∆Vwould (by definition of δ) be simply

∆m = δ∆V.

Since the density is not uniform, then we approximate the total mass of thesphere by slicing it into thin concentric, spherical shells, ri ≤ r ≤ ri+1 eachof volume ∆V (ri) where the density is constant there, say δ(r) ≈ δ(ri) forall r in the interval [ri, ri+1].

Then the mass of the ith spherical shell is

∆mi ≈ δ(ri)∆V (ri).

But the volume of a thin spherical shell with inner radius ri and thickness∆r is

∆Vi(ri) = 43π(ri + ∆r)3 − 4

3πr3

i

= 43π[r3

i + 3r2i ∆r + 3ri(∆r)2 + (∆r)3 − r3

i ]≈ 4

3π3r2

i ∆r = 4πr2i ∆r

since (∆r)2 ≈ 0 and (∆r)3 ≈ 0. Thus,

∆mi ≈ 4πr2i δ(ri)∆r

and

M =n−1∑i=0

∆mi ≈n−1∑i=0

4πr2i δ(ri)∆r.

Taking the limit n →∞ (infinitely many thin shells) we obtain

M =∫ R

04πr2δ(r)dr.

Center of MassThe center of mass is the so-called ”balancing point” of an object (or sys-tem.) For example, when two children are sitting on a seesaw, the point atwhich the seesaw balances, i.e. becomes horizontal is the center of mass ofthe seesaw.

Discrete Point Masses: One Dimensional Case

84

Consider again the example of two children of mass m1 and m2 sitting oneach side of a seesaw. It can be shown experimentally that the center of massis a point P on the seesaw such that

m1d1 = m2d2

where d1 and d2 are the distances from m1 and m2 to P respectively.In order to generalize this concept, we introduce an x-axis with points m1

and m2 located at points with coordinates x1 and x2. See Figure 136.

Figure 136

Since P is the balancing point then we must have

m1(x− x1) = m2(x2 − x).

Solving for x we find

x =m1x1 + m2x2

m1 + m2

.

The product m1x1 is called the moment of m1 about the origin.The above result can be extended to a system with many points as follows:

The center of mass of a system of n point-masses m1, m2, · · · , mn locatedat x1, x2, · · · , xn along the x-axis is given by the formula

x =

∑ni=1 mixi∑ni=1 mi

.

Continuous System:One Dimensional CaseNext we consider a continuous system. Suppose that we have an object lyingon the x-axis between x = a and x = b. At point x, suppose that the objecthas mass density (mass per unit length) of δ(x). To calculate the center of

85

mass, we divide the object into n pieces, each of length ∆x. On each piece,the density is nearly constant, so the mass of the ith piece is

mi ≈ δ(xi)∆x.

The center of mass is then

x =

∑ni=1 mixi∑ni=1 mi

≈∑n

i=1 xiδ(xi)∆x∑ni=1 δ(xi)∆x

.

Letting n →∞ we obtain

x =

∫ ba xδ(x)dx∫ ba δ(x)dx

.

Example 50.4Find the center of mass of a 2-meter rod lying on the x-axis with its left endat the origin if its density is δ(x) = 15x2 kg/m.

Solution.The total mass is

M =∫ 2

015x2dx = 5x3

∣∣∣20

= 40 kg.

The center of mass is

x =

∫ 20 15x3dx

40=

1

40

[x4

4

]2

0

= 1.5 m.

Two Dimensional SystemThe concept of center of mass can be applied to two dimensional objects aswell.The determination of the center of mass in two dimensions is done in asimilar manner. If a mass m is located at a point (x, y) then we define themoment of m about the x-axis to be the product my and the monentof m about the y-axis to be the product mx. Let (x, y) be the center ofmass. The procedure of finding formulas for x and y is the same as the onedimensional case. Add up the masses times their x-locations then divide bytotal mass to get x. Next, add up the masses times their y-locations thendivide by total mass to get y. Hence the two formulas:

x =

∑ni=1 ximi∑ni=1 mi

, y =

∑ni=1 yimi∑ni=1 mi

.

86

In the continuous case with uniform density δ we have

x =

∫xδA(x)dx

M, y =

∫yδA(y)dy

M

where A(x) and A(y) are the lengths of strips perpendicular to the x− andy−axes, respectively. Note that, for variable density, finding the center ofmass requires the use of double and multiple integrals, topics that are dis-cussed in Calculus III.

Example 50.5Point-masses of 4, 8, 3, and 2 kilograms are located at (−2, 3), (2,−6), (7,−3),and (5, 1) respectively. Find the coordinates of the center of mass.

SolutionApplying the formula above we find

x =(4)(3) + (8)(−6) + (3)(−3) + (2)(1)

4 + 8 + 3 + 2=

39

17

and

y =(4)(−2) + (8)(2) + (3)(7) + (2)(5)

4 + 8 + 3 + 2= −43

17.

Example 50.6Suppose an isosceles triangle with uniform density, altitude a, and base b isplaced in the xy-plane as shown in Figure 137.

Figure 137

Show that the center of mass is at x = a3, y = 0. Hence, show that the center

of mass is independent of the triangle’s base.

87

Solution.

Because the mass of the triangle is symmetrically distributed with respect tothe x-axis, then y = 0. If δ is the density of the triangle and m is its massthen

δ =m

A=

mab2

=2m

ab.

We partition the triangle into vertical strips as shown in Figure 138.

Figure 138

Let hi be the length of the base of the triangle with vertex at (a, 0) andpassing through (xi, 0). Using similar triangles we find that

hi

b=

a− xi

a.

That is, hi = ba(a−xi). Hence, the area of the ith strip is hi∆x = b

a(a−xi)∆x

and the approximate mass of the ith strip is

mi ≈b(a− xi)

a

2m

ab∆x =

2m(a− xi)

a2∆x.

The approximate moment is

n∑i=1

xi2m(a− xi)

a2∆x

and the exact moment is

limn→∞∑n

i=1 xi2m(a−xi)

a2 ∆x =∫ a0

2mx(a−x)a2 dx

= 2ma2

∫ a0 (ax− x2)dx = 2m

a2

[ax2

2− x3

3

]a0

= ma3

.

88

Hence,

x =ma3

m=

a

3.

Finally, he center of mass is the point (a3, 0).

Remark 50.1The center of mass of a body need not be within the body itself; the centerof mass of a ring or a hollow cylinder of unifrom density is located in theenclosed space, not in the object itself.

89

Practice Problems

Exercise 50.1Find the mass of a rod of length 10 cm with density δ(x) = e−x gm/cm at adistance of x cm from the left end.

Exercise 50.2A rod has length 2 meters. At a distance x meters from its left end, thedensity of the rod is given by

δ(x) = 2 + 6x gm/m.

(a) Write a Riemann sum approximating the total mass.(b) Find the exact mass by converting the sum into a definite integral.

Exercise 50.3Find the total mass of the triangular region shown below which has densityδ(x) = 1 + x grams/cm2.

Exercise 50.4The density of oil in a circular oil slick on the surface of the ocean at a dis-tance r meters from the center of the slick is given by δ(r) = 50

1+rkg/m2.

(a) If the slick extends from r = 0 to r = 10, 000 m, find a Riemann sumapproximating the total mass of oil in the slick.(b) Find the exact value of the mass of oil in the slick by turning your suminto an integral an evaluating it.(c) Within what distance r is half the oil in the slick contained?

90

Exercise 50.5An exponential model for the density of the earth’s atmosphere says that ifthe temperature of the atmosphere were constant, then the density of theatmosphere as a function of height h (in meters), above the surface of theearth would be given by

δ(h) = 1.28e−0.000124h kg/m3.

(a) Write (but do not evaluate) a sum that approximates the mass of theportion of the atmosphere from h = 0 to h = 100 m (i.e. the first 100 metersabove sea level). Assume the radius of the earth is 6400 km.(b) Find the exact answer by turning your sum in part (a) into an integral.Evaluate the integral.

Exercise 50.6Three point masses of 4 gm each are placed at x = −6, 1, 4 and 3. Whereshould a fourth point of mass 4 gm be placed to make the center of mass atthe origin?

Exercise 50.7A rod of length 3 meters with density δ(x) = 1+x2 gm/m is positioned alongthe positive x-axis, with its left end at the origin. Find the total mass andthe center of mass of the rod.

Exercise 50.8A rod with density δ(x) = 2 + sin x lies on the x-axis between x = 0 andx = π. Find the center of mass of the rod.

91

51 Applications to Physics

It has been shown how calculus can be applied to find solutions to geometricproblems such as problems concerned with computing area, volume, and arclength. In this section calculus is used to solve problems that arise fromPhysics.

• The Concept of WorkThe work done by a constant force, F , in moving an object a distance , d, isequal to the product of the force and the distance moved. That is,

W = F · d.

The SI (international) unit of work is the joule (J), which is the work doneby a force of one Newton (N) pushing a body along one meter (m). Thus, 1joule = 1 N-m. In the British system, a unit work is the foot-pound. Since1N = 0.224809 lb(1 lb = 4.45 N) and 1m = 3.28084 ft (1 ft = 0.305 m)then 1J = 0.737561 ft− lb(1 ft− lb = 1.36 J).Now, in most cases the applied force is not constant, but varies over thestraight line of motion. For example, suppose that the force, F (x), acting ona particle as it moves along the straight line from a to b varies continuously.In order to find the total work done by the force we divide the interval [a, b]into n small equal subintervals [xi−1, xi] so that the change in F is smallalong each subinterval, i.e approximately constant. Then the work done bythe force in moving the body from xi−1 to xi is approximately:

∆Wi ≈ F (xi)∆x.

So the total work is

W =n∑

i=1

∆Wi ≈n∑

i=1

F (xi)∆x.

As n →∞ the Riemann sum at the right converges to the following integral:

W =∫ b

aF (x)dx.

Example 51.1Consider a spring on the x-axis so that its right end is at x = 0 when the

92

spring is at its rest position. According to Hooke’s Law, the force needed tostretch the spring from 0 to x is proportional to x, i.e. F (x) = kx where kis called the spring constant. Find the work done in stretching the springa length of a.

Solution.

The work needed to stretch the spring from 0 to a is then the integral

W =∫ a

0kxdx =

ka2

2.

• Work Done Against GravityMass and weight are often confused. Weight is the force of gravity on anobject. The mass of an object is the quantity of matter it comprises. Anobject’s weight will vary depending on a given gravity. For example an objectthat weighs 10 pounds on earth is weightless in interstellar space. On thecontrary, an object will have the exact same mass. Gravity causes objects tofree fall with a constant acceleration (9.8meters/second2 on earth).In the SI system, the unit of mass is the kilograms. Thus, 1 Kg of iron standsfor the mass of iron present. Its weight is the amount of force exerted onit by gravity. Since force = mass × acceleration then 1 Kg of iron weighs1Kg · 9.8m/sec2 = 9.8kg·m

sec2= 9.8N. (A newton is a unit of force or weight).

The unit of mass in the British system is slug.Now, according to Newton’s Law, the force of gravity at a distance r fromthe center of the earth is

F (r) =k

r2

where k is some positive constant.The work needed to lift a body from a point at distance R1 from the centerof the Earth to another point at distance R2 is given by the integral

W =∫ R2

R1

k

r2dr = −k

r

∣∣∣∣∣R2

R1

= k(

1

R1

− 1

R2

).

Example 51.2Find the work needed to lift a body of weight 1N , 1000 km from the surfaceof the Earth. The Earth radius is 6378 Km.

93

Solution.First we will find the value of k. Since F (6378) = 1 N then k

63782 = 1, sok = 63782. Next we have R1 = 6378, R2 = 6378 + 1000 = 7378km, hence

W = 63782(1

6378− 1

7378) = 864.462N −Km.

Since 1Km = 1000m, the result in joule is

864.462N −Km = 864.462N × 1000m = 864462J.

• Work Done Filling (or Emptying) a Tank

Example 51.3A tank in the shape of a right circular cone of height 12 ft and radius 3ft is inserted into the ground with its vertex pointing down and its top atground level as shown in Figure 139. If the tank is filled with water (densityρ = 62.4 lb/ft3) to a depth of 6 ft, how much work is performed in pump-ing all the water in the tank to ground level? What changes if the water ispumped to a height of 3 ft above ground level?

Figure 139

Solution.Set up a coordinate system with the origin at the vertex of the cone and they-axis as the axis of symmetry as shown in Figure 140.

94

Figure 140

Consider a layer of distance yi from the vertex of the cone and with thickness∆y. The volume of such a circular layer is

Vi = πx2i ∆y.

Using similar triangles we find that

xi

3=

yi

12

and consequently xi = yi

4. Thus,

Vi =π

16y2

i ∆y.

Hence its weight is mi = 62.4 π16

y2i ∆y. The work done to raise it to the top of

the tank is

Wi =62.4π

16y2

i (12− yi)∆y.

Adding the works done to raise these slices we obtain the total work done toempty the tank:

W =∫ 60

62.4π16

y2(12− y)dy = 3.9π∫ 60 (12y2 − y3)dy

= 3.9π[4y3 − 1

4y4]60

= 2106π ft− lb.

Now if the water is pumped to a height of 3 ft above ground level, all thatchanges is the distance moved by the layer of water. It becomes 12+3−yi =15− yi and the work is given by

W = 62.4π∫ 6

0

y2

16(15− y)dy ≈ 9263ft− lb.

95

• Force and PressurePressure is the force per unit area acting on an object. You measure air,steam, gas pressure, and the fluid pressure in hydraulic systems in poundsper square inch (psi). However, you measure water pressure in pounds persquare foot. So the pressure in a liquid is the force per unit area exerted bythe liquid. The pressure is exerted equally in all directions and it increasesin depth.The pressure p of a liquid at a given depth h is given by the formula

p = δ · g · h

where g is the acceleration due to gravity and δ is the density of the liquid.For a constant pressure over a given area, the force on the surface is givenby

Force = Pressure · Area.

If the pressure is variable then the force is found by dividing the surface intosmall pieces in such a way that the pressure is nearly constant and then writea Riemann sum that yields a definite integral giving the total force. Sincethe pressure varies with depth, we divide the surface into horizontal strips,each of which is at an approximately constant depth. The following exampleillustrates these concepts.

Example 51.4Set up and calculate a definite integral giving the total force on the damshown in Figure 141. The density of water is δ = 1000kg/m3.

Figure 141

96

Solution.We divide the dam into horizontal strips in which the pressure is almostconstant. (See Figure 142). Let’s find the area of a strip. The equation ofthe line going through the points (1500, 0) and (1800, 100) is y = x

3− 500.

Thus,Ai = 2xi∆y

where xi = 3yi + 1500. Hence

Ai = (6yi + 3000)∆y.

The pressure is given by

pi = δgyi = 1000 · 9.8yi = 9800yi.

Thus, the total force is

F =∫ 100

09800y(6y + 3000)dy = 9800

[2y3 + 1500y2

]100

0= 1.666 · 1011.

Figure 142

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Practice Problems.

In Exercises 1 - 2, the force, F , required to compress a spring by a distancex meters is given by F (x) = 3x newtons.

Exercise 51.1Find the work done in compressing the spring from x = 1 to x = 2.

Exercise 51.2Find the work done to compress the spring to x = 3 starting at the equilibriumposition ,x = 0.

Exercise 51.3The gravitational force on a 1 kg object at a distance r meters from the centerof the earth is F (r) = 4·1014

r2 newtons. Find the work done in moving the objectfrom the surface of the earth to a height of 106 meters above the surface. Theradius of the earth is 6.4 · 106 meters.

Exercise 51.4A rectangular water tank has length 20 ft, width 10 ft, and depth 15 ft. If thetank is full, how much work does it take to pump all the water out?

Exercise 51.5A water tank is in the form of a right circular cylinder with height 20 ft andradius 6 ft. If the tank is half full of water, find the work required to pumpall of it over the top rim.

Exercise 51.6Suppose the tank in the previous problem is full of water. Find the workrequired to pump all of it to a point 10 ft above the top of the tank.

98

52 Applications to Economics

In this section we consider some applications of definite integrals used in eco-nomics such as the present and future values of a continuous income streamand the consumers’ and producers’ surplus.

Discrete Present and Future ValueMany business deals involve payments in the future. For example, when acar or a home is bought on credits, payments are made over a period of time.The future value, FV, of a payment P is the amount to which P would havegrown if deposited today in an interest bearing bank account. The presentvalue, PV, of a future payment FV, is the amount that would have to bedeposited in a bank account today to produce exactly FV in the account atthe relevant time future.If interest is compounded n times a year at a rate r for t years, then therelationship between FV and PV is given by the formula

FV = PV (1 +r

n)nt.

In the case of continuous compound interest, the forumla is given by

FV = PV ert.

Example 52.1You need $10,000 in your account 3 years from now and the interest rate is8% per year, compounded continuously. How much should you deposit now?

Solution.We have FV = $10, 000, r = 0.08, t = 3 and we want to find PV. Solving theformula FV = PV ert for PV we find PV = FV e−rt. Substituting to obtain,PV = 10, 000e−0.24 ≈ $7, 866.28.

Present and Future Value of a Continuous Income StreamWhen an income stream flows into an investment, the investment grows be-cause of the continuous flows of money and the interest compounded on themoney invested. Thus, two functions are required: a function defining theflow of money, and a function defining a function multiplier. In this section,we will find the present value and the future value of a continuous incomestream.

99

Let S(t) be the flow rate in dollars per year. To find the present value of acontinuous income stream over a period of M years we divide the interval[0, M ] into n equal subintervals each of length ∆t = M

nand with division

points 0 = t0 < t1 < · · · < tn = M. That is, over each time interval weare assuming a single payment is made. Assuming interest r is compoundedcontinuously, the present value of the total money deposited is approximatedby the following Riemann sum:

PV ≈ S(t1)e−rt1∆t + S(t2)e

−rt2∆t + · · ·S(tn)e−rtn∆t =n∑

i=1

S(ti)e−rti∆t.

Letting ∆t → 0, i.e. n →∞, we obtain

PV =∫ M

0S(t)e−rtdt.

The future value is given by

FV = erM∫ M

0S(t)e−rtdt.

Example 52.2An investor is investing $3.3 million a year in an account returning 9.4%APR. Assuming a continuous income stream and continuous compoundingof interest, how much will these investments be worth 10 years from now?

Solution.Using the formula for the future value defined above we find

FV = e.94∫ 10

03.3e−0.094tdt ≈ $54.8million.

Example 52.3At what canstant, continuous rate must money be deposited into an accountif the account contain $20,000 in 5 years? The account earns 6% interestcompounded continuously.

Solution.Given FV = $20, 000, M = 5, r = 0.06. Since S is assumed to be constantthen we have

20, 000 = S∫ 5

0e−0.06tdt.

100

Solving for S we find

S =20, 000∫ 5

0 e−0.06tdt≈ $4, 630 per year.

Supply and Demand CurvesThe quantity q manufactured and sold depends on the unit price p. In gen-eral, when the price goes up then manufacturers are willing to supply moreof the product whereas consumers are going to reduce their buyings. Sinceconsumers and manufacturers react differently to changes in price, there aretwo curves relating p and q.

The supply curve is the quantity that producers are willing to make ata given price. Thus, increasing price will increase quantity.The demand curve is the amount that will be bought by consumers at agiven price. Thus, decreasing price will increase quantity.Even though quantity is a function of price, it is the tradition to use thevertical y-axis for the variable p and the horizontal x-axis for the variable q.The supply and demand curves intersect at point (q∗, p∗) called the point ofequilibrium. We call p∗ the equilibrium price and q∗ the equilibriumquantity.

Example 52.4Find the equilibrium point for the supply function S(p) = 3p − 50 and thedemand function D(p) = 100− 2p.

Solution.Setting the equation S(p∗) = D(p∗) to obtain 3p∗−50 = 100−2p∗. By adding2p∗ + 50 to both sides we obtain 5p∗ = 150. Solving for p∗ we find p∗ = 30.Substituting this value in S(p) we find q∗ = 3(30)− 50 = 40.

Consumer and Producer SurplusThe definitions of demand and supply must be remembered:

Demand tells us the price that consumers would be willing to pay for each dif-ferent quantity. According to the law of demand, when the price increases thedemand decreases and when the price decreases the demand increases. Thegraphical representation of the relationship between the quantity demanded

101

of a good and the price of the good is known as the demand curve.Supply tells us the price that producers would be willing to charge in orderto sell the different quantities. The law of supply asserts that as the priceof a good rises, teh quantity supplied rises, and as the price of a good fallsthe quantity supplied falls. The graphical representation of the relationshipbetween the quantity supplied of a good and the price of the good is knownas the supply curve.The demand and supply curve intersects at the point of equilibrium(q∗, p∗). We call p∗ the equilibrium price and the q∗ the equilibriumquantity. See Figure 143.

Figure 143

102

Consumers’ SurplusAt the equilibrium level, the consumers’ surplus is the difference betweenwhat consumers are willing to pay and their actual expenditure: It thereforerepresents the total amount saved by consumers who were willing to paymore than p∗ per unit.To calculate the consumers’ surplus, we first calculate the consumers’ totalexpenditure. Divide the interval [0, q∗] into n equal pieces each of length ∆q.According to Figure 144, the consumers’ total expenditure is given by thesum

D(q1)∆q + D(q2)∆q + · · ·+ D(qn)∆q =n∑

i=1

D(qi)∆q.

