MATH 1014 3.0 W2016 APPLIED CALCULUS I - SECTION P

Stewart Chapter 6 Applications of Integration

REMEMBER

TABLE OF INDEFINITE INTEGRALS

See also Reference Pages 6-10. And remember or review

1) The Riemann Sum (Section 5.2) and 2) The Fundamental Theorem of Calculus (Parts 1 and 2) - Section 5.3. The link between antiderivatives and definite integrals.

1

1

( ) ( ) [ ( ) ( )] ( ) ( )

1( 1) ln | |

1

ln

nn

xx x x

cf x dx c f x dx f x g x dx f x dx g x dx

k dx kx C

xx dx C n dx x C

n x

ae dx e C a dx C

a

+

= + = +

= +

= + ¹ - = ++

= + = +

ò ò ò ò òò

ò ò

ò ò

2 2

1 12 2

sin cos cos sin

sec tan csc cot

sec tan sec csc cot csc

1 1tan sin

1 1

sinh cosh cosh sinh

x dx x C x dx x C

x dx x C x dx x C

x x dx x C x x dx x C

dx x C dx x Cx x

x dx x C x dx x C

- -

= - + = +

= + = - +

= + = - +

= + = ++ -

= + = +

ò òò òò ò

ò ò

ò ò

6.1 Areas between curves.

Consider the region S that lies between two curves y = f(x) and y = g(x) and between the vertical lines x = a and x = b. (Here, f and g are continuous functions and f(x) ≥ g(x) for all x in [a, b]. )

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The Riemann sum

is therefore an approximation to what we intuitively think of as the area of S.

The area A of the region bounded by the curves y = f(x), y = g(x), and the lines x = a, x = b, where f and g are continuous and for all x in [a, b], is:

f(x)>g(x)>0

What if g(x) < 0?3

[ ]1

( *) ( *)n

i ii

f x g x x=

- Då

( ) ( )f x g x³

( ) ( )b

aA f x g x dx= -é ùë ûò

Another type of problem asks for areas between two curves,

Area = 1/3

If we actually want AREA, we must split into 3 sections.

4

( ) ( )b

aA f x g x dx= -ò

Example 5

A1 + A2 = 2√2 – 2

An alternative slicing, if curves can be written as x = g(y) etc.

5

Example 6. Find the area enclosed by the line y = x - 1 and the parabola y2 = 2x + 6.

6

( )

( ) ( )( )

( )

4

2

4 2122

4 2122

43 2

2

1 46 3

1 3

4

14

2 3 2

(64) 8 16 2 8 18

R LA x x dy

y y dy

y y dy

y yy

-

-

-

-

= -

é ù= + - -ë û

= - + +

ùæ ö= - + + úç ÷

è ø û= - + + - + - =

òòò

Alternative: It would need splitting the region in two and computing the areas labeled A1 and A2.

Can we write out the integral? And evaluate? No reason why not, just a little more work than method #1.

PS 6.1. Try a selection from 5-28, 44 (numerical), 48, 49, 51.

Area of a circle: Can we determine this by integration?

x2+y2=a2; A = 4 dxxaa

0

22 substitution x = a cosθ (or a sinθ)

….Remember cos 2θ = cos2θ - sin2θ = 2 cos2θ – 1 = 1 - 2 sin2θ

--------------------------------

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6.2 Volumes

Basic idea is to slice the solid into elemental pieces or slabs. approximate the slabs by cylinders of cross sectional area A and height h, (sides perpendicular to base), Elemental volume is Ah. Then sum over all slabs and approximate the volume of the solid. take the limit as |P| → 0 and you have the volume.

We divide S into n ‘slabs’ of equal width ∆x using the planes Px1, Px2, . . . to slice the solid. A(xi*) is X-sectional area (perpendicular to the x-axis) of the slice at xi*, the sample point between xi-1 and xi.

DEFINITION OF VOLUME

8

1

lim ( *) ( )n b

i axi

V A x x A x dx®¥

=

= D =å òn

When we use the volume formula it is important to rememberthat A(x) is the area of a cross-section obtained by slicing through x perpendicular to the x-axis.

Notice that, for a cylinder, the cross-sectional area is constant: A(x) = A for all x. So, our definition of volume gives V = Ah

Volume of a SphereArea of circle, radius y, A= πy2

Using the definition of volume with a = -r and b = r, we have:

Use symmetry

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343 .V rp=

( )2 2

2 2

0

3 32 3

0

343

( )

2 ( )

2 23 3

r r

r r

r

r

V A x dx r x dx

r x dx

x rr x r

r

p

p

p p

p

- -= = -

= -

é ù æ ö= - = -ç ÷ê ú

ë û è ø=

ò òò

Example 2 : Find the volume of the solid obtained by rotating about the x-axis the region under the curve y=√x from 0 to 1.

Example 3 : Find the volume of the solid obtained by rotatingthe region bounded by y = x3, y = 8, and x = 0 about the y-axis.

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Those above are Solids of Revolution. Obtained from the basic formulae,

with the As being areas of circles of varying radius.

Other examples involve “washers’,

A = π(outer radius)2 – π(inner radius)2

But some volumes are not solids of revolution.

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( )( ) orb d

a cV A x dx V A y dy= =ò ò

Example 7 : The figure shows a solid with a circular base of radius 1. Parallel cross-sections perpendicular to the base are equilateral triangles. Find the volume of the solid.

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1

1

1 12 2

1 0

13

0

( )

3(1 ) 2 3(1 )

4 32 3

3 3

V A x dx

x dx x dx

xx

-

-

=

= - = -

é ù= - =ê ú

ë û

òò ò

Example 8

Find the volume of a pyramid whose base is a square with sideL and whose height is h.

Variations, circular base, triangular base …. Always V = (1/3) Base Area x Height.

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Example 8: A wedge is cut out of a circular cylinder of radius 4 by two planes. One plane is perpendicular to the axis of the cylinder. The other intersects the first at an angle of 30° along a diameter of the cylinder. Find the volume of the wedge.

A cross-section perpendicular to the x-axis at a distance x fromthe origin is a triangle ABC, whose base is y = (16-x2)1/2 and whose height is |BC| = y tan 30° = (16-x2)1/2/√3. Area A(x). Then volume,

14

2 212

2

1( ) 16 16

3

16

2 3

A x x x

x

= - × -

-=

( )

4

4

24 4 2

4 0

43

0

( )

16 116

2 3 3

1 12816

33 3 3

V A x dx

xdx x dx

xx

-

-

=

-= = -

é ù= - =ê ú

ë û

ò

ò ò

PS 6.2. Lots of good examples. In particular #s 50, 51, 52, 53, 71(a wine barrel).

Qu 50.

Qu 51,

15

Qu 52.

Qu 54.

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6.3 Volumes by Cylindrical Shells

Some volume problems are very difficult to handle by the methods discussed in Section 6.2.

Let’s consider the problem of finding the volume of the solid obtained by rotating about the y-axis the region bounded byy = 2x2 - x3 and y = 0. Cannot easily find the inverse x = f-1(y).

Is there another way? Another partitioning or slicing of the body? Try cylindrical shells.

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CYLINDRICAL SHELLS METHODThe figure shows a cylindrical shell with inner radius r1, outer radius r2, and height h.

dV = [circumference] [height] [thickness]Elemental Volume dV ≈ 2πr h dr with dr = r2-r1.

Now, let S be the solid obtained by rotating about the y-axis the region bounded by y = f(x) [where f(x) ≥ 0], y = 0, x = a and x = b, where b > a ≥ 0.

So, an approximation to the volume V of S is given by the sumof the volumes of these shells:

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1

lim 2 ( ) 2 ( )n b

i i ani

x f x x x f x dxp p®¥

=

D =å ò

The volume of the solid obtained by rotating about the y-axis the region under the curve y = f(x) from a to b, is:

where 0 ≤ a < b.

Think of a typical shell, cut and flattened, with radius x, circumference 2πx, height f(x), and thickness ∆x or dx:

19

2 ( )b

aV xf x dxp= ò

( ){

[ ] {2 ( )b

athicknesscircumference height

x f x dxpò 123

So back to the problem we started with,

Find the volume of the solid obtained by rotating about the y-axis the region bounded by y = 2x2 - x3 and y = 0.

We see that a typical shell has radius x, circumference 2πx, and height f(x) = 2x2 - x3.