Letting ∆q → 0 to obtain (See Figure 144)

Total Expenditure =∫ q∗

0D(q)dq.

Figure 144

Thus,

Consumers′ Surplus =∫ q∗

0(D(q)− p∗)dq.

Graphically, it is the area between demand curve and the horizontal line atp∗. See Figure 145.

103

Figure 145

Producers’ SurplusThe producers’ surplus is the extra amount earned by producers who werewilling to charge less than the selling price of p∗ per unit, and is given by

Producers′ Surplus =∫ q∗

0(p∗ − S(q))dq.

Graphically, it is the area between suuply curve and the horizontal line atp∗. See Figure 146.

Figure 146

Example 52.5The demand and supply equations are given by D(q) = 60− q2

10and S(q) =

30+ q2

5. Find the consumers’ and producers’ surplus at the equilibrium price.

104

Solution.To find the Consumers and Producers surplus under equilibrium we first needto find the equilibrium point by setting supply=demand and solving for q:

30 +q2

5= 60− q2

10implies q∗ = 10.

Substituting this into the supply (or demand) equation we find the equilib-rium price p∗ = 50. Now we use formulas of the Consumers and Producerssurplus:

Consumers′ Surplus :∫ 100

[(60− q2

10

)− 50

]dq = 10q − q3

30

∣∣∣100≈ 66.67

Producers′ Surplus :∫ 100

[50−

(30 + q2

5

)]dq = 20q − q3

15

∣∣∣100≈ 133.33

105

Practice Problems

Exercise 52.1Find the future value of an income stream of $1,000 per year, deposited intoan account paying 8% interest, compounded continuously, over a 10-year pe-riod.

Exercise 52.2Find the present and future values of an income stream of $2,000 a year, fora period of 5 years, if the continuous interest rate is 8%.

Exercise 52.3A person deposits money into a retirement account, which pays 7% interestcompounded continuously, at a rate of $1,000 per year for 20 years. Calcu-late:

(a) The balance in the account at the end of the 20 years.(b) The amount of money actually deposited into the account.(c) The interest earned during the 20 years.

Exercise 52.4(a) A bank account earns 10% interest compounded continuously. At what(constant, continuous) rate must a parent deposit money into such an accountin order to save $100,000 in 10 years for a child’s college expenses?(b) If the parents decide instead to deposit a lump sum now in order to attainthe goal of $100,000 in 10 years, how much must be deposited now?

Exercise 52.5Sales of Version 6.0 of a computer software package start out high and de-crease exponentially. At time t, in years, the sales are s(t) = 50e−t thousandsof dollars per year. After two years, Version 7.0 of the software is releasedand replaces Version 6.0. Assume that all income from software sales isimmediately invested in government bonds which pay interest at a 6% ratecompounded continuously, calculate the total value of sales of Version 6.0over the two year period.

Exercise 52.6An oil company discovered an oil reserve of 100 million barrels. For time

106

t > 0, in years, the company’s extraction plan is a linear declining functionof time as follows:

q(t) = a− bt,

where q(t) is the rate of extraction of oil in millions of barrels per year attime t and b = 0.1 and a = 10.

(a) How long does it take to exhaust the entire reserve?(b) The oil price is a constant $20 a barrel, the extraction caost per barrelis a constant $10, and the market interest rate is 10% per year, compoundedcontinuously. What is the present value of company’s profit?

Exercise 52.7The dairy industry is an example of acartel pricing: the government has setmilk prices artificially high. On a supply and demand graph, label p+, a priceabove the equilibrium price. Using the graph, describe the effect of forcingthe price up to p+ on:

(a) The consumer surplus.(b) The producer surplus.(c) The total gains from trade(Consumer surplus + producer surplus).

107

53 Continuous Random Variables: Distribu-

tion Function and Density Function

Statistics is one of the major topics of mathematics. However, the study ofstatistics requires the study of probability theory.What is probability? Before answering this question we start with some basicdefinitions.

An experiment is any operation whose outcome cannot predicted with cer-tainty. The sample space S of an experiment is the set of all possibleoutcomes for the experiment. For example, if you roll a die one time thenthe experiment is the roll of the die. A sample space for this experimentcould be S = {1, 2, 3, 4, 5, 6} where each digit represents a face of the die.An event is any subset of the sample space.

Probability is the measure of occurrence of an event. It is a number be-tween 0 and 1. If the event is impossible to occur then its probability is 0.If the occurrence is certain then the probability is 1. The closer to 1 theprobability is, the more likely the event is to occur.

A random variable α is a numerical valued function defined on a sam-ple space. For example, in rolling two dice α might represent the sum of thepoints on the two dice. Similarly, in taking samples of college students αmight represent the number of hours per week a student studies, or a stu-dent’s GPA.Random variables may be divided into two types: discrete random vari-ables and continuous random variables. A discrete random variable is onethat can assume only a countable number of values. It is usually the resultof counting. A continuous random variable can assume any value in one ormore intervals on a line. It is usually the result of a measurement.

Example 53.1State whether the random variables are discrete or continuous:

(a) The height of a student in your class.(b) The number of left-handed students in your class.

108

Solution.(a) The randon variable in this case is a result of measurement and so it is acontinuous random variable.(b) The random variable takes whole positive integers as values and so is adiscrete random variable.

In this section, we limit our discussion to continuous random variables.

Cumulative Distribution FunctionLet S be a sample space and α : S → IR be a continuous random variable.Then the cumulative distribution function (cdf) F (t) of the variable αis defined as follows

F (t) = P (α ≤ t) = P ({s ∈ S : α(s) ≤ t})

i.e., F (t) is equal to the probability that the variable α assumes values, whichare less than or equal to t.

Example 53.2If we think of an electron as a particle, the function

P (r) = 1− (2r2 + 2r + 1)e−2r

is the cumulative distribution function of the distance, r, of the electron ina hydrogen atom from the center of the atom. The distance is measured inBohr radii. (1 Bohr radius = 5.29×10−11m.) Interpret the meaning of P (1).

Solution.P (1) = 1−5e−2 ≈ 0.32. This numbers says that the electron is within 1 Bohrradius from the center of the atom 32% of the time.

Next, we discuss the properties of the cumulative distribution function F (t)for a continuous random variable α.

Theorem 53.1The cumulative distribution function of a continuous random variable α sat-isfies the following properties:

(a) 0 ≤ F (t) ≤ 1.

109

(b) P (a ≤ α ≤ b) = F (b)− F (a).(c) F (t) is a non-decreasing function, i.e. if a ≤ b then F (a) ≤ F (b).(d) F (t) → 0 as t → −∞ and F (t) → 1 as t →∞.

Proof.We will prove (a) - (c).(a) Since F is defined in terms of a probability then 0 ≤ F (t) ≤ 1.(b) Let a and b be two real numbers with a < b. Then

P (a ≤ α ≤ b) = P (α ≤ b)− P (α ≤ a)= F (b)− F (a).

(c) This is a result of (b).

Figure 147 illustrates a representative cdf.

Figure 147

Probability Density FunctionIf F (t) is the cumulative distribution function for a continuous random vari-able α then the probability density function (pdf) f(t) for α satisfies

f(t) = F (t),

i.e., f(t) is the derivative of the cumulative distribution function F (t).It follows from the definition of density function and the Fundamental The-orem of Calculus that

F (t) =∫ t

−∞f(x)dx,

110

i.e. for every real number t, F (t) is the area under the graph of f to the leftof t. Moreover,

P (a ≤ α ≤ b) = P (α ≤ b)− P (α ≤ a)= F (b)− F (a)

=∫ b−∞ f(t)dt−

∫ a−∞ f(t)dt

=∫ ba f(t)dt

Theorem 53.2 (More Properties of f(t))(a) f(t) ≥ 0 for all t.(b)

∫−∞∞f(t)dt = 1.

(c) limt→−∞ f(t) = 0 = limt→∞ f(t).

Proof.(a) Since f(t) = F ′(t) and F (t) is nondecreasing then F ′(t) ≥ 0 and sof(t) ≥ 0.(b) Since limt→∞ F (t) = 1 then

∫∞−∞ f(t)dt = 1.

(c) This follows from the fact that the graph of F (t) levels off when t → ±∞.

The density function for a continuous random variable , the model for somereal-life population of data, will usually be a smooth curve as shown in Fig-ure 148.

Figure 148

Example 53.3Suppose that the function f(t) defined below is the density function of some

111

random variable α.

f(t) =

{e−t t ≥ 0,0 t < 0.

Compute P (−10 ≤ α ≤ 10).

Solution.

P (−10 ≤ α ≤ 10) =∫ 10−10 f(t)dt

=∫ 0−10 f(t)dt +

∫ 100 f(t)dt

=∫ 100 e−tdt

= −e−t|100 = 1− e−10.

112

Practice ProblemsDecide if the function graphed in Exercises 1 - 4 is a probability densityfunction or a cumulative distribution function. Give reasons. Find the valueof c. Sketch and label the other function. That is, sketch and label the pdfif the problem shows a cdf, and the cdf if the problem shows a pdf.

Exercise 53.1

Exercise 53.2

Exercise 53.3

113

Exercise 53.4

Exercise 53.5A large number of people take a standardized test, receiving scores describedby the density function p graphed in Figure 149. Does the density functionimply that most people receive a score near 50? Explain why or why not?

Figure 149

Exercise 53.6Suppose F (x) is the cumulative distribution function for heights (in meters)of trees in a forest.

(a) Explain in terms of trees the meaning of the statement F (7) = 0.6.(b) Which is greater, F (6) or F (7)? Justify your answer in terms of trees.

Exercise 53.7An experiment is done to determine the effect of two new fertilizers A andB on the growth of a species of peas. The cumulative distribution functions

114

of the heights of the mature peas without treatment and treated with each ofA and A are graphed in Figure 150.

(a) About what height are most of the unfertilized plants?(b) Explain in words the effect of the fertilizerd A and B on the mature heightof the plants.

Figure 150

Exercise 53.8Figure 151 shows a density function and the corresponding cumulative distri-bution function. Which curve represents the density function and which rep-resents the cumulative distribution function? Give a reason for your choice.

Figure 151

Exercise 53.9After measuring the duration of many telephone calls, the telephone com-pany found their data was well approximated by the density function p(x) =

115

0.4e−0.4x, where x is the duration of call, in minutes.

(a) What percentage of calls last between 1 and 2 minutes?(b) What percentage of calls last 1 minute or less?(c) What percentage of calls last 3 minutes or more?(d) Find the cumulative distribution function.

Exercise 53.10Students at the University of California were surveyed and asked their gradepoint average.(The GPA ranges from 0 to 4, where 2 is just passing.) Thedistribution of GPAs is shown in Figure 152.

(a) Roughly what fraction of students are passing?(b) Roughly what fraction of students have honor grades (i.e. GPA above 3)?(c) Why do you think there is a peak around 2?(d) Sketch the cumulative distribution function.

Figure 152

116

54 The Median and Mean

In this section we discuss two ways of measuring the ”average” value for adistribution function, namely, the median and the mean.

The MedianThe median income in the U.S. is the income M such that half the popu-lation earn incomes ≤ M (so the other half earn incomes ≥ M). In termsof probability, we can think of income as a random variable α. Then theprobability that α ≤ M is 1/2, and the probability that α ≥ M is also 1/2.In general, if α is a continuous random variable then the median of α is anumber M such that

P (α ≤ M) =1

2.

If p(x) is the probability density function of α then we can calculate themedian by solving the equation∫ M

−∞p(x)dx =

1

2.

Graphically, half the area under the graph of p(x) is to the left of M. SeeFigure 153.

Figure 153

Example 54.1The time in minutes between individuals joining the line at an Ottawa PostOffice is a random variable with the density function

p(x) =

{2e−2x, x ≥ 0

0, x < 0.

117

Find the median time between individuals joining the line and interpret youranswer.

Solution.To find the median, we must solve∫ M

−∞p(x)dx =

1

2

or ∫ M

02e−2xdx =

1

2.

Thus,

−e−2x|M0 = 12

1− e−2M = 12

e−2M = 12

2M = ln 2M = ln 2

2≈ 0.3466

This means that half the people get in line less than 0.3466 minutes (about21 seconds) after the previous person, while half arrive more than 0.3466minutes later.

Sometimes we cannot solve the equation∫M−∞ p(x)dx = 1/2 for M analyt-

ically, as the next example shows.

Example 54.2Find the median for the random variable with density function

p(x) =

{30x4(1− x), x ≥ 0

0, x < 0

Solution. ∫M−∞ p(x)dx = 1

2∫M0 30x4(1− x)dx = 1

2∫M0 (30x4 − 30x5)dx = 1

2

6x5 − 5x6|M0 = 12

12M5 − 10M6 − 1 = 0.

118

There is no general analytical method for obtaining the solution, the onlymethod we can use is numerical. Using a graphing calculator we find M ≈0.735.

The MeanLet S be a sample space and α : S −→ IR be a continuous random variable.The mean of α is the average value of all the values of α(s) where s is anelement of S. That is,

E(α) =

∑s∈S α(s)

|S|where |S| denotes the number of elements of S. Finding the above sum whenS has a large number of elements is tedious. Instead, we will try to replacethe sum by a definite integral.Let p(x) be the density function (for a continuous random variable α) definedon a finite interval [a, b]. We will find the mean by the method of slicing.Break up the interval into n subintervals [xi−1, xi], each of length ∆x, as wedid for Riemann sums. From the previous section, we have

P (xi−1 ≤ α ≤ xi) ≈ p(xi)∆x,

i.e., the approximate area under the graph of p over [xi−1, xi]). If we let Si

be the set of elements in S such that xi−1 ≤ α(s) ≤ xi then

P (xi−1 ≤ α ≤ xi) =|Si||S|

.

(This a result from probability theory, if A is a subset of S then P (A) = |A||S| .)

Thus, |Si| = |S|p(xi)∆x. Since for each s in Si we have α(s) ≈ xi thenxip(xi)|S|∆x represents the sum of all values satisfying xi−1 ≤ α(s) ≤ xi.Adding together all of the values of α(s) and averaging we find

E(X) ≈n∑

i=1

xip(xi)∆x.

Now these approximations get better as n →∞, and we notice that the sumabove is a Riemann sum converging to

E(α) =∫ b

axp(x)dx.

The above argument applies if either a or b are infinite. In this case, one hasto make sure that all improper integrals in question converge.

119

Example 54.3Let α be a continuous random variable with density function

p(x) =

{2e−2x, x ≥ 0

0, x < 0.

Find the mean of α.

Solution.Using the integral formula for the mean and integration by parts we find

E(α) =∫∞0 2xe−2xdx

= limb→∞∫ b0 2xe−2xdx

= limb→∞[−xe−2x − 1

2e−2x

]b0

= limb→∞(

12− 1

2e−2b − be−2b

)= 1

2

Normal DistributionWhen the graph of the density function p(x) of a random variable is bell-shaped and symmetric about the line going through its central peak then wesay that α has a normal distribution. Its density function has the formula

p(x) =1

σ√

2πe−

(x−µ)2

2σ2

where µ > 0 and σ > 0 called the standard deviation.

Example 54.4Consider the normal distribution, p(x).

(a) Show that p(x) has maximum when x = µ.(b) Show that p(x) has points of inflection at x = µ + σ and x = µ− σ.(c) Describe in your own words what µ and σ tell you about the distribuion.(d) Represents P (µ− σ ≤ α ≤ µ + σ) graphically.

Solution (a) Finding the derivative p′(x) we obtain

p′(x) = −(x− µ)

σ3√

2πe−

(x−µ)2

2σ2 .

120

Thus, x = µ is the only critical number. Finding the second derivative weobtain

p′′(x) =1

σ3√

2πe−

(x−µ)2

2σ2

((x− µ)2

σ2− 1

).

Hence, p′′(µ) = −1 < 0 so that x = µ is a maximum.(b) Solving the equation p′′(x) = 0 we find x = µ± σ. If µ− σ < x < µ + σ

then |x−µ| < σ and therefore (x−µ)2

σ2 < 1 so that p′′(x) < 0 and the graph ofp is concave down. Similarly, we see that p′′(x) > for x < µ−σ or x > µ+σso that the graph of p(x) is concave up. It follows that p(x) has points ofinflection at x = µ± σ.(c) It can be shown that E(x) = µ so that µ represents the mean of thedistribution. Statisticians use the standard deviation of a continuous randomvariable α as a way of measuring its dispersion, or the degree to which is it”scattered.” The larger σ is, the wider the bell.(c) The shaded region in Figure 154 represents P (µ− σ ≤ α ≤ µ + σ).

Figure 154

121

Practice Problems

Exercise 54.1(a) Using a calculator or a computer, sketch graphs of the density functionof the normal distribution

p(x) =1

σ√

2πe−

(x−µ)2

2σ2

(i) For µ = 5 and σ = 1, 2, 3.(ii) For σ = 1 and µ = 4, 5, 6.(b) Explain how the graphs confirm that µ is the mean of the distribution andthat σ is a measure of how close;y the data is clustered around the mean.

Exercise 54.2Suppose that x measures the time (in hours) it takes for a student to completean exam. Assume that all students are done within two hours and the densityfunction for x is given by

p(x) =

{14x3, if 0 < x < 20, otherwise.

(a) What proportion of students take between 1.5 and 2.0 hours to finish theexam.(b) What is the mean time for students to complete the exam?(c) Compute the median of this distribution.

Exercise 54.3In 1950 an experiment was done observing the time gaps between successivecars on the Arroyo Seco Freeway. The data show that the density functionof these time gaps was given approximately by

p(x) = ae−0.122x

where x is the time in seconds and a is a constant.

(a) Find the value of a.(b) Find P, the cumulative distribution function.(c) Find the median and mean time gap.(d) Sketch rough graphs of p and P.

122

Exercise 54.4The distribution of IQ scores is often modeled by the normal distribution withmean 100 and standard deviation 15.

(a) Write a formula for the density distribution if IQ scores.(b) Estimate the fraction of the population with IQ between 115 and 120.

Exercise 54.5The speeds of cars on a road are approximately normally distributed withmean 58 km/hr and standard deviation 4 km/hr.

(a) What is the probability that a randomly selected car is going between60 and 65 km/hr.(b) What fraction of all cars are going slower than 52 km/hr?

Exercise 54.6Let P (x) be the cumulative distribution function for the income distributionin the US in 1973 (income is measured in thousands of dollars). Some valuesof P (x) are int the following table:

Income x (thousands) 1 4.4 7.8 12.6 20 50P(x)(%) 1 10 25 50 75 99

(a) What fraction of the population made between $20,000 and $50,000.(b) What was the median income?(c) Sketch a density function for this distribution. Where, approximately,does your density function have a maximum? What is the significance of thispoint, in terms of income distribution? How can you recognize this point onthe graph of the density function and on the graph of the cumulative distri-bution?

123

55 Geometric Series

Geometric series are frequently used in mathematics. They provide a goodintroduction of infinite series which we will discuss in the next section.A finite geometric series is a finite sum that starts with an initial value aand then obtain each new term of the series by multiplying by a commonratio r. Thus, an example of such a series is

a + ar + ar2 + ar3 + · · ·+ arn−1.

Example 55.1A defense contractor doubles its production every year after producing 1000units in the first year. Let S10 be the production after 10 years. Write S10

as a geometric series.

Solution.The sum is given by the finite geometric series

S10 = 1000 + 1000(2) + 1000(22) + · · ·+ 1000(29).

Note that the above sum becomes extremely difficult if the number of yearsis very large. Thus, one might asks for a simple formula to find the sum forany number of years. We will derive next a formula for any finite geometricseries.Let

Sn = a + ar + ar2 + · · ·+ arn−1.

Multiply both sides of the above equation by r to obtain

rSn = ar + ar2 + ar3 + · · ·+ arn.

Now subtracting this equation from the above equation, we get

(1− r)Sn = a− arn = a(1− rn).

Finally, dividing both sides of the previous equation by (1− r) to obtain

Sn = a1− rn

1− r.

Provided that r 6= 1. If r = 1 then Sn = na.

124

Example 55.2Use the above sum to find the number of units the contractor produces after10 years.

Solution.Using the above formula with n = 10, r = 2, and a = 1000 to obtain

S10 = 10001− 210

1− 2= 1, 023, 000.

A finite arithmetic series is a series that starts with initial value a andsuch that the new term is obtained by adding a constant r. Thus, an exampleof a finite arithmetic series is the series

Sn = a + (a + r) + (a + 2r) + · · ·+ (a + (n− 1)r).

Example 55.3Find a formula for the series Sn. Hint: Write Sn in reverse order and thenadd this to the initially given form of Sn.

Solution.Write the sum in reverse order, i.e.

Sn = (a + (n− 1)r) + (a + (n− 2)r) + (a + (n− 3)r) + · · ·+ a.

Adding this expression of Sn with the initially given expression to obtain

2Sn = (2a + (n− 1)r) + (2a + (n− 1)r) + · · ·+ (2a + (n− 1)r).

That is,2Sn = n[2a + (n− 1)r]

orSn =

n

2[2a + (n− 1)r].

Example 55.4A military unit purchases 10 spare parts during the first month of a contract,15 spare parts during the second month, 20 spare parts during the thirdmonth, 25 spare parts during the fourth month and so on. The acquisitionofficer wants to know the total number of spare parts the unit will haveacquired after 50 months.

125

SolutionLet S50 be the total number of spare parts after 50 months. Then S50 is afinite arithmetic sum with a = 10, n = 50, and r = 5. Using the formula foran arithmetic finite sum we find

S50 =50

2[2(10) + (50− 1)2] = 6, 625 spare parts.

Example 55.5Let Sn be the sum of the first n positive integers, i.e.

Sn = 1 + 2 + 3 + · · ·+ n.

Find a formula for Sn.

Solution.According to the formula for a finite arithmetic sum with a = 1 and r = 1we have

Sn =n

2(n + 1).

Example 55.6Find the sum of the series

3 +3

2+

3

4+ · · ·+ 3

210.

Solution.We have

3 + 32

+ 34

+ · · ·+ 3210 = 3(1 + 1

2+ 1

22 + · · ·+ 1210 )

= 31− 1

211

1− 12

= 6(1− 2−11).

Now, an infinite geometric series with initial value a and ratio r is anyinfinite sum of the form

a + ar + ar2 + · · ·+ arn + · · ·

We define the following finite sums

S1 = aS2 = a + arS3 = a + ar + ar2

...Sn = a + ar + ar2 + · · ·+ arn−1.

126

We call the above sums, partial sums. Note that Sn is a finite geometricseries so that

Sn =a(1− rn)

1− rr 6= 1.

Theorem 55.1(a) If r = 1 then Sn = na and Sn becomes large when n is large if a > 0 andnegatively large if a < 0.(b) If r = −1 then rn alternates between −1 and 1 so that rn has no limitwhen n is large.(c) If |r| > 1 then the geometric series is divergent. (d) If |r| < 1 thenlimn→∞ rn = 0 and therefore limn→∞ Sn = a

1−r. In this case we say that the

infinite geometric series converges with sum equals to a1−r

and we write

a + ar + ar2 + · · ·+ arn + · · · = a

1− r.

If a geometric series does not converge then we say that it diverges.

Proof.The prove is given in Theorem 56.4 of the next section.

Example 55.7Determine whether the geometric series given below is convergent or diver-gent.

1000 + 1000(1.05) + 1000(1.05)2 + · · ·

Solution.Finding the nth partial sum we have

Sn =1000(1− (1.05)n

1− 1.05.

But (1.05)n −→∞ as n −→∞ so that the given series is divergent.

Example 55.8Determine whether the geometric series given below is convergent or diver-gent.

1 +1

2+

1

4+

1

8+ · · ·

127

Solution.Finding the nth partial sum we have

Sn =1−

(12

)n

1− 12

.

But(

12

)n−→ 0 as n −→∞ so that the series converges to 2.

128

Practice Problems

In Exercises 1 - 7, decide which of the following are geometric series. Forthose which are, give the first term and the ratio and the value of the sum.For those which are not, explain why not.

Exercise 55.11− 1

2+ 1

4− 1

8+ · · ·

Exercise 55.21 + 1

2+ 1

3+ · · ·

Exercise 55.32 + 1 + 1

2+ 1

4+ 1

8+ · · ·

Exercise 55.4y2 + y3 + y4 + y5 + · · ·

Exercise 55.51− x + x2 − x3 + x4 − · · ·

Exercise 55.61− y2 + y4 − y6 + · · ·

Exercise 55.71 + (2z) + (2z)2 + (2z)3 + · · ·

Exercise 55.8Find the sum of the series

∑∞n=4(

13)n.

Exercise 55.9A ball is dropped from a height of 10 feet and bounces. Each bounce is 3

4

of the height of the bounce before. Thus after the ball hits the floor the firsttime, the ball rises to a height of 10(3

4) = 7.5 feet, and after it hits the floor

for the second time, it rises to a height of 7.5(34) = 10(3

4)2 = 5.625 feet.

(a) Find an expression for the height to which the ball rises after it hits thefloor for the nth time.(b) Find an expression for the total vertical distance the ball has traveledwhen it hits the floor for the first, second, third, and fourth times.(c) Find an expression for the total vertical distance the ball has traveledwhen it hits the floor for the nth time. Express your answer in closed-form.

129

56 Convergence of Sequences and Series

We start this section by introducing the concept of a sequence and study itsconvergence.

Convergence of Sequences.An infinite sequence is a list of numbers a1, a2, a3, · · ·.

Example 56.1(1) 1, 1

2, 1

3, · · · .

(2) 0, 2, 0, 2, 0, · · · .

Formally, a sequence is a function f with domain the set of nonnegative in-tegers IN. The image of n is denoted by f(n) = an. We represent a sequenceby the notation {an}∞n=1 and we call an the nth term of the sequence. Forexample, the nth term of the sequence (1) above is an = 1

n, n ≥ 1 while the

nth term of (2) is an = 1 + (−1)n.

We say that a sequence {an}∞n=1 converges to a number L if and only ifan approaches L as n gets larger and larger, i.e if the difference |an − L|can be made as small as we wish by taking n large enough. Using the ε− δargument, if ε > 0 is given then we can find a positive integer N such that

|an − L| < ε for n ≥ N.

We writelim

n→∞an = L.

If a sequence does not converge then we say that it diverges.

Example 56.2(1) Since limn→∞

3n2+5n2+n+1

= 3 then the sequence { 3n2+5n2+n+1

}∞n=1 converges to 3.(2) Since (−1)n alternates between -1 and 1 then the sequence {1+(−1)n}∞n=1

alternates between 0 and 2 and so it diverges.

Theorem 56.1 (Convergence Properties of Sequences)(a) limn→∞(an + bn) = limn→∞ an + limn→∞ bn.(b) limn→∞ kan = k limn→∞ an, where k is a constant.(c) limn→∞ anbn = (limn→∞ an)(limn→∞ bn).(d) limn→∞

an

bn= limn→∞ an

limn→∞ bnprovided that limn→∞ bn 6= 0.

(e) Squeeze Principle: If an ≤ bn ≤ cn for all n and if limn→∞ an =limn→∞ cn = L then limn→∞ bn = L.

130

Proof.We will prove (a) and leave the rest as an exercise. Suppose that limn→∞ an =L and limn→∞ bn = L′. We will show that limn→∞(an + bn) = L+L′. Indeed,let ε > 0 be given. There there exist poisitve integers N1 and N2 such that

|an − L| < ε

2for n ≥ N1

and|bn − L′| < ε

2for n ≥ N2.

Let N = N1 + N2. Then for n ≥ N we have

|(an + bn)− (L + L′)| ≤ |an − L|+ |bn − L′| < ε

2+

ε

2< ε.

Next, we discuss a very useful theorem that establishes the convergence of agiven sequence (without, however, revealing the limit of the sequence). Butfirst we introduce a couple of definitions:We say that a sequence {an}∞n=1 is increasing if an ≤ am whenever n ≤ m.A sequence {an}∞n=1 is bounded from above if there is a positive constantC such that an ≤ C for all n ≥ 1.

Theorem 56.2 (Monotone Convergence Theorem)An increasing sequence that is bounded from above is always convergent.

Proof.Since {an}∞n=1 is bounded from above then there exists a C > 0 such thatan ≤ C for all n ≥ 1. Let c be the smallest positive constant such that an ≤ cfor all n ≥ 1. We will show that limn→∞ an = c.Let ε > 0. Then there is an integer N such that (See Figure 155).

aN > c− ε.

Figure 155

131

Since the sequence {an}∞n=1 is increasing then

an > c− ε

for all n ≥ N. Thus,|an − c| = |c− an| < ε

for all n ≥ N. This shows that the sequence is convergent.

The following theorem shows the connection between the convergence of afunction and the convergence of a sequence.(See Figure 156.) This basicallyallows us to replace limits of sequences with limits of functions. In particular,this is useful for using L’Hopital’s rule in computing limits of sequences.

Theorem 56.3If limx→∞ f(x) = L and an = f(n) then limn→∞ an = L.

Figure 156

Proof.Let ε > 0 be given. Since limx→∞ f(x) = L then we can find a δ > 0 suchthat

|f(x)− L < ε if |x| ≥ δ.

Let N = (integer part of δ) + 1. Then N is a positive integer greater thanor equal to δ. Thus, if n ≥ N then

|an − L| = |f(n)− L| < ε.

This shows that limn→∞ an = L.

132

Example 56.3(a) Find limn→∞

ln (n+1)n+1

.

(b) Find limn→∞sin n√n+1

.

Solution.(a) Applying L’Hopital’s rule we have

limn→∞ln (n+1)

n+1= limx→∞

ln (x+1)x+1

= limx→∞1

x+1

1= 0.

(b) Since −1 ≤ sin n ≤ 1 then − 1√n+1

≤ sin n√n+1

≤ 1√n+1

. But limn→∞1√n+1

=

limx→∞1√x+1

= 0. By the squeeze rule we have

limn→∞

sin n√n + 1

= 0.

The following is a list of useful sequences.

Theorem 56.4(a) If r = 1 then limn→∞ rn = 1.(b) If r = −1 then the sequence {rn} is divergent.(c) If |r| > 1 then the sequence {rn} is divergent.(d) If |r| < 1 then limn→∞ rn = 0.

(e) For r > 0, limn→∞ r1n = 1.

Proof.(a) Trivial.(b) If r = −1 then the sequence {rn} alternates between the two values -1and 1 and so the sequence is divergent.(c) |r| > 1 implies r > 1 or r < −1. Suppose first that r > 1. Let ε > 0. LetN be a positive integer greater than ε

r−1. Then for n ≥ N we have

rn = (1 + (r − 1))n

≥ 1 + n(r − 1)(by the binomial formula)> 1 + N(r − 1)> ε.

This shows that for any given positive number we can find a term in the se-quence {rn} which is greater than the number. This means that {rn} → ∞

133

as n →∞.If r < −1 then rn = (−1)n(−r)n with −r > 1. Thus, as n becomes large,rn alternates between large positive numbers and large negative numbers sothat again the limit rn does not exist.(d) If 0 < r < 1 then rn = 1

(r−1)n with (r−1)n → ∞ as n → ∞. (See (c)).Hence, rn → 0 as n →∞.If −1 < r < 0 then 0 < −r < 1. In this case, rn = (−1)n(−r)n → 0 asn →∞.(e) Let d = r

1n . Then ln d = ln c

n→ ∞ as n → ∞. Hence, d = eln d → e0 = 1

as n →∞.

Convergence of SeriesThe greek letter Σ is used to denote summations such as

Σni=1ai = a1 + a2 + · · ·+ an

orΣn

i=mai = am + am+1 + · · ·+ an.

The sum of a sequence {an}∞n=1 is called a series, denoted by

Σ∞n=1an = a1 + a2 + · · ·+ an + · · ·

To determine whether this series converges or not we consider the sequenceof partial sums defined as follows:

S1 = a1

S2 = a1 + a2...

Sn = a1 + a2 + · · ·+ an.

We say that a series Σ∞n=1an converges to a number L if and only if the

sequence {Sn}∞n=1 converges to L and we write

Σ∞n=1an = lim

n→∞Sn = L.

A series which is not convergent is said to diverge.

Example 56.4(a) The geometric series Σ∞

n=1xn converges for |x| < 1 with sum equals to

x1−x

.(b) The series Σ∞

n=1(−1)n diverges since the sequence of partial sums alter-nates between the values -1 and 0.

134

Example 56.5Show that the series Σ∞

n=11

n(n+1)converges to 1.

Solution.Using partial fractions we can write

1

n(n + 1)=

1

n− 1

n + 1.

Thus,S1 = 1− 1

2

S2 = (1− 12) + (1

2− 1

3) = 1− 1

3

S3 = S2 + (13− 1

4) = (1− 1

3) + (1

3− 1

4) = 1− 1

4...

Sn = 1− 1n+1

.

It follows that limn→∞ Sn = 1.

Using Theorem 56.1 (a) and (b) we have the following properties of con-vergent series.

Theorem 56.5If Σ∞

n=1an and Σ∞n=1bn are two convergent series and k is a constant then

(a) Σ∞n=1(an ± bn) = Σ∞

n=1an ± Σ∞n=1bn

(b) Σ∞n=1kan = kΣ∞

n=1an.

Theorem 56.6Let N be a positive integer. Suppose that an = bn for all n ≥ N. Thenthe series

∑∞n=1 an and

∑∞n=1 bn either both converge or both diverge. Thus,

changing a finite number of terms in a series does not change whether or notit converges, although it may change the value of its sum if it does converge.

Proof.The proof follows from the equality

∑∞n=1 an =

∑Nn=1(an − bn) +

∑∞n=1 .

Improper integrals can be used to determine the convergence or divergenceof some series as shown by the following theorem.

Theorem 56.7 (The Integral Test)Suppose that f(x) is a positive and decreasing function. Assume that an =

135

f(n).(a) If

∫∞1 f(x)dx converges, then the series Σ∞

n=1an converges.(b) If

∫∞1 f(x)dx diverges, then the series Σ∞

n=1an diverges.

Proof.Since f is decreasing then for any i ≥ 1

f(i + 1) ≤∫ i+1

if(x)dx ≤ f(i).

See Figure157.

Figure 157

But ai = f(i) so that

ai+1 ≤∫ i+1

if(x)dx ≤ ai.

Adding these inequalities for i = 1 to i = n to obtain

a2 + a3 + ·an+1 ≤∫ n+1

1f(x)dx ≤ a1 + a2 + ·an.

If {Sn} is the sequence of partial sums associated to the series∑∞

n=1 an , thenwe have

Sn+1 − a1 ≤∫ n+1

1f(x)dx ≤ Sn.

or equivalently ∫ n+1

1f(x)dx ≤ Sn ≤ a1 +

∫ n

1f(x)dx.

136

(a) If∫∞1 f(x)dx converges then the se quence {Sn} converges since

Sn ≤ a1 +∫ n

1f(x)dx.

Therefore∑∞

n=1 an converges.(b) If

∫ n1 f(x)dx is divergent then {Sn} is divergent since∫ n+1

1f(x)dx ≤ Sn.

Thus,∑∞

n=1 an is divergent.

Remark 56.1Note that if c ≥ 0 then∫ ∞

cf(x)dx =

∫ 1

cf(x)dx +

∫ ∞

1f(x)dx.

Example 56.6Show that the series Σ∞

n=11np converges for p > 1 and diverges for p ≤ 1. This

series is referred to as p-series.

Solution.We already know that the improper integral

∫∞1

1xp dx converges for p > 1

and diverges for p ≤ 1. By the Integral Test the series∑∞

n=11np converges for

p > 1 and diverges for p ≤ 1.

The following result provides a procedure for testing the divergence of aseries. This is known as the the nth term test for convergence.

Theorem 56.8If the series Σ∞

n=1an is convergent then limn→∞ an = 0.

Solution.We know that Sn = a1 + a2 + · · ·+ an and Sn+1 = a1 + a2 + · · ·+ an + an+1 =Sn + an so it follows that Sn+1 − Sn = an. Suppose that the series convergesto a number L. Then limn→∞ Sn = limn→∞ Sn+1 = L. Thus, limn→∞ an =limn→∞(Sn+1 − Sn) = L− L = 0.

Remark 56.2The theorem states that if we know that the series is convergent then limn→∞ an =0. The converse is not true in general. That is, the condition limn→∞ an = 0does not necessarily imply that the series

∑∞n=1 an is convergent. By Exam-

ple 56.6, the series∑∞

n=11n

is divergent even though limn→∞1n

= 0.

137

Practice Problems

Exercises 1 - 6 give expressions for the nth term an of a sequence. Findthe limit of each sequence, if it exists.

Exercise 56.1an = (0.2)n.

Exercise 56.2an = 2n.

Exercise 56.3an = 3 + e−2n.

Exercise 56.4an = cos (πn).

Exercise 56.5an = 3+4n

5+7n.

Exercise 56.6an = 2n+(−1)n5

4n+(−1)n3.

Exercise 56.7Use the integral test to decide whether the series Σ∞

n=1n

n2+1converges or di-

verges.

Exercise 56.8Use the integral test to decide whether the series Σ∞

n=1ne−n2converges or

diverges.

Do the series in Exercises 9 - 12 converge or diverge?

Exercise 56.9Σ∞

n=1((34)n + 1

n).

Exercise 56.10Σ∞

n=13

n+2

138

Exercise 56.11Σ∞

n=1n

n+1.

Exercise 56.12Σ∞

n=13

(2n−1)2.

Exercise 56.13Find the sum Σ∞

n=13n+54n .

139

57 Tests for Convergence

The Comparison Test

Going back to our discussion of improper integrals, we recall the comparisontheorem which tells that an improper integral is convergent if it is smallerthan a known convergent improper integral and is divergent if it is largerthan a known divergent improper integral. A similar result holds for series.

Theorem 57.1 (Comparison test)Let {an}∞n=1 and {bn}∞n=1 be two sequences of nonnegative terms.(i) If an ≤ bn for n ≥ N and the series

∑∞n=1 bn is convergent then the series∑∞

n=1 an is convergent as well.(ii) If an ≤ bn for n ≥ N and the series

∑∞n=1 an is divergent then the series∑∞

n=1 bn is also divergent.

Proof.Let a′n = aN+n−1 and b′n = bN+n−1. Since the convergence or divergence ofa series is not affected by deleting a finite number of terms then either theseries

∑∞n=1 a′n and

∑∞n=1 an both converge or both diverge. The same applies

fo the series with terms bn and b′n.(a) Let Sn and Tn be the nth partial sums of

∑∞n=1 a′n and

∑∞n=1 b′n. Then

Sn ≤ Tn. If∑∞

n=1 bn converges, say to T, then Tn → T as n →∞ and Tn < Tfor all n since Tn is increasing. Thus, Sn ≤ Tn < T. So we have that Sn isincreasing and bounded from above. By Theorem 56.2, {Sn} is convergentand therefore

∑∞n=1 a′n is convergent. Hence,

∑∞n=1 an converges.

(b) Now suppose that∑∞

n=1 an diverges. Then limn→∞ Sn = ∞. But Tn ≥ Sn

so that limn→∞ Tn = ∞. This shows that∑∞

n=1 b′n is divergent and conse-quently the series

∑∞n=1 bn is divergent.

Keep in mind that the major purpose of this comparison test is that someseries are very difficult to test for convergence directly. Instead we comparethem to well known series such as the p-series.

Example 57.1Show that the series

∑∞n=1

1√n2−n+1

is divergent.

140

Solution.Indeed, for n ≥ 1 we have n2 +(1−n) ≤ n2 so that

√n2 − n + 1 ≤

√n2 = n.

This implies that 0 ≤ 1n≤ 1√

n2−n+1. Since the series

∑∞n=1

1n

is divergent (har-

monic series) then the comparison test asserts that the series∑∞

n=11√

n2−n+1is divergent.

Example 57.2Show that the series

∑∞n=1

n−1n3+n

is convergent.

Solution.To see this, note that n3 + n ≥ n3 so that 1

n3+n≤ 1

n3 . Thus, n−1n3+n

≤ n−1n3 ≤

nn3 = 1

n2 . Since the series∑∞

n=11n2 is convergent then by the comparison test

the series∑∞

n=1n−1n3+n

is also convergent.

The difficulty with the comparison test is that when the nth term of a se-ries

∑∞n=1 an is complicated then it might be difficult to figure out the series∑∞

n=1 bn that need to be compared with. The following comparison test isoften easier to apply, because after deciding on

∑∞n=1 bn we need only take a

limit of the quotient an

bnas n →∞.

Theorem 57.2Let

∑∞n=1 an and

∑∞n=1 bn be two series with positive terms. If

limn→∞

an

bn

= L > 0

then either both series converge or both diverge.

Proof.Since L > 0 and limn→∞

an

bn= L then there exists a positive integer N such

that for n ≥ N we have

|an

bn

− L| < L

2

i.e.

L− L

2<

an

bn

< c +c

2or

L

2<

an

bn

<3

2L.

141

This is equivalent to

L

2bn < an <

3

2Lbn, n ≥ N.

If the series∑∞

n=1 an converges then by the comparison test the series∑∞

n=1L2bn

is convergent. By Theorem 56.5(b), the series∑∞

n=1 bn is also convergent.Conversely, if

∑∞n=1 bn is convergent then

∑∞n=1

32Lbn is convergent and by the

comparison test∑∞

n=1 an is convergent. Similarly,∑∞

n=1 an is divergent if andonly if

∑∞n=1 bn is divergent.

Example 57.3Determine whether the series

∑∞n=1

3n+14n3+n2−2

converges or diverges.

Solution.For large n we have an = 3n+1

4n3+n2−2≈ 3n

4n3 = 34n2 . So let bn = 1

n2 . Then

limn→∞

an

bn

= limn→∞

3n3 + n2

4n3 + n2 − 2=

3

4.

Since the series∑∞

n=11n2 is convergent then so does the series

∑∞n=1

3n+14n3+n2−2

.

Absolute Convergence

Consider a series∑∞

n=1 an which has both positive and negative terms. Wesay that this series is absolutely convergent if the series of absolute values∑∞

n=1 |an| is convergent. The following theorem provides a test of convergencefor series of the above type.

Theorem 57.3If∑∞

n=1 |an| is convergent then∑∞

n=1 an is convergent.

Proof.The proof of this result is quite simple. Let bn = |an| ≥ 0. By assumption, theseries

∑∞n=1 bn =

∑∞n |an| is convergent. But |an| ≤ bn for all n. Now, part (i)

of the comparison test asserts that the series∑∞

n=1 an must be convergent.