20

( )( )( )

( )

2 2 3

0

2 3 4

0

24 51 12 5 0

32 165 5

2 2

2 (2 )

2

2 8

p

p

p

p p

= -

= -

é ù= -ë û= - =

òò

V x x x dx

x x x dx

x x

Example 3

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve y = √x from 0 to 1.

To use shells, we relabel the curve y = √x as x = y2.

In this problem, the disk method was simpler, but both are fine.

Try volume of a sphere of radius a using cylindrical shells. Could rotate about either x or y axis.

Try problems from PS 6.3

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( )( )1 2

0

1 3

0

12 4

0

2 1

2 ( )

22 4 2

V y y dy

y y dy

y y

p

p

pp

= -

= -

é ù= - =ê ú

ë û

òò

6.4 Work = Force x distance (briefly)

If F is measured in Newtons and d in meters, then the unit for W is a Newton-meter called a Joule (J).

Work = Force x distance defines work as long as the force is constant. However, what happens if the force f(x) is variable?Assume force is in x direction. As with other integrals

for a force f(x) in the x direction moving from x=a to x=b.

In an example, we use a law from physics: Hooke’s Law.

The force required to maintain a spring stretched x units beyond its natural length is proportional to x

f(x) = kxwhere k is a positive constant (called the spring constant).

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1

lim ( *) ( )n b

i ani

W f x x f x dx®¥

=

= D =å ò

A force of 40 N is required to hold a spring that has been stretched from its natural length of 10 cm to a length of 15 cm. How much work is done in stretching the spring from 15 cm to 18 cm?

According to Hooke’s Law, the force required to hold the spring stretched x meters beyond its natural length is f(x) = kx.When the spring is stretched from 10 cm to 15 cm, the amount stretched is 5 cm = 0.05 m.

This means that f(0.05) = 40, so 0.05k = 40.Therefore, k = 40/0.5 = 800 Nm-1.

Thus, f(x) = 800x and the work done in stretching the spring from 15 cm to 18 cm is:

Do a few PS 6.4 problems. Bike Tire example (c.f 6.4 - 27, 28)

6.5 Average Value of a FunctionIt is easy to calculate the average value of finitely many numbers y1, y2 , . . . , yn :

or better in Σ notation - …….23

( ) ( )

0.0820.08

0.050.05

2 2

800 8002

400 0.08 0.05 1.56J

xW xdx

ù= = ú

û

é ù= - =ë û

ò

1 2 nave

y y yy

n

+ + ××× +=

However, how do we compute the average temperature duringa day if infinitely many temperature readings are possible?

AVERAGE VALUE OF A FUNCTIONIn general, let’s try to compute the average value of a function y = f(x), a ≤ x ≤ b.

We start by dividing the interval [a, b] into n equal sub-intervals, each with length Δx= (b-a)/n.

Then, we choose points x1*, . . . , xn* in successive sub-intervals and calculate the average of the numbers f(x1*), . . . , f(xn*): i.e. with n = (b-a)/Δx

favg

If we let n increase, we would be computing the average valueof a large number of closely spaced values.

By the definition of a definite integral, the limiting value for the average is:

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[ ]

1

1

1

( *) ( *)

1( *) ( *)

1( *)

n

n

n

ii

f x f xb a

x

f x x f x xb a

f x xb a =

+ × × × +-D

= D + × × × + D-

= D- å

1

1 1lim ( *) ( )

n b

i ani

f x x f x dxb a b a®¥

=

D =- -å ò

So, we define the average value of f on the interval [a, b] as:

Examples: f(x) = x on [0,1], f(x) = xn on [0,1], n > 0, How about f(x) = 1?

MEAN VALUE THEOREM

In general, is there a number c at which the value of a functionf is exactly equal to the average value of the function—that is, f(c) = fave?

The mean value theorem for integrals states that this is true for continuous functions.

The geometric interpretation of the Mean Value Theorem for Integrals is as follows.

For ‘positive’ functions f, there is a number c such that the rectangle with base [a, b] and height f(c) has the same area as the region under the graph of f from a to b.

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1( )

b

ave af f x dx

b a=

- ò

An example, y = 1+x2 on [-1,2]. f(-1) and f(1) = favg

Do some examples from PS 6.5.-------------------------------------------------

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