Example 57.4

Show that the series∑∞

n=1(−1)n−1

n2 is convergent.

142

Solution.Indeed, the series of absolute values

∑∞n=1

1n2 is convergent (p-series with

p = 2) so by the above theorem, the series∑∞

n=1(−1)n−1

n2 is also convergent.

Example 57.5Show that the series 1−x+x2−x3 + · · · is absolutely convergent for |x| < 1.

Solution.Since the geometric series 1+x+x2 +x3 + · · · converges for |x| < 1 then thegiven series is absolutely convergent.

Remark 57.1It is very important to be very careful with the statement of the abovetheorem. The theorem says that if we know that the series

∑∞n=1 |an| is

convergent then the series∑∞

n=1 an is definitely convergent. However, it ispossible that

∑∞n=1 |an| is divergent and still

∑∞n=1 an is convergent. The

following example illustrates this situation.

Example 57.6


n=1(−1)n−1

nis convergent but the series

∑∞n=1

1n

is di-vergent.

Solution.The alternating series test (see discussion below) asserts that the series∑∞

n=1(−1)n 1n

is convergent. However, the series∑∞

n=1 |(−1)n−1

n| =

∑∞n=1

1n

is divergent (harmonic series)

When a series is such that∑∞

n=1 |an| is divergent but∑∞

n=1 an is conver-gent then we say that the series

∑∞n=1 an is conditionally convergent.

The Ratio Test

The integral test is hard to apply when the integrand involves factorialsor complicated expressions. We shall now introduce a test that can be usedto help determine convergence or divergence of series when other tests arenot applicable.

Theorem 57.4Let

∑∞n=1 an be a series and suppose that limn→∞

∣∣∣an+1

an

∣∣∣ = L ≥ 0.

143

(a) If L < 1 then the series∑∞

n=1 an converges.(b) If L > 1 then the series

∑∞n=1 an diverges.

(c) If L = 1 then the test fails, that is the test does not tell us anythingabout the convergence of the series.

Proof.(a) Suppose that 0 ≤ L < 1. Since limn→∞

∣∣∣an+1

an

∣∣∣ = L then we can find apositive integer N such that∣∣∣∣an+1

an

∣∣∣∣− L ≤∣∣∣∣∣∣∣∣an+1

an

∣∣∣∣− L∣∣∣∣ < 1− L

2, for n ≥ N.

This is equivalent to ∣∣∣∣an+1

an

∣∣∣∣ < L + 1

2, for n ≥ N.

Let r = L+12

. Clearly, L < r < 1. Thus,

|an+1| < r|an|, for n ≥ N.

Hence,|aN+1| < r|aN ||aN+2| < r|aN+1| < r2|aN ||aN+3| < r|aN+2| < r3|aN |

...

Since the series∑∞

n=1 rn|aN | is a geometric series with r < 1 then it is conver-gent. By the comparison test the series

∑∞n=1 |aN+n| is also convergent. By

Theorem 57.3, the series∑∞

n=1 aN+n is convergent. Since the convergence ordivergence is unaffected by deleting a finite number of terms then the series∑∞

n=1 an is convergent.(b) Suppose now that L > 1. Then there is a positive integer N such thatfor n ≥ N ∣∣∣∣∣∣∣∣an+1

an

∣∣∣∣− L

∣∣∣∣ < L− 1

2or

−L− 1

2<

∣∣∣∣an+1

an

∣∣∣∣− L <L− 1

2

i.e., ∣∣∣∣an+1

an

∣∣∣∣ > L + 1

2, for n ≥ N.

144

Let r = L+12

. Then L > r > 1. Thus, for n ≥ N we have∣∣∣an+1

an

∣∣∣ > 1 or

|an+1| > |an|. This implies that limn→∞ an 6= 0 and by Theorem 56.8, theseries

∑∞n=1 an is divergent.

(c) Consider the series∑∞

n=11n. Then this series is divergent and limn→∞

1n+1

1n

=

1. On the other hand, the series∑∞

n=11n2 is convergent with limn→∞

1(n+1)2

1n2

=

1. Thus, when L = 1 the test is inconclusive.

Example 57.7

1. The series∑∞

n=1(−1)n−1

nis convergent by the alternating series test. Note

that limn→∞ |an+1

an| = n

n+1= 1 i.e., L = 1 in the previous theorem.

2. The series∑∞

n=1(−1)n−1 is divergent. Also, note that limn→∞ |an+1

an| = 1,

i.e. L = 1.3. The series

∑∞n=1

(−1)n−1

n!is convergent since limn→∞ |an+1

an| = limn→∞

n!(n+1)!

=

limn→∞1

n+1= 0 so L = 0 < 1 in the above theorem.

4. The series∑∞

n=1(−2)n−1 is divergent since limn→∞ |an+1

an| = 2 > 1.

Remark 57.2When testing a series for convergence, normally concentrate on the nth termtest and the ratio test. Use the comparison test only when both tests fail.

Alternating Series TestBy an alternating series we mean a series of the form

∑∞n=1(−1)n−1an where

an > 0. For instance, the series∑∞

n=1(−1)n−1

n. Here an = 1

n. The following

theorem provides a way for testing alternating series for convergence.

Theorem 57.5 (Alternating Series Test)An alternating series

∑∞n=1(−1)n−1an is convergent if and only if:

(i) The sequence {an}∞n=1 is decreasing, i.e. an+1 < an for all n;(ii) limn→∞ an = 0.

Proof.Let {Sn} be the sequence of partial sums of the series

∑∞n=1(−1)n−1an. Notice

the followingS4 − S2 = a3 − a4 ≥ 0S6 − S4 = a5 − a6 ≥ 0

S8 − S6 = a7 − a8 ≥ 0...

145

Thus,S2 ≤ S4 ≤ S6 ≤ S8 ≤ · · · ;

It follows that the sequence {S2n} is increasing. Moreover, for all n ≥ 1 wehave

S2n = a1 − (a2 − a3)− (a4 − a5)− · · · (a2n−2 − a2n−1)− a2n ≤ a1.

Hence, the sequence {S2n} is bounded from above. By Theorem 56.2, thereis an S > 0 such that limn→∞ S2n = S.Next, we consider the terms of {Sn} with odd subscripts:

S1 − S3 = a2 − a3 ≥ 0S3 − S5 = a4 − a5 ≥ 0

S5 − S7 = a6 − a7 ≥ 0...

Thus,S1 ≥ S3 ≥ S5 ≥ S7 ≥ · · · ;

It follows that the sequence {S2n+1} is decreasing. Moreover, S2n+1 = S2n +a2n+1. Thus, limn→∞ S2n+1 = limn→∞ S2n + limn→∞ a2n+1 = S + 0 = S. Itfollows that

limn→∞

Sn = S.

Figure 158 shows how the terms of {Sn} and S are ordered on a line.

Figure 158

Example 57.8


n=1(−1)n−1

nis convergent.

Solution.To see this, let an = 1

n. Then n < n+1 implies that 1

n+1< 1

nthat is an+1 < an.

Also, limn→∞ an = limn→∞1n

= 0. Hence, by the previous theorem the givenseries is convergent.

146

Remark 57.3(a)It follows from the above theorem that if

∑∞n=1(−1)n−1an converges then∑∞

n=1(−1)n−1an ≤ a1.(b) From Figure 158, for each n ≥ 1, we have S is between Sn and Sn+1 sothat the distance between S and Sn is less than the distance between Sn andSn+1. Thus, we have an upper bound for the error:

|Sn − S| < |Sn+1 − Sn| = an+1.

(c) Keep in mind that the tests used in this section are basically used totest for convergence. However, when a series is convergent these tests do notprovide a value for the sum.

147

Practice Problems

Use the comparison test to determine whether the series in Exercises 1-7converge.

Exercise 57.1∑∞n=1

e−n

n2 .


1ln (n+1)

.


13n+1

.


1n4+en .

Exercise 57.5∑∞n=1 2−n n+1

n+2.


2n+1n2n−1

.


n sin nn3+1

.

Exercise 57.8Use the ratio test to show that the series

∑∞n=1

(n!)2

(2n)!is convergent.

Exercise 57.9Use the ratio test to show that the series

∑∞n=1

2n

n3+1is divergent.

Exercise 57.10Use the alternating series test to show that the series

∑∞n=1

(−1)n−1

2n+1is conver-

gent.

Determine which of the series in Exercises 11 - 14 converge.

Exercise 57.11∑∞n=1 e−n.

148

Exercise 57.12∑∞n=1 en.


(n−1)!n2 .


(−1)n−1√

3n−1.

Exercise 57.15Show that if

∑∞n=1 |an| converges then

∑∞n=1(−1)n−1an converges.

149

58 Power Series

Let {an}∞n=0 be a sequence of numbers. Then a power series about x = ais a series of the form

∞∑n=0

an(x− a)n = a0 + a1(x− a) + a2(x− a)2 + · · ·

Note that a power series about x = a always converges at x = a with sumequals to a0.

Example 58.11. A polynomial of degree m is a power series about x = 0 since

p(x) = a0 + a1x + a2x2 + · · ·+ amxm.

Note that an = 0 for n ≥ m + 1.2. The geometric series 1 + x + x2 + · · · is a power series about x = 0 withan = 1 for all n.3. The series 1

x+ 1

x2 + 1x3 + · · · is not a power series since it has negative

powers of x.4. The series 1 + x + (x− 1)2 + (x− 2)3 + (x− 3)4 + · · · is not a power seriessince each term is a power of a different quantity.

Convergence of Power SeriesTo study the convergence of a power series about x = a one starts by fixingx and then constructing the partial sums

S0(x) = a0,S1(x) = a0 + a1(x− a),S2(x) = a0 + a1(x− a) + a2(x− a)2,

...Sn(x) = a0 + a1(x− a) + a2(x− a)2 + · · ·+ an(x− a)n.

Thus obtaining the sequence {Sn(x)}∞n=0. If this sequence converges to anumber L, i.e. limn→∞ Sn(x) = L, then we say that the power series con-verges to L for the specific value of x. Otherwise, we say that the powerseries diverges.Power series may converge for some values of x and diverge for other values.The following theorems provides a tool for determining the values of x forwhich the power series converges and those for which it diverges.

150

Theorem 58.1Suppose that

∑∞n=0 an(x− a)n is a power series that converges for x = c + a

and diverges for x = d + a. Then

(i)∑∞

n=0 an(x− a)n converges absolutely for |x− a| < |c|; and(ii)

∑∞n=0 an(x− a)n diverges for |x− a| > |d|.

Proof.(i) There is nothing to show if x = a, i.e c = 0. So we assume that

∑∞n=0 an(c)n

converges with c 6= 0. By the nth term test, limn→∞ ancn = 0. Thus, there

exists a positive integer N such that |ancn| < 1 for all n ≥ N. Let M =∑N−1

n=0 |ancn|+ 1. Then, |anc

n| ≤ M for all n ≥ 0. Now, for n ≥ 0 we have

|an(x− a)n| = |ancn| ·

∣∣∣∣x− a

c

∣∣∣∣n ≤ M∣∣∣∣x− a

c

∣∣∣∣n .

But the series∑∞

n=0 M∣∣∣x−a

c

∣∣∣n is a convergent geometric series since∣∣∣x−a

c

∣∣∣ < 1.

Hence, by the comparison test∑∞

n=0 |an(x− a)n| converges.(ii) Suppose that |x− a| > |d|. If

∑∞n=0 an(x− a)n is convergent then by (i),∑∞

n=0 andn is absolutely convergent. This contradicts the fact that

∑∞n=0 and

n

is divergent. Hence,∑∞

n=0 an(x− a)n diverges.

Theorem 58.2Given a power series

∞∑n=0

an(x− a)n = a0 + a1(x− a) + a2(x− a)2 + · · ·

Then one of following is true:(i) The series converges only at x = a;(ii) the series converges for all x;(iii) There is some positive number R such that the series converges absolutelyfor |x − a| < R and diverges for |x − a| > R. The series may or may notconverge for |x− a| = R. That is for the values x = a−R and x = a + R.

Proof.Let C be the collection of all real numbers at which the series

∑∞n=0 an(x−a)n

converges. If C = {a} the series converges only at x = a and diverges for allx 6= a. This establishes (i). If C = (−∞,∞) then the series converges for all

151

values of x. This establishes (ii). To prove (iii), we assume that C 6= {a} andC 6= (−∞,∞). The condition C 6= (−∞,∞) guarantees the existence of areal number d such that

∑∞n=0 and

n diverges. Hence, by applying the previ-ous theorem, |x− a| < |d| whenever x is in C. So C is bounded from above.Let R > 0 be the smallest upper bound of C. Thus, R < |d|. If |x − a| > Rthen x is not in C and so

∑∞n=0 an(x − a)n divverges. If |x − a| < R then

we can find an x0 is C such that |x− a| < x0 ≤ R, by the way R is defined.Since

∑∞n=0 anx

n0 converges then by the previous theorem

∑∞n=0 an(x − a)n

converges. Since C 6= {a} then there is an x in C such that x 6= a; hence,0 < |x − a| ≤ R. We cannot asserts what happens at either x = a − R orx = a + R.

The largest interval for which the power series converges is called the in-terval of convergence. If a power series converges at only the point x = athen we define R = 0. If a power series converges for all values of x then wedefine R = ∞. We call R the radius of convergence.

Finding the radius of convergenceThe next theorem gives a method for computing the radius of convergenceof many series.

Theorem 58.3Suppose that

∑∞n=0 an(x− a)n is a power series with an 6= 0 for all n ≥ 0.

(a) If limn→∞|an+1||an| = 0 then R = ∞ and this means that the series con-

verges for all x.(b) If limn→∞

|an+1||an| = L > 0 then R = 1

L.

(c) If limn→∞|an+1||an| = ∞ then R = 0 and this means that the series diverges

for all x 6= a.

Proof.Consider the ratio ∣∣∣∣∣an+1(x− a)n+1

an(x− a)n

∣∣∣∣∣ =∣∣∣∣an+1

an

∣∣∣∣ |x− a|.

Suppose that limn→∞

∣∣∣an+1

an

∣∣∣ = L ≥ 0. Then the sequence{∣∣∣an+1

an

∣∣∣ |x− a|}∞

n=0

converges to R · |x− a|.

152

(a) If L = 0 then |x − a| · L = 0 < 1 and the series converges for all realnumbers x. Thus, R = ∞.(b) Suppose that L > 0 and finite. If R · |x− a| < 1 or |x− a| < 1

Lthen by

the ratio test, the series∑∞

n=0 an(x − a)n is convergent. If |x − a| > 1L

thenby the ratio test, the series is divergent. Hence, R = 1

L.

(c) If L = ∞ then the series diverges for all x 6= a. Thus, R = 0.

Example 58.2The radius of convergence of the power series

∑∞n=0

xn

n!is R = ∞ since 1

R=

limn→∞|an+1||an| = limn→∞

n!(n+1)!

= limn→∞1

n+1= 0. Thus, the series converges

everywhere.

Example 58.3Consider the power series

∑∞n=1(−1)n−1 (x−1)n

n. Then by the previous theorem

1R

= limn→∞n

n+1= 1 so that R = 1. Hence, the series converges for |x−1| < 1

and diverges for |x − 1| > 1. So the interval of convergence is the interval0 < x < 2. What about the endpoints x = 0 and x = 2? If we replacex by 0 we obtain the series −∑∞

n=11n

which is divergent (harmonic series).If we replace x by 2 we obtain the alternating series

∑∞n=1(−1)n−1 1

nwhich

converges by the alternating series test. Thus, the interval of convergence is0 < x ≤ 2.

153

Practice Problems

Exercise 58.1Find the radius of convergence of the power series

∑∞n=1(−1)n−1 x2n−1

(2n−1)!.

Exercise 58.2Find the radius and the interval of convergence of the series

∑∞n=0 22nx2n.

Write each of the series in Exercises 3 - 5 using sigma notation, i.e.∑

cn(x−a)n.

Exercise 58.31− (x−1)2

2!+ (x−1)4

4!− (x−1)6

6!+ · · ·

Exercise 58.4(x− 1)3 − (x−1)5

2!+ (x−1)7

4!− (x−1)9

6!+ · · ·

Exercise 58.52(x + 5)3 + 3(x + 5)5 + 4(x+5)7

2!+ 5(x+5)9

3!+ · · ·

Use Theorem 58.3 to find the radius of convergence of the power series inExercises 6 - 10.

Exercise 58.6∑∞n=0(5x)n.


(n+1)2n+n

xn.

Exercise 58.8x + 4x2 + 9x3 + 16x4 + · · ·

Exercise 58.91 + 2x + 4x2

2!+ 8x3

3!+ · · ·

Exercise 58.10x− x3

3+ x5

5− x7

7+ · · ·

154

Exercise 58.11(a) Determine the radius of convergence of the series

x− x2

2+

x3

3+ · · ·+ (−1)n−1xn

n+ · · ·

What does this tell us about the interval of convergence of this series?(b) Investigate convergence at the endpoints of the interval of convergence ofthis series.

Exercise 58.12Show that the series

∑∞n=1

(2x)n

nconverges for |x| < 1

2. Investigate whether the

series converges for x = 12

and x = −12.

155

59 Approximations by Taylor’s Polynomials

In this and the next section we are interested in approximating function val-ues by using polynomials which are easy to compute. The polynomials usedin the process are referred to as Taylor’s polynomials.

Let f(x) be a function which has derivatives f ′(a), f ′′(a), f ′′′(a), · · · , f (n)(a).We want to approximate f(x) by a polynomial of degree n as follows.

f(x) ≈ c0 + c1(x− a) + c2(x− a)2 + c3(x− a)3 + · · ·+ cn(x− a)n.

Note that f(a) ≈ c0. Now if we take the first derivative of f(x) we find

f ′(x) ≈ c1 + 2c2(x− a) + 3c3(x− a)2 + 4c4(x− a)3 + · · ·+ ncn(x− a)n.

From this we see that f ′(a) ≈ c1. Next, take the second derivative of f(x) toobtain

f ′′(x) ≈ 2 · 1c2 + 3 · 2c3(x− a) + 4 · 3c4(x− a)2 + · · ·+ n · (n− 1)cn(x− a)n−2

This implies that f ′′(a) ≈ 2 · 1c2 = 2!c2 or c2 ≈ f ′′(a)2!

. Again, taking the thirdorder derivative of f(x) we find

f ′′′(x) ≈ 3 · 2 · 1c3 + 4 · 3 · 2c4(x− a) + · · ·+ n · (n− 1) · (n− 2)cn(x− a)n−3.

It follows that f ′′′(a) ≈ 3 ·2 ·1c3 = 3!c3 or c3 ≈ f ′′′(a)3!

. Continuing this process

of taking successive derivatives we find c4 ≈ f (4)(a)4!

, · · · , cn ≈ f (n)(a)n!

. Hence,f(x) can be approximated by the polynomial

f(x) ≈ f(a)+f ′(a)

1!(x−a)+

f ′′(a)

2!(x−a)2+

f ′′′(a)

3!(x−a)3+· · ·+f (n)(a)

n!(x−a)n.

We define the nth Taylor polynomial of f(x) at x = a to be the polynomial

Pn(x) = f(a)+f ′(a)

1!(x−a)+

f ′′(a)

2!(x−a)2+

f ′′′(a)

3!(x−a)3+· · ·+f (n)(a)

n!(x−a)n.

Example 59.1Use the process of successive differentiation to write the polynomial

p(x) = x2 + x + 2

in the formp(x) = c0 + c1(x− 2) + c2(x− 2)2.

156

Solution.We have

p(x) = x2 + x + 2, c0 = p(2) = 8p′(x) = 2x + 1, c1 = p′(2) = 5

p′′(x) = 2, c2 = p′′(2)2!

= 1

Thus,p(x) = 8 + 5(x− 2) + (x− 2)2

Example 59.2Find the 4th degree Taylor polynomial approximating

f(x) =1

1 + x

near a = 0.

Solution.We have

f(x) = 1x+1

, c0 = f(0) = 1

f ′(x) = −(x + 1)−2, c1 = f ′(0) = −1

f ′′(x) = 2(x + 1)−3, c2 = f ′′(0)2!

= 22

= 1

f ′′′(x) = −6(x + 1)−4, c3 = f ′′′(0)3!

= −1

f (4)(x) = 24(x + 1)−5, c4 = f (4)(0)4!

= 1

Thus,P4(x) = 1− x + x2 − x3 + x4.

Example 59.3Find the third degree Taylor polynomial approximating

f(x) = arctan x,

near a = 0.

Solution.We have

f(x) = arctan x, c0 = f(0) = 0f ′(x) = 1

1+x2 , c1 = f ′(0) = 1

f ′′(x) = −(1 + x2)−2(2x), c2 = f ′′(0)2!

= 0

f ′′′(x) = −2(1 + x2)−3(4x2)− 2(1 + x2)−2, c3 = f ′′′(0)3!

= −13

157

Thus,

P3(x) = x− 1

3x3.

Example 59.4Find the fifth degree Taylor polynomial approximating

f(x) = ln (1 + x),

near a = 0.

Solution.We have

f(x) = ln (1 + x), c0 = f(0) = 0f ′(x) = 1

1+x, c1 = f ′(0) = 1

f ′′(x) = − 1(1+x)2

, c2 = f ′′(0)2!

= −12

f ′′′(x) = 2(1+x)3

, c3 = f ′′′(0)3!

= 13

f (4)(x) = − 6(1+x)4

, c4 = f (4)(0)4!

= −14

f (5)(x) = 24(1+x)5

, c5 = f (5)(0)5!

= 15

Thus,

P5(x) = x− 1

2x2 +

1

3x3 − 1

4x4 +

1

5x5.

Example 59.5Find the fourth degree Taylor polynomial approximating

f(x) = sin x,

near a = π2.

Solution.We have

f(x) = sin x, c0 = f(π2) = 1

f ′(x) = cos x, c1 = f ′(π2) = 0

f ′′(x) = − sin x, c2 =f ′′(π

2)

2!= −1

2

f ′′′(x) = − cos x, c3 =f ′′′(π

2)

3!= 0

f (4)(x) = sin x, c4 =f (4)(π

2)

4!= 1

24

Thus,

P4(x) = 1− 1

2(x− π

2)2 +

1

24(x− π

2)4.

158

Example 59.6Suppose that the function f(x) is approximated near x = 0 by a sixth degreeTaylor polynomial

P6(x) = 3x− 4x3 + 5x6.

Find the value of the following:(a) f(0) (b) f ′(0) (c) f ′′′(0) (d) f (5)(0) (e) f (6)(0)

Solution.If

P6(x) = c0 + c1x + c2x2 + c3x

3 + c4x4 + c5x

5 + c6x6

then c0 = 0, c1 = 3, c2 = 0, c3 = −4, c4 = c5 = 0, and c6 = 5.

(a) f(0) = c0 = 0, (b) f ′(0) = c1 = 3, (c) f ′′′(0) = 3!c3 = −24, (d)f (5)(0) = 5!c5 = 0, (e) f (6)(0) = 6!c6 = 3600.

Example 59.7Let g(x) be a function such that g(5) = 3, g′(5) = −1, g′′(5) = 1 and g′′′(5) =−3.(a) What is the Taylor polynomial of degree 3 for g(x) near 5?(b) Use (a) to approximate g(4.9).

Solution(a) We have: c0 = g(5) = 3, c1 = g′(5) = −1, c2 = g′′(5)

2!= 1

2, and c3 = g′′′(5)

3!=

−12. Thus, P3(x) = 3− (x− 5) + 1

2(x− 5)2 − 1

2(x− 5)3.

(b) g(4.9) = 3− (4.9− 5) + 12(4.9− 5)2 − 1

2(4.9− 5)3 = 3.1675.

159

Practice Problems

Exercise 59.1Use the process of successive differentiation to write the polynomial

p(x) = x2 + x + 2

in the formp(x) = c0 + c1(x− 2) + c2(x− 2)2.

Exercise 59.2Write the polynomial p(x) = x3 − x2 + 1 in powers of x− 1

2.

Exercise 59.3Let p(x) = x3 − x2 + 1. Compute p(0.50028) to five decimal places.

In Exercises 4 - 9, compute Pn(x) for the given f, n and a.

Exercise 59.4f(x) = 1

1+x, n = 4, a = 0.

Exercise 59.5f(x) = arctan x, n = 3, a = 0.

Exercise 59.6f(x) = 3

√1− x, n = 3, a = 0.

Exercise 59.7f(x) = ln (1 + x), n = 5, a = 0.

Exercise 59.8f(x) = sin x, n = 4, a = π

2.

Exercise 59.9f(x) = ex, n = 4, a = 1.

Exercise 59.10Suppose that the function f(x) is approximated near x = 0 by a sixth degreeTaylor polynomial

P6(x) = 3x− 4x3 + 5x6.

Find the value of the following:(a) f(0) (b) f ′(0) (c) f ′′′(0) (d) f (5)(0) (e) f (6)(0)

160

Exercise 59.11Let g(x) be a function such that g(5) = 3, g′(5) = −1, g′′(5) = 1 and g′′′(5) =−3.(a) What is the Taylor polynomial of degree 3 for g(x) near 5?(b) Use (a) to approximate g(4.9).

Exercise 59.12Find the second-degree Taylor polynomial for f(x) = 4x2−7x+2 about x = 0.What do you notice?

161

60 The Error in Taylor Polynomial Approxi-

mations: Taylor’s Theorem

In this section we would like to see how good the Taylor polynomials approx-imation discussed in the previous section is.To this end, let En(x) be the error between the true value of f(x) and theapproximated value Pn(x). That is, En(x) = f(x) − Pn(x). The followingtheorem provides an explicit formula for En(x).

Theorem 60.1 (Taylor’s Theorem)Suppose that f(x) has derivatives f ′(x), f ′′(x), · · · , f (n)(x), f (n+1)(x) near x =a. Then there is a c between x and a such that

f(x) = Pn(x) + En(x)

where

Pn(x) = f(a) +f ′(a)

1!(x− a) +

f ′′(a)

2!(x− a)2 + · · ·+ f (n)(a)

n!(x− a)n

and

En(x) =f (n+1)(c)

(n + 1)!(x− a)n+1.

Proof.The theorem is trivially true if x = a. In this case, we let c = a and En(x) = 0.So suppose that x 6= a.Let En(x) = f(x)− Pn(x). Define the function

F (y) =n∑

i=0

f (i)(y)

i!(x− y)i +

En(x)

(x− a)n+1(x− y)n+1

where y 6= a. Then F is continuous on [a, b] and differentiable in (a, b) withderivative

F ′(y) =f (n+1)(y)

n!(x− y)n − (n + 1)En(x)

(x− a)n+1(x− y)n.

By the Mean Value Theorem, there is a < c < b such that

F ′(c) =F (b)− F (a)

b− a.

162

But F (a) = F (x) = f(x) so that F ′(c) = 0. This implies that

f (n+1)(c)

n!(x− c)n =

(n + 1)En(x)

(x− a)n+1(x− c)n,

which reduces to En(x) = f (n+1)(x)(n+1)!

(x− a)n+1.

Example 60.1We will find En(x) for f(x) = sin x about x = 0. Using the idea of successivedifferentiation we find the following

f ′(x) = cos x, f ′(c) = cos cf ′′(x) = − sin x, f ′′(c) = − sin cf ′′′(x) = − cos x, f ′′′(c) = − cos cf (4)(x) = sin x, f (4)(c) = sin cf (5)(x) = cos x, f (5)(c) = cos cf (6)(x) = − sin x, f (6)(c) = − sin cf (7)(x) = − cos x, f (7)(c) = − cos cf (8)(x) = sin x, f (8)(c) = sin cf (9)(x) = cos x, f (9)(c) = cos c

...

It follows that

f (n+1)(c) =

cos c if n is a multiple of 4− sin c if n− 1 is a multiple of 4− cos c if n− 2 is a multiple of 4sin c if n− 3 is a multiple of 4.

Hence, using the above theorem we see that

En(x) =

cos c(n+1)!

xn+1 if n is a multiple of 4− sin c(n+1)!

xn+1 if n− 1 is a multiple of 4− cos c(n+1)!

xn+1 if n− 2 is a multiple of 4sin c

(n+1)!xn+1 if n− 3 is a multiple of 4.

Note that for all n we have |En(x)| ≤ |x|n+1

(n+1)!. But we know that limn→∞

xn+1

(n+1)!=

0 ( since the series∑∞

n=0xn

n!is convergent by the ratio test so its nth term

converges to zero). Hence, limn→∞ En(x) = 0.

163

Remark 60.1Note that from the above theorem, we have an upper bound for the error:

|f(x)− Pn(x)| = |En(x)| ≤ M

(n + 1)!|x− a|n+1,

where M is the maximum value of f (n+1)(x) on the interval between a andx.

Example 60.2Use Taylor’s theorem to approximate sin 3◦ to four decimal places accuracy;that is, the magnitude of the error is less than 0.5× 10−4.

Solution.Since ◦ is close to 0 then we approximate sin x by a Taylor polynomial nearx = 0. Thus, by Taylor Theorem we have sin x = Pn(x) + En(x). Replace xby π

60= 3◦ to obtain

∣∣∣∣En

(π

60

)∣∣∣∣ ≤ (π

60

)n+1

· 1

(n + 1)!< 0.5× 10−4.

Solving for n we find n = 3. Thus,

sin 3◦ = sin(

π

60

)≈ π

60−

(π60

)3

3!≈ 0.0523.

164

Practice Problems

Exercise 60.1Find En(x) for the function f(x) = cos x about x = 0. Show that limn→∞ En(x) =0.

Exercise 60.2Find En(x) for the function f(x) = ex about x = 0. Show that limn→∞ En(x) =0.

Exercise 60.3Give a bound on the error E4, when ex is approximated by its fourth-degreeTaylor polynomial for −0.5 ≤ x ≤ 0.5.

165

61 Taylor Series

Let f(x) be a function with derivatives of any order at x = a, that is, f isan infinitely differentiable function. Fix a value of x near a and considerthe sequence of Taylor ploynomials {Pn(x)}∞n=0 where

Pn(x) = f(a) +f ′(a)

1!(x− a) +

f ′′(a)

2!(x− a)2 + · · ·+ f (n)(a)

n!(x− a)n.

If limn→∞ Pn(x) exists and is equal to f(x) then we write

f(x) =∞∑

n=0

f (n)(a)

n!(x− a)n.

The right-hand series is called the Taylor series expansion of f(x) about

x = a. We call f (n)(a)n!

(x−a)n the general term of the series. It is a formmulathat gives any term in the series.

Example 61.1Let f(x) = 1

1−x. Finding successive derivatives we have

f ′(x) = (1− x)−2, f ′(0) = 1f ′′(x) = 2(1− x)−3, f ′′(0) = 2 = 2!f ′′′(x) = 3 · 2(1− x)−4, f ′′′(0) = 3!

...f (n)(x) = n · (n− 1) · · · 3 · 2(1− x)−(n+1), f (n)(0) = n!

Thus,

Pn(x) = 1 + x + x2 + x3 + · · ·+ xn =1− xn+1

1− x.

This is the nth partial sum of a geometric series that converges for |x| < 1.Moreover,

1 + x + x2 + x3 + · · · = 1

1− x.

This shows thatf(x) = 1 + x + x2 + · · ·

for all −1 < x < 1.

166

Remark 61.1For a given function f at a given x, it is possible that the Taylor seriesconverges to a value different from f(x). However, the Taylor series of mostof the functions discussed in this section do converge to the original function.

Example 61.2Show that the Taylor polynomials sequence {Pn(x)}∞n=0 of the function f(x)defined below about x = 0 converges to 0 for all x. Thus, if x 6= 0 thenlimn→∞ Pn(x) 6= f(x).

f(x) =

{0 if x = 0

e−2

x2 if x 6= 0

Solution.Using a graphing calculator, one can see that the function f and all itsderivatives are flat at 0. That is, the derivatives of f(x) of all orders arezero. Hence, Pn(x) = 0 for all x and for all n. This shows that if x 6= 0 thenlimn→∞ Pn(x) = 0 6= f(x).

Taylor Series Expansion of f(x) = cos x About x = 0Let f(x) = cos x. We will find the Taylor series expansion of f(x) aboutx = 0. Indeed,

f ′(x) = − sin xf ′′(x) = − cos xf ′′′(x) = sin xf (4)(x) = cos x

...

We see that the derivatives go through a cycle of length 4 and then repeatthat cycle forever. It follows that

f (k)(0) =

{0 if k is odd

(−1)k2 if k is even.

Hence,

P2n(x) = P2n+1 = 1− 1

2!x2 +

1

4!x4 − · · ·+ (−1)n x2n

(2n)!=

n∑k=0

(−1)k x2k

(2k)!

. Now, consider the series

∞∑n=0

(−1)n x2n

(2n)!= 1− x2

2!+

x4

4!− · · ·

167

and let an = (−1)n

(2n)!. Then

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

(2n + 2)(2n + 1)= 0 < 1.

This shows that the series is convergent for all values of x. It remains to showthat the series converges to cos x. For this purpose, we need to apply Taylor’sTheorem discussed in the previous section. Write

cos x = Pn(x) + En(x).

Then cos x−Pn(x) = En(x). By the previous section we showed that limn→∞ En(x) =0. This implies that

cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · ·

Taylor Series Expansion of f(x) = sin x About x = 0Let f(x) = sin x. We will find the Taylor series expansion of f(x) aboutx = 0. Indeed,

f ′(x) = cos xf ′′(x) = − sin xf ′′′(x) = − cos xf (4)(x) = sin x

...

We see that the derivatives go through a cycle of length 4 and then repeatthat cycle forever. It follows that

f (k)(0) =

{0 if k is even

(−1)k−12 if k is odd.

Hence, for n ≥ 1,

P2n(x) = P2n1(x) = x− x3

3!+

x5

5!− · · ·+ (−1)n

(2n + 1)!x2n+1 =

n∑k=0

(−1)k x2k+1

(2k + 1)!

. Now, let an = (−1)n

(2n+1)!. Then

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

(2n + 3)(2n + 2)= 0 < 1.

168

This shows that the series∑∞

n=0(−1)n

(2n+1)!x2n+1 is convergent for all values of x.

It remains to show that the series converges to sin x. For this purpose, weneed to apply Taylor’s Theorem discussed in the previous section. Write

sin x = Pn(x) + En(x).

From the previous section we see that En −→ 0 as n −→∞. Thus,

sin x =∞∑

n=0

(−1)n

(2n + 1)!x2n+1.

Taylor Series Expansion of f(x) = ex About x = 0For all nonnegative integer k we have f (k)(x) = ex and f (k)(0) = 1. Thus,the nth Taylor polynomial is given by the expression

Pn(x) = 1 +x

1!+

x2

2!+ · · ·+ xn

n!.

Let an = 1n!

. Then by the ratio test we have

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

1

n + 1= 0 < 1.

Thus, the series∑∞

n=0xn

n!converges for all values of x. It remains to show that

the series converges to ex. For that, we need to use Taylor Theorem. Writef(x) = Pn(x) + En(x) where

|En(x)| =∣∣∣∣∣f (n+1)(c)

(n + 1)!xn+1

∣∣∣∣∣ = ec

(n + 1)!|x|n+1 → 0, as n →∞,

since limn→∞xn

n!= 0. Hence,

ex = 1 +x

1!+

x2

2!+ · · · =

∞∑n=0

xn

n!.

Taylor Series Expansion of f(x) = ln (1 + x) About x = 0Taking derivatives:

f ′(x) = (1 + x)−1, f ′(0) = 1 = 0!f ′′(x) = −(1 + x)−2, f ′′(0) = −1!f ′′′(x) = 2(1 + x)−3, f ′′′(0) = 2!

f (4)(x) = −6(1 + x)−4, f (4)(0) = −3!...

f (n)(x) = (−1)n−1(n− 1)!(1 + x)−n, f (n)(0) = (−1)n−1(n− 1)!.

169

Hence,

Pn(x) = x− x2

2+

x3

3− · · ·+ (−1)n−1xn

n=

n∑k=1

(−1)k−1xk

k!.

Letting an = (−1)n−1 1n

and applying the ratio test we find that

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = 1.

Hence, the series∑∞

n=1(−1)n−1 xn

nconverges for all −1 < x < 1. By the

alternating series test we know that∑∞

n=1(−1)n−1

nis convergent so the interval

of convergence of the previous series is −1 < x ≤ 1. It remains to show thatthe series converges to ln (1 + x).Using Taylor theorem, we can write f(x) = Pn(x) + En(x), where

|En(x)| =∣∣∣∣∣f (n+1)(c)

(n + 1)!xn+1

∣∣∣∣∣ ≤ 1

n!|1 + c|n+1· |x|

n+1

(n + 1)!→ 0, as n →∞.

Hence,

ln (1 + x) =∞∑

n=1

(−1)n−1xn

n, − 1 < x ≤ 1.

Taylor Series Expansion of f(x) = (1 + x)p About x = 0Finding successive derivatives:

f(x) = (1 + x)p, f(0) = 1f ′(x) = p(1 + x)p−1, f ′(0) = pf ′′(x) = p(p− 1)(1 + x)p−2, f ′′(0) = p(p− 1)f ′′′(x) = p(p− 1)(p− 2)(1 + x)p−3, f ′′′(0) = p(p− 1)(p− 2)

...f (n)(x) = p(p− 1) · · · (p− n + 1)(1 + x)p−n, f (n)(0) = p(p− 1) · · · (p− n + 1).

Hence,

Pn(x) = 1 +n∑

k=1

p(p− 1) · · · (p− k + 1)

k!xk.

Consider the series 1 +∑∞

n=1p(p−1)···(p−n+1)

k!xn. Letting an = p(p−1)···(p−n+1)

n!

and applying the ratio test we find

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣∣ p(p− 1) · · · (p− n)n!

(n + 1)!p(p− 1) · · · (p− n + 1)

∣∣∣∣∣ = limn→∞

∣∣∣∣p− n

n + 1

∣∣∣∣ = 1.

170

Hence, the radius of convergence is 1 and therefore the series 1+∑∞

n=1p(p−1)···(p−n+1)

n!xn

converges for all −1 < x < 1. It remains to show that the series converges to(1 + x)p.

f(x) = 1 +∞∑

n=1

p(p− 1) · · · (p− n + 1)

n!xn, − 1 < x < 1.

According to Section 63, the function f(x) is differentiable and has powerseries expansion

f ′(x) =∞∑

n=1

np(p− 1) · · · (p− n + 1)

n!xn−1, − 1 < x < 1.

Thus,

xf ′(x) + f ′(x) =∑∞

n=1np(p−1)···(p−n+1)

n!xn +

∑∞n=1

np(p−1)···(p−n+1)n!

xn−1

=∑∞

n=1np(p−1)···(p−n+1)

n!xn + p +

∑∞n=1

(n+1)p(p−1)···(p−n)(n+1)!

xn

= p +∑∞

n=1

[(n+1)p(p−1)···(p−n)

(n+1)!+ np(p−1)···(p−n+1)

n!

]xn

= p + p∑∞

n=1p(p−1)···(p−n+1)

n!xn

= pf(x),

where we have used the fact that

(n + 1)p(p− 1) · · · (p− n)

(n + 1)!+

np(p− 1) · · · (p− n + 1)

n!= p·p(p− 1) · · · (p− n + 1)

n!.

It follows that f(x) satisfies the differential equation

xf ′(x) + f(x) = pf(x)

or equivalently,(1 + x)f ′(x)− pf(x) = 0.

Now, define the function g(x) = f(x)(1+x)p ,−1 < x < 1. Then g′ is differentiable

with derivativeg′(x) = (1+x)pf ′(x)−f(x)p(1+x)p−1

(1+x)2p

= (1+x)f ′(x)−pf(x)(1+x)p+1 = 0.

It follows that g(x) = C for all −1 < x < 1. But g(0) = f(0) = 1 so thatC = 1. This implies that f(x) = (1 + x)p.

171

Example 61.3Use the Binomial series expansion to find the Taylor series expansion of 1

1+x

about x = 0.

Solution.Use the binomial series with p = −1 to obtain

11+x

= (1 + x)−1

= 1− x + x2 − x3 + · · ·

By the ratio test, this series converges for all −1 < x < 1.

172

Practice Problems

Exercise 61.1Consider the function f(x) = sin x.(a) Show that f (k)(0) = 0 for k even.(b) Find the Taylor polynomials P2n(x) and P2n+1(x).(c) Show that the series

x− x3

3!+

x5

5!− · · ·+ (−1)n x2n+1

(2n + 1)!+ · · ·

converges for all x.(d)Use Taylor’s Theorem to show that limn→∞ En(x) = 0.

Exercise 61.2Consider the function f(x) = ex.(a) Show that f (k)(0) = 1 for all values of k.(b) Find the Taylor polynomial Pn(x).(c) Show that the series

1 + x +x2

2!+

x3

3!+ · · ·+ xn

n!+ · · ·

converges for all x.(d)Use Taylor’s Theorem to show that limn→∞ En(x) = 0.

Exercise 61.3Find the Taylor series of f(x) = ln (1 + x) about x = 0, and calculate itsinterval of convergence.

Exercise 61.4Find the Taylor series of the function f(x) = (1 + x)p about x = 0, andcalculate its interval of convergence.

Exercise 61.5Find the Taylor series about x = 0 for 1

1+x.

In Exercises 6 - 9 find the general term of the Taylor series.

173

Exercise 61.6

1

1 + x= 1− x + x2 − x3 + x4 − · · ·

Exercise 61.7

ln (1− x) = −x− x2

2− x3

3− x4

4− · · ·

Exercise 61.8

ex2

= 1 + x2 +x4

2!+

x6

3!+ · · ·

Exercise 61.9

x2 cos (x2) = x2 − x6

2!+

x10

4!− x12

6!+ · · ·

Exercise 61.10(a) Find the terms up to degree 6 of the Taylor series for f(x) = sin (x2)about x = 0 by taking derivatives.(b) Compare your result in part (a) to the series for sin x. How could youhave obtained your answer to part (a) from the series for sin x?

Exercise 61.11Find the radius of convergence of the Taylor series around x = 0 for ln (1− x).

Exercise 61.12Find x such that

1 + x + x2 + x3 + · · · = 5.

Exercise 61.13Find the sum of the following converging series

1 +2

1!+

4

2!+

8

3!+ · · ·+ 2n

n!+ · · ·


1− 1

3!+

1

5!− 1

7!+ · · ·+ (−1)n

(2n + 1)!+ · · ·

174


1− 1

2!+

1

4!− 1

6!+ · · ·+ (−1)n

(2n)!+ · · ·

175

62 Constructing New Taylor Series from Known

Ones

Recall that the question of finding Taylor series of a function f(x) about apoint x = a amounts to finding the coefficients of a power series

∑∞n=0 cn(x−

a)n where cn = f (n)(a)n!

.For some functions finding f (n)(a) is very tedious. Instead, we use other wayswhich are considered easier.We first list five series that we discussed in the previous section:

ex = 1 +x

1!+

x2

2!+ · · ·+ xn

n!+ · · · (5)

This series converges for all values of x.

ln x = (x− 1)− (x− 1)2

2+

(x− 1)3

3− · · ·+ (−1)n−1 (x− 1)n

n+ · · · (6)

This series converges for 0 < x ≤ 2.

sin x = x− x3

3!+

x5

5!− · · ·+ (−1)n x2n+1

(2n + 1)!+ · · · (7)


cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · · (8)


1

1− x= 1 + x + x2 + x3 + · · ·+ xn + · · · (9)

This series converges for −1 < x < 1.

(1 + x)p = 1 + px +p(p− 1)

2!x2 + · · ·+ p(p− 1)(p− 2) · · · (p− n + 1)

n!xn + · · ·(10)

176


Next, we discuss ways of constructing new series from the above listed series.

New Series by SubstitutionOne can derive new series from known series as suggested by the followingexamples.

Example 62.1Find the Taylor series of x

ex about x = 0.

Solution.Replacing x by −x in Formula 5 we find

e−x = 1− x +x2

2!− · · ·+ (−1)n xn

n!+ · · ·

Now, multiplying both sides of this equality by x to obtain

x

ex= xe−x = x− x2 +

x3

2!+ · · ·+ (−1)n xn+1

n!+ · · ·

Example 62.2Find the Taylor series of f(x) = 1

1+x2 about x = 0.

Solution.Replacing x by −x2 in Formula 9 we can write

1

1 + x2= 1− x2 + x4 − x6 + · · ·+ (−1)nx2n + · · ·


New Series by Differentiation and IntegrationThe following result allows us to construct new series from old ones usingthe processes of differentiation and integration.

Theorem 62.1Suppose that the power series

c0 + c1(x− a) + c2(x− a)2 + · · ·

177

converges for all x such that |x− a| < R. Then this series defines a function

f(x) = c0 + c1(x− a) + c2(x− a)2 + · · ·

with domain the set of all x such that |x− a| < R.

(a) The power series

c1 + 2c2(x− a) + 3c3(x− a)2 + · · ·

obtained by differentiating the original series term by term also has radiusof convergence R and sums to f ′(x), i.e.

f ′(x) = c1 + 2c2(x− a) + 3c3(x− a)2 + · · ·

(b) The power series

C + c0(x− a) + c1(x− a)2

2+ c2

(x− a)3

3+ · · ·

obtained by integrating the original series term by term also has radius ofconvergence R and sums to

∫f(x)dx, i.e.

∫f(x)dx = C + c0(x− a) + c1

(x− a)2

2+ c2

(x− a)3

3+ · · ·

The proof of this theorem is beyond the scope of calculus and therefore isomitted.

Example 62.3Find the Taylor series about x = 0 of cos x from the series of sin x.

Solution.We know that d

dx(sin x) = cos x, so we start with Formula 7 and differentiate

this series term by term we get the series

cos x = 1− x2

2!+

x4

4!− · · ·+ (−1)n x2n

(2n)!+ · · ·

Example 62.4Find the Taylor’s series about x = 0 for arctanx from the series for 1

1+x2 .

178

SolutionIntegrating term by term of the series obtained in Example 62.2 we find

arctan x =∫ dx

1 + x2= C + x− x3

3+

x5

5− · · ·

where −1 < x < 1. Since arctan 0 = 0 then C = 0 and therefore

arctan x = x− x3

3+

x5

5− · · ·

Applications Of Taylor Series

Example 62.5Use the series of arctan x to estimate the numerical value of π.

Solution.Substituting x = 1 in to the series for arctan x gives

π = 4 arctan 1 = 4(1− 1

3+

1

5− 1

7+

1

9− · · ·)

By using 500 terms of this series one finds

π ≈ 3.140.

Remember what we’ve said that some functions have no antiderivative whichcan be expressed in terms of familiar functions. This makes evaluating defi-nite integrals of these functions difficult because the Fundamental Theoremof Calculus cannot be used. However, if we have a series representation of afunction, we can oftentimes use that to evaluate a definite integral.

Example 62.6Estimate the value of

∫ 10 sin (x2)dx.

Solution.The integrand has no antiderivative expressible in terms of familiar functions.However, we know how to find its Taylor series: we know that

sin t = x− t3

3!+

t5

5!− · · ·

179

Now if we substitute t = x2, we have

sin (x2) = x2 − x6

3!+

x10

5!− · · ·

In spite of the fact that we cannot antidifferentiate the function, we canantidifferentiate the Taylor series:∫ 1

0 sin (x2)dx =∫ 10 (x2 − x6

3!+ x5

5!− · · ·)dx

= (x3

3− x7

7·3! + x11

11·5! − · · ·)|10

= 13− 1

7·3! + 111·5! −

115·7! + · · ·

≈ 0.31026

Practice Problems

Exercise 62.1Find the first four nonzero terms of the Taylor series about x = 0 of thefunction f(x) = ln (1− 2x).

Exercise 62.2Find the first four nonzero terms of the Taylor series about x = 0 of thefunction f(x) = arcsin x.

Exercise 62.3Find the first four nonzero terms of the Taylor series about x = 0 of thefunction f(x) = 1√

1−x2 .

Exercise 62.4Find the first four nonzero terms of the Taylor series about x = 0 of thefunction f(x) = x3 cos (x2).

Exercise 62.5Find the first four nonzero terms of the Taylor series about x = 0 of thefunction f(x) = ex cos x.

Exercise 62.6Find the Taylor series about 0 for the function f(x) = x sin (x2)−x3, includ-ing the general term.

Exercise 62.7Expand the function f(x) = 1

2+xabout x = 0 in terms of x

2.

180

Exercise 62.8Consider the two functions f(x) = e−x2

and g(x) = 11+x2 .

(a) Write the Taylor series for the two functions about x = 0. What is similarabout the two series? What is different?(b) Looking at the series, which function do you predict will be larger overthe interval (−1, 1)? Graph both and see.(c) Are the functions even, odd or neither?How might you see this by lookingat the series expansions?

181

63 Introduction to Fourier Series

The study of Fourier series is of fundamental importance in both theoreti-cal and applied mathematics, and they are used for modelling phenomenain engineering and physics as well as in areas such as computer science andmedical research.Unlike the approximation by Taylor ploynomials, which requires the approx-imated function to be continuous and the approximation is near the valuewhere the Taylor polynomial is centered, Fourier polynomials can be used torepresent functions, both continuous and discontinuous, and approximationis done over larger intervals. Thus, Fourier approximations are rather globalwhereas approximations by Taylor polynomials are local.

To start with, suppose that f is a function of period 2π, that is, f(x+2π) =f(x) for all x. Now, recall the trigonometric identities

cos (mx) sin (nx) =1

2[sin (m + n)x + sin (m− n)x] (11)

cos (mx) cos (nx) =1

2[cos (m + n)x + cos (m− n)x] (12)

sin (mx) sin (nx) =1

2[cos (m− n)x− cos (m + n)x] (13)

For n > 0 we have∫ π

−πsin (nx)dx =

[− 1

ncos (nx)

]π−π

= − 1

n(cos (nπ)− cos (−nπ)) = 0 (14)

and∫ π

−πcos (nx)dx =

[1

nsin (nx)

]π−π

= − 1

n(sin (nπ)− sin (−nπ)) = 0. (15)

Next, for any m 6= n we have∫ π

−πsin (mx) cos (nx)dx = −1

2

[1

m + ncos ((m + n)x) +

1

m− ncos ((m− n)x)

]π−π

= 0 (16)

182

and if n = m 6= 0 then∫ π

−πsin (mx) cos (mx)dx =

1

2

∫ π

−πsin (2mx)dx = − 1

4m[cos (2mx)]π−π

= 0.

If m = 0 then sin (mx) = 0 and the above integral is zero. If n = 0 then∫ π

−πsin (mx) cos (nx)dx =

∫ π

−πsin (mx)dx = −1

2

[1

mcos (mx)

]π−π

= 0.

Next, for n 6= m we have∫ π

−πcos (mx) cos (nx)dx =

1

2

[1

m + nsin ((m + n)x)) +

1

m− nsin ((m− n)x)

]π−π

= 0 (17)

and∫ π

−πsin (mx) sin (nx)dx = −1

2

[1

m + nsin ((m + n)x))− 1

m− nsin ((m− n)x)

]π−π

= 0. (18)

Finally, for n ≥ 1 we have∫ π

−πcos2 (nx)dx =

∫ π

−π

1 + cos 2nx

2dx

=1

2

[x +

sin 2nx

2n

]π−π

= π (19)

and similarly, ∫ π

−πsin2 (nx)dx =

∫ π

−π

1− cos 2nx

2dx

=1

2

[x− sin 2nx

2n

]π−π

= π. (20)

We want to approximate f(x) with a sum of trigonometric functions of theform

f(x) ≈ Fn(x) = a0 +n∑

k=1

ak cos (kx) +n∑

k=1

bk sin (kx). (21)

183

We call Fn(x) the Fourier polynomial of degree n. The constants a0, a1, · · · , an

and b1, b2, · · · , bn are called the Fourier coefficents.The main question is to select the Fourier coefficients in such a way thatFn(x) is a good approximation to f(x).Integrating both sides of (21) and using (14)-(20) we find that

a0 =1

2π

∫ π

−πf(x)dx.

Now, multiplying both sides of (21) by cos x and then integrating over theinterval [−π, π] we find that

a1 =1

π

∫ π

−πf(x) cos xdx.

Similarly, for k = 2, 3, · · · , n we have

ak =1

π

∫ π

−πf(x) cos (kx)dx.

In a similar way we show that, for 1 ≤ k ≤ n,

bk =1

π

∫ π

−πf(x) sin (kx)dx.

Example 63.1Find the trigonometric polynomial of degree two of the function

f(x) =

{0, −π ≤ x < 01, 0 ≤ x < π

Solution.Indeed, we have

a0 = 12π

∫ 0−π 0dx + 1

2π

∫ π0 1dx = 0 + 1

2π(π)

= 12

a1 = 1π

∫ π0 cos xdx = 1

π[sin x]π0

= 0b1 = 1

π

∫ π0 sin xdx = − 1

π[cos x]π0

= 2π

a2 = 1π

∫ π0 cos 2xdx = 1

2π[sin 2x]π0

= 0b2 = 1

π

∫ π0 sin 2xdx = − 1

2π[cos 2x]π0

= 0.

184

Thus, the trigonometric polynomial of degree 2 is

F2(x) = F1(x) =1

2+

2

πsin x.

Figure 159, shows the graphs of the function f together with F2(x).

Figure 159

Without going through the details, we calculate the coefficients for the thirddegree Fourier polynomial to obtain

F3(x) =1

2+

2

πsin x +

2

3πsin (3x).

Graphing f(x) and F3(x) we see

185

Figure 160

We conclude that a better approximation is found by taking Fourier polyno-mials of high degree. This leads to the introduction of the so called Fourierseries.Fix x and construct the sequence of trigonometric polynomials Fn(x). Iflimn→∞ Fn(x) exists and is equal to f(x) then we write

f(x) = a0 +∞∑

n=1

(an cos (nx) + bn sin (nx)).

We call the right hand side series the Fourier series of f(x).

Example 63.2Find the Fourier series of the function

f(x) = x, − π ≤ x ≤ π.

Solution.

Since the function is odd then an = 0 for all n ≥ 0. Thus, we look forbn for n ≥ 1. Indeed,

bn = 1π

∫ pi−pi x sin (nx)dx = 1

π

[−x cos (nx)

n+ sin (nx)

n2

]π−π

= − 2n

cos (nπ) = 2n(−1)n+1.

Thus, the Fourier series is given by

f(x) = 2

(sin x− sin (2x)

2+

sin (3x)

2− · · ·

).

So far we have been assuming that the underlying function f(x) is 2π −periodic. We consider next the case of a function that is b − periodic, i.e.f(x + b) = f(x) for all x in [−π, π]. Letting x = bt

πthen for x in the interval

[−π, π] the variable t is in the interval [− b2, b

2]. Moreover, the new function

g(t) = f

(bt

2π

)= f(x)

186

is defined on the interval [−π, π] and is 2π − periodic since

g(t + 2π) = f

(b(t + 2π)

2π

)= f(x + b) = f(x) = g(t).

In this case, we can find the Fourier series of g(t) as before obtaining

g(t) = a0 +∞∑

n=1

(an cos (nt) + bn sin (nt)).

Substituting t = 2πxb

we get

f(x) = a0 +∞∑

n=1

(an cos

(2πnx

b

)+ bn sin

(2πnx

b

))

where

a0 = 1b

∫ b2

− b2

f(x)dx;

ak = 2b

∫ b2

− b2

f(x) cos (2πnxb

)dx;

bk = 2b

∫ b2

− b2

f(x) sin (2πnxb

)dx.

187

Practice Problems

Which of the series in Exercises 1 - 4 are Fourier series?

Exercise 63.11 + cos x + cos2 x + cos3 x + · · ·

Exercise 63.2sin x + sin (x + 1) + sin (x + 2) + · · ·

Exercise 63.3cos x

2+ sin x− cos (2x)

4− sin (2x)

2+ cos (3x)

8+ sin (3x)

8+ · · ·

Exercise 63.412− 1

3sin (2x) + 1

4sin (2x)− 1

5sin (3x) + · · ·

Exercise 63.5Repeat Exercise 5 with the function

f(x) =

{−x, −π ≤ x < 0x, 0 ≤ x < π

Use a graphing calculator to draw the graph of each approximation.

In Exercises 6 - 8, find the nth Fourier polynomial for the given functions,assuming them to be periodic with period 2π. Graph the first three approx-imations with the original function.

Exercise 63.6f(x) = x2, − π < x ≤ π.

Exercise 63.7

h(x) =

{0, −π < x ≤ 0x, 0 < x ≤ π

Exercise 63.8h(x) = x, − π < x ≤ π.

188

64 The Definition of a Differential Equation

The purpose of this and the coming sections is to introduce the concept ofdifferential equation, explain what a solution is, and discuss graphical (slopefields) numerical (Euler’s method) and analytical (seperation of variables)methods for finding the solutions.

• Basic DefinitionsA differential equation ( or DE) is an equation involving an unknown func-tion and its derivatives. The order of the differential equation is the orderof the highest derivative of the unknown function involved in the equation.We say that a differential equation is linear if and only if the powers of theunknown function and all its derivatives are either 0 or 1. Thus, the equation

a(x)y′ + b(x)y = c(x)

is a first order linear differential equation while the equation

a(x)y′′ + b(x)y′ + c(x)y = d(x)

is a second order linear differential equation.If the righthand side of the above equations is zero then we say that theequation is homogeneous otherwise the equation is said to be nonhomo-geneous.By a solution to a differential equation we mean a function that satisfiesthe differential equation.

Example 64.1Show that the function y = 100 + e−t is a solution to the DE

y′ = 100− y.

Solution.Indeed, finding the first order derivative of y we have y′ = −e−t. Also,100− y = 100− (100+ e−t) = −e−t. Thus, y′ = 100− y so that y = 100+ e−t

is a solution to the given DE.

The process of the above example applies also with the function y = 100 +Ce−t where C is any constant. Such a solution is referred to as the general

189

solution of the DE. The solution in the above example is obtained fromthe general solution by replacing C by 1. We call this solution a particularsolution. Also note that this particular solution satisfies the two equations{

y′ = 100− yy(0) = 101.

A differential equation together with a set of conditions, known as initialconditions, is called an initial value problem (in short IVP). To solve adifferential equation means to find its general solution.

Example 64.2Show that y = sin (2t) is a particular solution to the equation

d2y

dt2+ 4y = 0.

Solution.Differentiating the function y twice we obtain y′′ = −4 cos (2t). Thus y′′ +4y = −4 cos (2t) + 4 cos (2t) = 0.

Remark 64.1The general solution of a first order differential equation has only one constantsince the solution is derived from integrating once. On the other hand, asecond order differential equation contains two constants since the solutioninvolves antidifferentiation twice.

190

Practice Problems

Exercise 64.1Match the graphs in Figure 161 with the following descriptions.

(a) The population of a new species introduced onto a tropical island.(b) The temperature of a metal ingot placed in a furnace and then removed.(c) The speed of a car traveling at uniform speed and then breaking uniformly.(d) The mass of carbon-14 in a historical specimen.(e) The concetration of tree pollen in the air over the course of a year.

Figure 161

Exercise 64.2Fill in the missing values in the table given if you know that dy

dx= 0.5y.

Assume the rate of growth given by dydx

is approximately constant over eachunit time interval and that the initial value of y is 8.

x 0 1 2 3 4y 8

Exercise 64.3Is y = x3 a solution to the differential equation

xy′ − 3y = 0?

191

Exercise 64.4Find the values of k for which y = x2 + k is a solution to the differentialequation

2y − xy′ = 10.

Exercise 64.5For what values of k (if any) does y = 5+3ekx satisfy the differential equation:

dy

dx= 10− 2y?

Exercise 64.6Find the value(s) of ω for which y = cos ωt satisfies

d2y

dx2+ 9y = 0.

Exercise 64.7(a) Show that P = 1

(1+e−t satisfies the logistic equation

dP

dt= P (1− P ).

(b) What is the limiting value of P as t →∞?

Exercise 64.8Match the following differential equations and possible solutions. (Note thegiven functions may satisfy more tan one equation or none, and some equa-tions may have more than one solution.)

(a) y′′ = y (I) y = cos x(b) y′ = −y (II) y = cos (−x)(c) y′ = 1

y(III) y = x2

(d) y′′ = −y (IV ) y = ex + e−x

(e) x2y′′ − 2y = 0 (V ) y =√

2x

192

65 Slope Fields

In this section we describe a graphical method for solving initial value prob-lems known as the slope fields or direction fields. This method is oftenpractical when analytical methods for solving a differential equation are notpossible. An array of short line segments in the plane having the propertythat the line plotted at a point (x, y) has slope f(x, y) is called a slope fieldfor the differential equation dy

dx= f(x, y). Slope fields are basically used to

visualize the family of solutions of a given differential equation.

Example 65.1Find the slope of the particular solution to the differential equation

dy

dx= 2x

passing through (0,−1). What is the general solution?

Solution.Figure 162 shows the slope field of the particular solution to the given DEpassing through the point (0,−1). The figure was plotted using the followingMAPLE commands:>with(plots):>with(DEtools):>slopeplot:= DEplot(diff(y(x),x)=2*x,y(x),x=-3..3,y=-3..3):>g:=plot(x2 − 1, x=-3..3, y=-3..3, color=black):>display([slopeplot,g]);

193

Figure 162

The solution curves look like parabolas. Thus, the general solution is givenby the equation y = x2 + C.

Example 65.2Using the slope fields, guess the form of the solution curves of the differentialequation

dy

dx= −x

y.

Solution.The slope fields (See Figure 163) is obtained by executing the following Maplecommands

> with(plots):> with (DEtools):> slopeplot := DEplot(diff(y(x), x) = −x/y, y(x), x = −2..2, y = −2..2):> display(slopeplot);

Figure 163

The solution curves look like circles centered at the origin. Thus, the generalsolution is given implicitly by the equation x2 +y2 = C where C is a positiveconstant.

Example 65.3Find the slope field of the differential equation y′ = 2 − y and then sketchthe solution curves with initial conditions y(0) = 1, y(1) = 0 and y(0) = 3.

194

Solution.The slope field and the three curves are shown in Figure 164:

Figure 164

Note that for each solution curve limx→∞ y(x) = 2.

Remark 65.1Finally, we point out here that even though one can draw solution curves,some do not have simple formula.

195

Practice Problems

Exercise 65.1The slope field for the equation y′ = x + y is shown in Figure 165.

(a) Carefully sketch the solutions that pass through the points(i) (0, 0) (ii) (−3, 1) (iii) (−1, 0)(b) From your sketch, write the equation of the solution passing through(−1, 0).(c) Check your solution to part (b) by substituting it into the differentialequation.

Figure 165

Exercise 65.2The slope field for the equation dP

dt= 0.1P (10 − P ), for P ≥ 0, is in Figure

166.

(a) Plot the solutions through the points(i) (0, 0) (ii) (1, 4) (iii) (4, 1)(iv) (−5, 1) (v) −2, 12) (vi) (−2, 10).(b) For which positive values of P are the solutions increasing?decreasing?What is the limiting value of P as t approaches infinity?

196

Figure 166

Exercise 65.3The slope field for the equation dy

dx= sin x sin y, is in Figure 167.

(a) Plot the solutions through the points(i) (0,−2) (ii) (0, π).(b) What is the equation of the solution that passes through (0, nπ), where nis an integer?

Figure 167

Exercise 65.4One of the slope fields in Figure 168 has the equation y′ = x+y

x−y. Which one?

197

Figure 168

Exercise 65.5

Exercise 65.6Match the slope fields in Figure 169 with their differential equations:

(a) y′ = 1 + y2 (b) y′ = x (c) y′ = sin x(d) y′ = y (e) y′ = x− y (f) y′ = 4− y.

198

Figure 169

199

66 Euler’s Method

In the previous section, we introduced direction fields as means for obtaininga picture of various solutions to a differential equation, but sometimes weneed more than a rough graph of a solution. Euler’s method is a numericalmethod for estimating the value of a solution of the initial value problem:{

dydx

= f(x, y)y(x0) = y0

at a given point.We now describe the way the method works. Recall the definition of thederivative of a function y(x) at a point x0 given by

y′(x0) = limx→x0

y − y0

x− x0

.

For x = x0 + h with h small, that is x is close to x0, we can write

y′(x0) ≈y − y0

x− x0

ory − y0 ≈ y′(x0)(x− x0)

i.e.,y ≈ y0 + hf(x0, y0).

Thus, we are approximating the solution curve y = y(x) near (x0, y0) by thetangent line to the curve at this point.Next, we let x1 = x0 + h. We estimate y(x1) = y1 using the equation

y1 ≈ y0 + hf(x0, y0)

Next, we let x2 = x1 + h and estimate y(x2) = y2 using the equation

y2 ≈ y1 + hf(x1, y1).

After n steps, we let xn+1 = xn + h and we estimate y(xn+1) = yn+1 usingthe equation

yn+1 ≈ yn + hf(xn, yn).

Now, if we want to use this method to estimate y(b) starting from y(a) andusing n steps then h = b−a

n.

Geometrically, we obtain a sequence of line segments (tangent lines) thatapproximates the shape of the solution curve. (See Figure 170).Here is an example to help you understand this process.

200

Example 66.1Suppose that y(0) = 1 and dy

dx= y. Estimate y(0.5) in 5 steps using Euler’s

method.

Solution.We have a = 0, b = 0.5, y(0) = 1 and n = 5. Therefore, h = b−a

n= 0.1. The

following chart lists the steps needed:

k xk yk f(xk, yk)h0 0 1 0.11 0.1 1.1 0.112 0.2 1.21 0.1213 0.3 1.331 0.13314 0.4 1.4641 0.146415 0.5 1.61051

Thus, y(0.5) ≈ 1.61051.

Figure 170

Remark 66.11. Euler’s method approximates the value of the solution at a given point; itdoes not give an explicit formula of the solution.

2. It can be shown that the error in Euler’s method is proportional to 1n.

Thus, doubling the number of mesh points will decrease the error by 12.

201

Practice Problems

Exercise 66.1Consider the differential equation y′ = x+y. Use Euler’s method with h = 0.1to estimate y when x = 0.4 for the solution curves satisfying

(a) y(0) = 1 (b) y(−1) = 0.

Exercise 66.2Consider the solution to the differential equation y′ = y passing throughy(0) = 1.

(a) Sketch the slope field for this differential equation, and sketch the so-lution through the point (0, 1).(b) Use Euler’s method with step size h = 0.1 to estimate the solution atx = 0.1, 0.2, · · · , 1.(c) Plot the estimated solution on the slope field; compare the solution andthe slope field.(d) Check that y = ex is the solution to y′ = y with y(0) = 1.

Exercise 66.3(a) Use ten steps of Euler’s method to determine an approximate solutionfor the differential equation y′ = x3, y(0) = 0, using h = 0.1.(b) What is the exact solution? Compare it to the computed approximation.(c) Use a sketch of the slope field for this equation to explain the results ofpart (b).

Exercise 66.4(a) Use Euler’s method to approximate the value of y at x = 1 on the solutioncurve to the differential equation

dy

dx= x3 − y3

that passes through (0, 0). Use h = 15(i.e., 5 steps).

(b) Using Figure 171, sketch the solution that passes through (0, 0). Show theapproximation you made in part (a).(c) Using the slope field, say whether your answer to part (a) is an overesti-mate or an underestimate.

202

Figure 171

Exercise 66.5Consider the differential equation y′ = (sin x)(sin y).

(a) Calculate approximate y-values using Euler’s method with three steps andh = 0.1, starting at each of the following points:(i) (0, 2) (ii) (0, π).(b) Find the slope field and explain your solution to part(a) (ii).

Exercise 66.6Consider the differential equation

dy

dx= 2x, y(0) = 1.

(a) Use Euler’s method with two steps to estimate y when x = 1. Now usefour steps.(b) What is the formula for the exact value of y?(c) Does the error in Euler’s approximation behave as predicted in Remark66.1 (2)?

203

67 Seperation of Variables

In this section, we remind the reader of the method of seperation of variablesdiscussed in Section Applications of Antiderivatives. This method providesan analytical expression for the solution of a differential equation.

Example 67.1Use the method of seperation of variables to find the solution to the differ-ential equation

dy

dx= 2x.

Solution.The given equation leads to dy = 2xdx. Now integrate both sides∫

dy =∫

2xdx

and this leads to y = x2 + C.

Example 67.2Solve by means of the method of seperation of variables the DE

dy

dx= ky

where k is any constant.

Solution.The given equation leads to dy

y= kdx. Integrating both sides of this equation

leads to ln |y| = kx + C. Thus, y = ±ekx+C = ±eC · ekx or y = Cekx. Amodel where k > 0 is said to represent an exponential growth whereas themodel represents an exponential decay for k < 0.

In general, consider the differential equation

dy

dx= f(x)g(y).

Multiply both sides by 1g(y)

to obtain

1

g(y)

dy

dx= f(x).

204

Integrate both sides with respect to x to obtain∫ 1

g(y)y′dx =

∫f(x)dx

But dy = y′dx so that ∫ 1

g(y)dy =

∫f(x)dx.

Example 67.3Solve the differential equation

dy

dx=

y ln y

x

Solution.Seperating the variables and integrating we find∫ dy

y ln y=∫ dx

x.

Letting u = ln y we find ∫ du

u=∫ dx

x.

Thus,ln |u| = ln |x|+ C

or

ln∣∣∣∣ux∣∣∣∣ = C

so thatln y = u = Cx

and this implies thaty = eCx.

205

Practice Problems

Find the solutions to the differential equations in Exercises 1 - 8, subjectto the given initial conditions.

Exercise 67.1y dy

dx= 1, y(0) = 1.

Exercise 67.2dydx

= y2, y(0) = 100.

Exercise 67.31y

dydx

= 5, y(1) = 5.

Exercise 67.4dydx

= xy, y(0) = 1.

Exercise 67.5dydx

= 2y − 4, y(2) = 5.

Exercise 67.6dydx

+ y = 1, y(1) = 0.1.

Exercise 67.7dydx

= xey, y(0) = 0.

Exercise 67.8x(x + 1) dy

dx= y2, y(1) = 1.

Solve the differential equations in Exercises 9 - 11. Assume a, b, and k areconstants.

Exercise 67.9

dy

dx− y

k= 0.

Exercise 67.10

dy

dx− ay = b.

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Exercise 67.11

dy

dx= ky2(1 + x2).

Solve the differential equations in Exercises 12 - 13. Assume x ≥ 0, y ≥ 0.

Exercise 67.12

xdy

dx= (1 + ln x) tan y.

Exercise 67.13

dy

dx= −y ln

(y

2

), y(0) = 1.

Exercise 67.14Find the general solution to the differential equation modeling how a personlearns:

dy

dx= 100− y.

(b) Plot the slope field of this differential equation and sketch solutions withy(0) = 5 and y(0) = 110.(c) For each of the initial conditions in part (b), find the particular solutionand add to your sketch.

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68 Differential Equations: Growth and De-

cay Models

Differential equations whose solutions involve exponential growth or decayare discussed. Everyday real-world problems involving these models are alsointroduced.Consider the differential equation

dP

dt= kP.

Using the method of seperation of variables we find dPP

= kdt. Integratingboth sides to obtain ln |P | = kt+C or P (t) = Cekt. Note that C = P (0) = P0.Thus, P (t) = P0e

kt. We say that the solution represents a growth modelwhen k > 0 and a decay model when k < 0.

Applications for Growth/Decay Models

Example 68.1 (Doubling Time)A certain population grows exponentially. The population grows from 3500people to 6245 people in 8 years. How long will it take for the originalpopulation to double ?This time is called the doubling time.

Solution.We want to find the value of ”t” that will yield a population of 7000 people.So if 7000 = 3500ert then ert = 2. To find r we use the equality 6245 =3500e8r. Taking the natual logarithm of both sides we find 8r = ln

(62453500

)or

r = 18ln(

62453500

)≈ 0.0724. Thus, t = ln 2

0.0724≈ 9.58 years.

Example 68.2 (Half-Life)A team of archaeologists thinks they may have discovered Fred Flintsone’sfossilized bowling ball. But they want to determine whether the fossil is au-thentic before they report their discover to ”ABC’s Nightline.” (Otherwisethey run the risk of showing up on ”Hard Copy” instead.) Fortunately, oneof the scientists is a graduate of ATU’s Math 2924, so he calls upon his ex-perience as follows:The radioactive substance (Carbon 14) has a half-life of 5730 years. By mea-suring the amount of present in a fossil, scientists can estimate how old the

208

fossil is.Analysis of the ”Flinstone bowling ball” determines that 15% of the radioac-tive substance has already decayed. How old is the fossil ?

Solution.”Radioactive decay” means that we have a function of the form A(t) = A0e

rt.Using the given information we can find r. Indeed, 0.5A0 = A0e

5730r. Solv-ing for r we find r ≈ −0.000120968094. Next, we want to find the de-sired t. Since A(t) = .85A0 then A0e

−0.000120968094t = 0.85A0. Thus, t =− 1

0.000120968094ln (0.85) ≈ 1343.5 years.

Continuous Compound InterestContinuously compounded interest is interest that is compounded an infinitenumber of times per year on a particular investment for a specific number ofyears. That is, we want n in the compound interest formula

P (t) = P0

(1 +

r

n

)nt

= P0

[(1 +

r

n

)n]tto be ”very large”. But it can be shown that

limn→∞

(1 +

r

n

)n

= er.

Thus, the formula for the continuous compound interest is given by

P (t) = P0ert.

We call r the continuous growth rate. The annual growth rate is foundby solving the equation er = k + 1 for k. That is, k = er − 1.

Example 68.3A certain investment grows from a balance of $2500 to $4132 in 12 years.The investment account offers continuously compounded interest. What isthe annual interest rate ?

Solution.

Since P (t) = 2500ert then 4132 = 2500e12r. Thus, er =(

41322500

) 112 so that the

annual interest rate is

er − 1 =(

4132

2500

) 112

− 1 ≈ 4.19%.

209

Newton’s Law of Heating or CoolingImagine that you are really hungry and in one minute the pizza that you arecooking in the oven will be finished and ready to eat. But it is going to bevery hot coming out of the oven. How long will it take for the pizza, which isin an oven heated to 450 degrees Fahrenheit, to cool down to a temperaturecomfortable enough to eat and enjoy without burning your mouth?

Have you ever wondered how forensic examiners can provide detectives witha time of death (or at least an approximation of the time of death) based onthe temperature of the body when it was first discovered?

All of these situations have answers because of Newton’s Law of Heatingor Cooling. The general idea is that over time an object will heat up or cooldown to the temperature of its surroundings. In terms of differential equa-tions, the temperature of an object changes, over time, at a rate proportionalto the difference between its temperature and that of its surroundings:

dH

dt= k(H − S)

where S is the temperature of the surroundings. The object is being heatedwhen k > 0 and is cooling down when k < 0.

Example 68.4The temperature of a cup of coffee is initially 150◦F. After two minutes itstemperature cools to 130◦F. If the surrounding temperature of the roomremains constant at 70◦F, how much longer must I wait until the coffee coolsto 110◦F?

Solution.The differential equation that to be solved is

dH

dt= −k(H − 70), k > 0.

Using the seperation of variables technique, we have

dHH−70

= −kdt∫ dHH−70

=∫−kdt

H − 70 = Ce−kt

210

Since H(0) = 150◦F then 150−70 = C so that H(t) = 70+80e−kt. To find kwe use the fact that T (2) = 130. In this case, 130 = 70 + 80e−2k or e−2k = 3

4.

Hence, k = −12ln 3

4= ln

√43.

To finish the problem we must solve for t in the equation

110 = 70 + 80e−kt.

From this equation, we find e−kt = 0.5 or t = − 1k

ln 0.5 = − 1

ln√

43

ln 0.5 ≈4.81 minutes. Thus, I need to wait an additional 2.81 minutes.

Equilibrium SolutionsIf y′ = f(y) then a solution y to the equation y′ = 0 is called an equilibriumsolution. The graph of such a solution is a horizontal line. An equilibriumsolution is said to be stable if a small change in the initial conditions resultsin a solution that approaches the equilibrium solution as t →∞. If the solu-tion veers away from the equilibrium solution as t →∞ then the solution issaid to be unstable. See Figure 172.

Figure 172

Example 68.5The equilibrium solution of the previous example is H(t) = 70. Now, ift →∞ then H(t) = 70 + 80e−kt −→ 70. That is, the equilibrium solution isa stable one.

211

Practice Problems

Exercise 68.1Each of the curves in Figure 173 represents the balance in a bank accountinto which a single deposit was made at time zero. Assuming continuouslycompounded interest, find:

(a) The curve representing the largest initial deposit.(b) The curve representing the largest interest rate.(c) Two curves representing the same initial deposit.(d) Two curves representing the same interest rate.

Figure 173

Exercise 68.2(a) What are the equilibrium solutions to the differential equation

dy

dt= 0.2(y − 3)(y + 2)?

(b) Use a slope field to determine whether each equilibrium solution is stableor unstable.

Exercise 68.3A yam is put in a 200◦C oven and heats up according to the differentialequation

dH

dt= −k(H − 200), k is a positive constant.

212

(a) If the yam is at 20◦C when it is put in the oven, solve the differentialequation.(b) Find k using the fact that after 30 minutes the temperature of the yam is120◦C.

Exercise 68.4The rate of growth of a tumor is proportional to the size of the tumor.

(a) Write a differential equation satisfied by S, the size of the tumor, inmm, as a function of time, t.(b) Find the general solution to the differential equation.(c) If the tumor is 5 mm across at time t = 0, what does that tell you aboutthe solution?(d) If, in addition, the tumor is 8 mm across at time t = 3, what does thattell you about the solution?

Exercise 68.5Hydrocodone bitartrate is used as a cough suppressant. After the drug is fullyabsorbes, the quantity of drug in the body decreases at a rate proportional tothe amount left in the body. The hal-life of hydrocodone bitartrate in the bodyis 3.8 hours, and the usual dose is 10 mg.

(a) Write a differential equation for the quantity, Q, of hydrocodone bitar-trate in the body at time t, inhourd, since the drug was fully absorbed.(b) Solve the differential equation given in part (a).(c) Use the half-life to find the constant of proportionality.(d) How much of the 10 mg dose is still in the body after 12 hours?

Exercise 68.6(a) If B = f(t) is the balance at time t of a bank account that earns inter-est at a rate r%, compounded continuously, what is the differential equationdescribing the rate at which the balance changes? What is the constant ofproportionality, in terms of r?(b) What is the solution to the differential equation?(c) Sketch the graph of B = f(t) for an account that starts with $1,000 andearns interest at the following rates:

)i) 4% (ii) 10% (iii) 15%

213

Exercise 68.7Morphine is often used as a pain-relieving drug. The half-life of morphinein the body is 2 hours. Suppose morphine is administered to a patient intra-venously at a rate of 2.5 mg per hour, and the rate at which the morphine iseliminated is proportional to the amount present.

(a) Show that the constant of proportionality for the rate at which morphineleaves the body (in mg/hour) is k = −0.347.(b) Write a differential equation for the quantity, Q, of morphine in the bloodafter t hours.(c) Use the differential equation to find the equilibrium solution.( This is thelong-term amount of morphine in the body, once the system is stabilized.)

Exercise 68.8A detective finds a murder victim at 9 am. The temperature of the body ismeasured at 90.3◦F. One hour later, the temperature of the body is 89.0◦F.The temperature of the room has been maintained at a constant 68◦F.

(a) Assuming the temperature, T, of the body obeys Newton’s Law of cooling,write a differential equation for T.(b) Solve the differential equation to estimate the time the murder occurred.

214

69 First Order Differential Equations Models

Since most physical situations involve something changing, derivatives comeinto play resulting in a differential equation. In this section, We will investi-gate examples of how differential equations can model such situations.

Example 69.1 (Compartmental Analysis)Consider a tank with volume 100 liters containing a salt solution. Supposea solution with 2kg/liter of salt flows into the tank at a rate of 5 liters/min.The solution in the tank is well-mixed. Solution flows out of the tank at arate of 5 liters/min. If initially there is 20 kg of salt in the tank, how muchsalt will be in the tank as a function of time?

Solution.Let y(t) denote the amount of salt in kg in the tank after t minutes. We usea fundament property of rates:

TotalRate = Rate in − Rate out.

To find the rate in we use

kg

min=

liters

min· kg

liter= (5)(2) = 10kg/min.

The rate at which salt leaves the tank is equal to the rate of flow of solutionout of the tank times the concentration of salt in the solution. Thus, the rateout is

kg

min=

liters

min· kg

liter= (5)(

y

100) =

y

20kg/min.

The differential equation for the amount of salt is{y′ = 10− y

20

y(0) = 20.

Using the method of seperation of variable we find

y′ = 10− 0.05ydy

10−0.05y= dt∫ dy

10−0.05y=

∫dt

−20 ln |10− 0.05y| = t + C10− 0.05y = Ce−0.05t

y = 200− Ce−0.05t.

215

But y(0) = 20 so that C = 180. Hence, the amount of salt in the tank aftert minutes is given by the formula

y(t) = 200− 180e−0.05t.

Example 69.2 (Falling Body)A 50 kg mass is shot from a cannon straight up with an initial velocity of10 m/sec off a bridge that is 100 feet above the ground. If air resistance isgiven by 5v, determine the velocity of the mass when it hits the ground.

Solution.Figure 174 shows a sketch of the situation.

Figure 174

Since the vast majority of the motion will be in the downward direction wewill assume that everything acting in the downward direction should be pos-itive.The motion of the mass consists of two phases. The initial phase in whichthe mass is rising in the air and the second phase when the mass is on itsway down. Figure 175 shows the forces that are acting on the object on theway up and on the way down.

216

Figure 175

By Newton’s Second Law of Motion

Force = Mass× Acceleration.

The initial value problem for the first phase is

mg − 5v = mdv

dt, v(0) = −10

and that for the second phase is

mg − 5v = mdv

dt, v(t0) = 0

where t0 is the time when the object is at the highest point and is ready tostart on the way down. Note that at this time the velocity would be zero.Note that the differential equation for both of the situations is identical. Thiswont always happen, but in those cases where it does, we can ignore thesecond initial value condition and just let the first govern the whole process.Thus, we will consider solving the linear first order differential equation

mg − 5v = mdv

dt, v(0) = −10.

This is done as follows:

mdvdt

= mg − 5vdvdt

= − 5m

(v − mg

5

)∫ dv

v−mg/5= − 5

m

∫dt

ln |v −mg/5| = − 5m

t + C

v −mg/5 = Ce−5m

t

v = mg5

+ Ce−5m

t.

217

But, m = 50, g = 9.8m/sec2, v(0) = −10 so that C = −108. Thus,

v(t) = 98− 108e−0.1t.

Now, to find the velocity of the mass when it hits the ground we must firstthe time when it hits the ground. Let s(t) be the position of the functionwith respect to the bridge. Then

s(t) =∫

(98− 108e−0.1t)dt = 98t + 1080e−0.1t + C.

Since s(0) = 0 then C = −1080. Thus,

s(t) = 98t + 1080e−0.1t − 1080.

Using a calculator, solve the equation s(t) = 100 to find t ≈ 5.98147. Thus,the mass hits the ground with a velocity of

v(5.98147) ≈ 38.61841.

Example 69.3 (Lottery Winnings)You just won the lottery. You put your $5,000,000 in winnings into a fundthat has a rate of return of 4%. Each year you use $300,000. How muchmoney will you have twenty years from now?

SolutionAgain, using a propert of rates we have

TotalRate = Rate in − Rate out.

where the rate out is $300,000 and the rate in is 0.04x where x(t) denotesthe balance in the account after t years. Thus,

dx

dt= 0.4x− 300, 000.

This is both linear first order and separable. We separate and integrate toobtain ∫ dx

0.4x−300,000=

∫dt

25 ln |0.4x− 300, 000| = t + C

0.4x− 300, 000 = Cet25

x = 7, 500, 00 + Cet25 .

Since x(0) = 5, 000, 000 we find C = −2, 500, 000. Finally,

x(20) = 7, 500, 000− 2, 500, 000e2025 = $1, 936, 148.

218

Practice Problems

Exercise 69.1The velocity, v, of a dust particle of mass m and acceleration a satisfies theequation

mdv

dt= mg − kv, where g and k are constants..

By differentiating this equation, find a differential equation satisfied by a.(Your may contain m, g, k, but not v.) Solve for a, given that a(0) = g.

Exercise 69.2A deposit is made to a bank account paying 8% interest compounded contin-uously. Payments totaling $2,000 per year are made from this account.

(a) Write a differential equation for the balance, B, in the account aftert years.(b) Write the solution to the differential equation.(c) How much is in the account after 5 years is the initial deposit is:(i) $20,000? (ii) $30,000?

Exercise 69.3At time t = 0, a bottle of juice at 90◦F is stood in a mountain stream whosetemperature is 50◦F. After 5 minutes, its temperature is 80◦F. Let H(t) de-note the temperature of the juice at time t in minutes.

(a) Write a differential equation for H(t), using Newton’s Law of Cooling.(b) Solve the differential equation.(c) When will the temperature of the juice have dropped to 60◦F?

Exercise 69.4According to a simple physiological model, an athletic adult needs 20 caloriesper day per pound of body weight to maintain his weight. If he consumesmore or fewer calories than those required to maitain his weight, his weightchanges at a rate proportional to difference between the number of caloriesconsumed and the number needed to maitain his current weight; the constantof proportionality is 1/3500 pounds per calorie. Suppose that a particularperson has a constant calorie intake of I calories per day. Let W (t) be theperson’s weight in pounds at time t (mesured in days).

219

(a) What differential equation has solution W (t)?(b) Solve this differential equation.(c) Graph W (t) if the person starts out weighing 160 pounds and consumes3000 calories a day.

Exercise 69.5Water leaks out of a barrel at a rate proportional to the square root of thedepth of the water at the time. If the water level starts at 36 inches and dropsto 35 inches in 1 hour, how long will it take for all of the water to leak outof the barrel?

Exercise 69.6A spherical snowball melts at a rate proportional to its surface area.

(a) Write a differential equation for its volume, V.(b) If the initial value is V0, solve the differential equation and graph the so-lution.(c) When does the snowball disappear?

Exercise 69.7When a course ends, students start to forget the material they have learned.One model assumes that the rate at which a student forgets material is pro-portional to the difference between the material currently remembered andsome positive constant, a.

(a) Let y = f(t) be the fraction of the original material remembered t weeksafter the course has ended. Set up a differential equation for y. Your equationwill contain two constants; the constant a is less than y for all t.(b) Solve the differential equation.(c) Describe the practical meaning (in terms of the amount remembered) ofthe constants in the solution y = f(t).

Exercise 69.8An aquarium pool has volume 2 · 106 liters. The pool initially contains purefresh water. At t = 0 minutes, water containing 10 grams/liter of salt ispoured into the pool at a rate of 60 liters/minute. The salt water instantlymixes with the fresh water, and the excess mixture is drained out of the pool

220

at eh same rate (60 liters/minute).

(a) Write a differential equation for S(t), the mass of salt in the pool attime t.(b) Solve the differential equation to find S(t).(c) What happens to S(t) as t →∞.

221

70 The Logistic Model

One of the consequences of exponential growth is that the output f(t) in-creases indefinitely in the long run. However, in some situations there is alimit L to how large f(t) can get. For example, the population of bacteriain a laboratory culture, where the food supply is limited. In such situations,the rate of growth slows as the population reaches the carrying capacity. Oneuseful model is the logistic growth model.Thus, logistic functions model resource-limited exponential growth.A logistic function involves three positive parameters L, C, k and has thefrom

f(t) =L

1 + Ce−kt.

We next investigate the meaning of these parameters. From our knowledge ofthe graph of e−x we can easily see that e−kt → 0 as t →∞. Thus, f(t) → Las t → ∞. It follows that the parameter L represents the limiting value ofthe output past which the output cannot grow. We call L the carryingcapacity.Now, to interpret the meaning of C, we let t = 0 in the formula for f(t)and obtain (1 + C)f(0) = L. This shows that C is the number of times thatthe initial output must grow to reach L. Finally, the parameter k affectsthe steepness of the curve, that is, as k increases, the curve approaches theasymptote y = L more rapidly.

Example 70.1Show that a logistic function is approximately exponential function withcontinuous growth rate k for small values of t.

Solution.Rewriting a logistic function in the form

f(t) =Lekt

ekt + C

we see that f(t) ≈ L1+C

ekt for small values of t.

Graphs of Logistic FunctionsGraphing the logistic function f(t) = 185

1+48e−0.032t (See Figure 175) we find

222

Figure 175

As is clear from the graph above, a logistic function shows that initial expo-nential growth is followed by a period in which growth slows and then levelsoff, approaching (but never attaining) a maximum upper limit.Notice the characteristic S-shape which is typical of logistic functions.

Point of Diminishing ReturnsAnother important feature of any logistic curve is related to its shape: everylogistic curve has a single inflection point which separates the curve into twoequal regions of opposite concavity. This inflection point is called the pointof diminishing returns.

Finding the Coordinates of the Point of Diminishing ReturnsTo find the point of inflection of a logistic function of the form P = f(t) =

L1+Ce−kt , we notice that P satisfies the equation

dP

dt= kP

(1− P

L

).

Using the product rule we find

d2P

dt2= k

dP

dt

(1− 2P

L

).

Since dPdt

> 0 we conclude that d2Pdt2

= 0 at P = L2.

To find y, we set y = L2

and solve for t :

L2

= L1+Ce−kt

2L

= 1+Ce−kt

L

2 = 1 + Ce−kt

1 = Ce−kt

ekt = Ckt = ln Ct = ln C

k

Thus, the coordinates of the diminishing point of returns are(

ln Ck

, L2

).

Logistic functions are good models of biological population growth in species

223

which have grown so large that they are near to saturating their ecosystems,or of the spread of information within societies. They are also common inmarketing, where they chart the sales of new products over time.

Example 70.2The following table shows that results of a study by the United Nations(New York Times, November 17, 1995) which has found that world popula-tion growth is slowing. It indicates the year in which world population hasreached a given value:

Year 1927 1960 1974 1987 1999 2011 2025 2041 2071Billion 2 3 4 5 6 7 8 9 10

(a) Construct a scatterplot of the data, using the input variable t is thenumber of years since 1900 and output variable P = worldpopulation (inbillions).(b) Using a logistic regression, fit a logistic function to this data.(c) Find the point of diminishing returns. Interpret its meaning.

Solution.(a) For this part, we recommand the reader to use a TI for the plot.(b) Using a TI with the logisitc regression we find L = 11.5, C = 12.8, k =0.0266. Thus,

P =11.5

1 + 12.8e−0.0266t.

(c) The inflection point on the world population curve occurs when t =ln Ck

= ln 12.80.0266

≈ 95.8. In other words, according to the model, in 1995 worldpopulation attained 5.75 billion, half its limiting value of 11.5 billion. Fromthis year on, population will continue to increase but at a slower and slowerrate.

224

Practice Problems

Exercise 70.1The Tojolobal Mayan Indian community in Southern Mexico has available afixed amount of land. The proportion, P, of land in use for farming t yearsafter 1935 is modeled with the logistic function

P (t) =1

1 + 2.968e−0.0275t.

(a) What proportion of the land was in use for farming in 1935?(b) What is the long run prediction of this model? (c) When was half theland in use for farming?(d) When is the proportion of land used for farming increasing most rapidly?

Exercise 70.2The total number of people infected with a virus often grows like a logisticcurve. Suppose that 10 people originally have the virus, and that in the earlystages of the virus (with time, t, measured in weeks), the number of peopleinfected is increasing exponentially with k1.78. It is estimated that, in thelong run, approximately 5000 people become infected.

(a) Use this information to find a logistic function to model this situation.(b) Sketch a graph of your answer to part (a).(c) Use your graph to estimate the length of time until the rate at which peo-ple are becoming infected starts to decrease. What is the vertical coordinateat this point?

Exercise 70.3The table below gives the percentage, P, of households with a VCR, as a func-tion of year.

Year ’78 ’79 ’80 ’81 ’82 ’83 ’84P% 0.3 0.5 1.1 1.8 3.1 5.5 10.6Year ’85 ’86 ’87 ’88 ’89 ’90 ’91P% 20.8 36.0 48.7 58.0 64.6 71.9 71.9

(a) Explain why a logistic model is reasonable one to use for this data.(b) Use the data to estimate the point of inflection of P. What limiting valueL does the point of inflection predict? Does this limiting value appear to be

225

accurate given the percentages for 1990 and 1991?(c) The best logistic equation for this data turns out to be the following. Whatlimiting value does this model predict?

P =75

1 + 316.75e−0.699t.

Exercise 70.4An alternative method of finding the analytic solution to the logistic equation

dP

dt= kP

(1− P

L

)uses the substitution P = 1/u.

(a) Show thatdP

dt= − 1

u2

du

dt.

(b) Rewrite the logistic equation in terms of u and t, and solve for u in termsof t.(c) Using your answer to part (b), find P as a function of t.

Exercise 70.5Any population, P, for which we can ignore immigration, satisfies

dP

dt= Birth rate− death rate.

For organisms which need a partner for reproduction but rely on a chanceencounter for meeting a mate, the birth rate is proportional to the square ofthe population. Thus, the population of such a type of organism satisfies adifferential equation of the form

dP

dt= aP 2 − bP, with a, b > 0.

Suppose that a = 0.02 and b = 0.08.

(a) Sketch the slope field for this differential equation for 0 ≤ t ≤ 50, 0 ≤P ≤ 8.(b) Use your slope field to sketch the general shape of the solutions to thedifferential equation satisfying the following initial conditions:(i) P (0) = 1 (ii) P (0) = 3 (iii) P (0) = 4 (iv) P (0) = 5.(c) Are there any equilibrium values of the population? If so, are they stable?

226

71 Second-Order Differential Equations Mod-

els

In this section we consider solving a second order differential equation of theform

y′′ + ω2y = 0. (22)

We begin with the observation that if y satisfies this equation, then y′′ isequal to a constant multiple of y. Hence it would be reasonable to begin withy = eλx, for some constant λ, as an initial guess. In that case, y′ = λeλx andy′′ = λ2eλx so y will be a solution to (22) if and only if

λ2eλx + ω2eλx = 0.

Dividing through by eλx to obtain

λ2 + ω2 = 0

This equation has complex solutions and can be factored as

(λ− iω)(λ + iω) = 0, where i =√−1.

Hence, λ = ±iω. But, by definition

e±iωx = cos ωx± i sin ωx.

Since eiωx and e−iωx are solutions then by the principle of superposition sodo

1

2(eiωx + e−iωx) = cos ωx

and1

2i(eiωx − e−iωx) = sin ωx.

Hence, a general solution to (22) is

y = C1 cos (ωx) + C2 sin (ωx). (23)

Oscillations of a Mass on a SpringConsider a mass m attached to the end of a spring hanging from a ceiling.We assume that the mass of the spring is negligible in comparison with the

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mass m. When the system is left undisturbed, the force of gravity is balancedby the force the spring exerts on the mass, and the spring is in equilibriumposition. If we pull down on the mass, we feel a force pulling upwards andif we push the mass upward we feel a force pushing the mass down. If wepush the mass up and then releases it then the mass oscillates up and downaround the equilibrium position.By Hooke’s Law, the net force, F, exerted on the mass is proportional to thedisplacement, s :

F = −ks

where k > 0 is the spring constant and depends on the physical propertiesof the spring. The negative sign is to indicate that the net force is in theopposite direction to the displacement.Now, by Newton’s Second Law of Motion, we have

Net Force = Mass× Acceleration.

Thus, the motion of a mass on a spring is described by the second orderdifferential equation

−ks = md2s

dt2

ord2s

dt2+

k

ms = 0

By (23), the general solution to this equation is given by

s(t) = C1 cos

√ k

mt

+ C2 sin

√ k

mt

.

Example 71.1Find the solution to the equation

d2s

dt2+ 4s = 0

satisfying the boundary conditions s(0) = 0 and s(π4) = 20.

SolutionSince ω2 = 4 then ω = 2 so that the general solution is given by

s(t) = C1 cos 2t + C2 sin 2t.

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The condition s(0) = 0 leads to C1 = 0 and s(π4) = 20 leads to C2 = 20.

Hence, s(t) = 20 sin 2t.

Now, let A =√

C21 + C2

2 , cos φ = C2

A, and sin φ = C1

A. Hence,

C1 cos (ωx) + C2 sin (ωx) = A sin φ cos (ωx) + A cos φ sin (ωx)= A sin (ωx + φ)

Since A is half the distance between the maximum and the minimum pointson the graph of A sin (ωx + φ) then it is called the amplitude. The angleφ is called the phase shift. It is assumed that −π < φ ≤ π. The period ofthe motion is given by T = 2π

ω.

Example 71.2Find the amplitude, period, and the phase shift of the solution to the equation

y′′ + 16y = 0, y(0) = 5, y′(0) = 0.

Solution.Since ω = 4 then the general solution is

y(x) = C1 cos (4x) + C2 sin (4x).

Since y(0) = 5 then C1 = 5. Since y′(0) = 0 then 4C2 = 0 so that C2 = 0.Hence, y = 5 cos (4x). Thus, y = 5 sin (4x + φ) where cos φ = 0 and sin φ = 1.The amplitude of this motion is A = 5, the period is T = π

2, and the phase

shift is φ = π2.

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Practice Problems

Exercise 71.1Check by differentiation that y(t) = 2 cos t+3 sin t is a solution to y′′+y = 0.

Exercise 71.2Check by differentiation that y(t) = A cos t+B sin t is a solution to y′′+y = 0for any constants A and B.

Exercise 71.3Check by differentiation that y(t) = A cos ωt + B sin ωt is a solution to y′′ +ω2y = 0 for all values of A and B.

Exercise 71.4What values of α and A make y(t) = A cos αt a solution to y′′ + 5y = 0 suchthat y′(1) = 3?

Exercise 71.5Write the function s(t) = cos t − sin t as a single sine function. Draw itsgraph.

Exercise 71.6Find the amplitude of the oscillation y(t) = 3 sin 2t + 4 cos 2t.

Exercise 71.7Write the function y(t) = 5 sin (2t)+12 cos (2t) in the form y(t) = A sin (ωt + α).

Exercise 71.8(a) Find the general solution of the differential equation

y′′ + 9y = 0.

(b) For each of the following initial conditions, find a particular solution.

(i) y(0) = 0, y′(0) = 1 (ii) y(0) = 1, y(1) = 0 (iii) y(0) = 0, y(1) = 1.(c) Sketch a graph of the solutions found in part (b).

230

Exercise 71.9Each graph in the figure below reprents a solution to one of the differentialequations:(a) x′′ + x = 0, (b) x′′ + 4x = 0 (c) x′′ + 16x = 0.Assuming the t-scales on the four graphs are the same, which graph repre-sents a solution to which equation? Find an equation for each graph.

Exercise 71.10The following differential equations represent oscillating springs.

(i) s′′ + 4s = 0 s(0) = 5, s′(0) = 0(ii) 4s′′ + s = 0 s(0) = 10, s′(0) = 0(iii) s′′ + 6s = 0 s(0) = 4, s′(0) = 0(iv) 6s′′ + s = 0 s(0) = 20, s′(0) = 0

Which differential equation represents:

(a) The spring oscillating most quickly (i.e. with the shortest period)?(b) The spring oscillating with the largest amplitude?(c) The spring oscillating most most slowly (i.e. with the longest period)?(d) The spring with largest maximum velocity?

231

Exercise 71.11A pendulum of length l makes an angle of x radians with the vertical (seefigure below). When x is small, it can be shown that, approximately:

d2x

dt2= −g

lx,

where g is the acceleration due to gravity.

(a) Solve this differential equation assuming that x(0) = 0 and x′(0) = v0.(b) Solve this equation assuming that the pendulum is let go from the positionwhere x = x0, i.e., the velocity of the pendulum is zero. Measure t from themoment when the pendulum is let go.

Exercise 71.12A brick of mass 3 kg hangs from the end of a spring. When the brick is atrest, the spring is stretched by 2 cm. The spring is then stretched an addi-tional 5 cm and released. Assume there is no air resistance.

(a) Set up a differential equation with initial conditions describing the mo-tion.(b) Solve the differential equation.

Exercise 71.13Consider the motion described by the differential equations:

(a) x′′ + 16x = 0, x(0) = 5, x′(0) = 0.(b) 25x′′ + x = 0, x(0) = −1, x′(0) = 2.

In each case, find a formula for x(t) and calculate the amplitude and periodof the motion.

232

72 Second-Order Homogeneous Linear Dif-

ferential Equations

By a homogeneous linear second order differential equation withconstant coefficients we mean an equation of the form

ay′′ + by′ + cy = 0 (24)

We begin with the observation that if y satisfies this equation, then y′′ is equalto a sum of constant multiples of y and y′. Hence it would be reasonable tobegin with y = eλx, for some constant λ, as an initial guess. In that case,y′ = λeλx and y′′ = λ2eλx so y will be a solution to (24) if and only if

aλ2eλx + bλeλx + ceλx = 0

Dividing by eλx gives the characteristic equation

aλ2 + bλ + c = 0 (25)

To solve this equation for λ we need to consider the following three cases:

• If b2 − 4ac > 0 then (25) has two distinct solutions C1eλ1x and C2e

λ2x,where C1 and C2 are arbitrary constants. By the principle of superposition,the general solution to (24) is given by

y = C1eλ1x + C2e

λ2x

If λ1 < 0 and λ2 < 0 then the equations are called overdamped(i.e. a lotof friction in the system).

• If b2 − 4ac = 0 then (25) has only one solution, λ = − b2a

. By substitu-tion, we can check that both y = C1e

λx and y = C2xeλx are solutions to (24)so that the general solution is given by

y = (C1 + C2x)eλx

If λ < 0 then the equation is said to be critically damped.

• If b2−4ac < 0 then (25) has complex conjugate roots λ1 = −b−i√

4ac−b2

2aand

λ2 = −b+i√

4ac−b2

2a. Writing λ1 = p− iq and λ2 = p+ iq and using the fact that

ex+iy = ex(cos y + i sin y)

233

we see that1

2(eλ1x + eλ2x) = epx cos (qx)

is a solution to (24) and

1

2i(eλ1x − eλ2x) = epx sin (qx)

so the general solution is given by

y = C1e− b

2ax cos (

√4ac− b2

2a)x + C2e

− b2a

x sin (

√4ac− b2

2a)x.

If − b2a

< 0 then the above oscillations are called underdamped

Example 72.1Find the solution of y′′ − 3y′ + 2y = 0 satisfying y(0) = 1 and y′(0) = 0.

Solution.Comparing the given equation with (24) we find a = 1, b = −3, and c = 2.Thus, b2 − 4ac = 1 > 0 and therefore

y = C1ex + C2e

2x

The conditions y(0) = 1 and y′(0) = 0 yield the linear system of equations{C1 + C2 = 1C1 + 2C2 = 0

Solving this system we find, C1 = −1 and C2 = 2. Thus,

y = −ex + 2e2x.

Example 72.2Find the solution of y′′ − 2y′ + y = 0 satisfying y(0) = 2 and y(1) = 0.

Solution.Comparing the given equation with (24) we find a = 1, b = −2, and c = 1.Thus, b2 − 4ac = 0 and therefore

y = (C1 + C2x)ex

The condition y(0) = 2 yields C1 = 2 and y(1) = 0 yields C2 = −2. Thus,

y = 2(1− x)ex.

234

Example 72.3Find the solution of y′′ + y = 0 satisfying y(0) = 1 and y′(0) = 0.

Solution.Comparing the given equation with (24) we find a = 1, b = 0, and c = 1.Thus, b2 − 4ac = −4 < 0 and therefore

y = C1 cos x + C2 sin x

The condition y(0) = 1 yields C1 = 1 and y′(0) = 0 yields C2 = 0. Thus,

y = cos x.

235

Practice Problems

Exercise 72.1Find the general solution to the given differential equation.

1. y′′ + 4y′ + 3y = 0.2. s′′ + 7s = 0.3. d2P

dt2+ dP

dt+ P = 0.

Exercise 72.2Solve the initial value problem.

1. y′′ + 5y′ + 5y = 0, y(0) = 1, y′(0) = 0.2. y′′ − 3y′ − 4y = 0, y(0) = 1, y′(0) = 0.3. y′′ + 6y′ + 5y = 0, y(0) = 1, y′(0) = 0.4. y′′ + 6y′ + 10y = 0, y(0) = 0, y′(0) = 2.

Exercise 72.3Solve the boundary value problem.

1. y′′ + 5y′ + 6y = 0, y(0) = 1, y(1) = 0.2. p′′ + 2p′ + 2p = 0, p(0) = 0, p(π

2) = 20.

Exercise 72.4Match the graphs of solutions in the figure with the differential equations be-low.

(a) x′′ + 4x = 0(b) x′′ − 4x = 0(c) x′′ − 0.2x′ + 1.01x = 0(d) x′′ + 0.2x′ + 1.01x = 0.

236

Exercise 72.5If y = 22t is a solution to the differential equation

d2y

dt2− 5

dy

dt+ ky = 0,

find the value of the constant k and the general solution to this equation.

Exercise 72.6For each of the differential equations given below, find the values of c thatmake the general solution:(a) overdamped (b) underdamped (c) critically damped.

1. s′′ + 4s′ + cs = 02. s′′ + 2

√2s′ + cs = 0

3. s′′ + 6s′ + cs = 0

Exercise 72.7Find a solution to the following differential equation which satisfies z(0) = 3and does not tend to infinity as t →∞ :

d2z

dt2+

dz

dt− 2z = 0.

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