111
Chapter 3 Overview: Indefinite Integrals As noted in the introduction, Calculus is essentially comprised of four operations.
Limits
Derivatives
Indefinite integrals (or Anti-Derivatives)
Definite Integrals There are two kinds of Integrals--the Definite Integral and the Indefinite Integral. For a long time in the mathematical world we did not know that integrals and derivatives were connected. But in the mid-1600s Scottish mathematician James Gregory published the first proof of what is now called the Fundamental Theorem of Calculus, changing the math world forever. The Indefinite Integral is often referred to as the Anti-Derivative, because, as an operation, it and the Derivative are inverse operations (just as squares and square roots, or exponential and log functions). In this chapter, we will consider how to reverse the differentiation process we have been employing. Anti-derivatives will also be used to handle differential equations. Differential
equations--equations that include
dy
dx as well as x and y--form a large subfield of
study in Calculus. We will explore the basics of differential equations through two lenses of analysis—algebraic and graphic. (There is a numerical lens, but we will not consider it in this class.) Algebraically, we will solve separable differential equations. These are relatively simple equations that will provide more practice for integration. Non-separable equations are beyond the scope of this class, but we will see what they look like and why they are so much more difficult to solve. The graphical approach to differential equations use a tool called a slope field. This is a visual representation of the tangent line slopes to a family of curves at a variety of points in the Cartesian plane. It looks, in many ways, like a magnetic field, and we may explore some of the applications of slope fields to physics, if time allows.
112
3.1: Anti-Derivatives--the Anti-Power Rule As we have seen, we can deduce things about a function if its derivative is know. It would be valuable to have a formal process to determine the original function from its derivative accurately. The process is called Anti-differentiation, or Integration.
Symbol: f (x)
dx ="the integral of f of x, d-x"
The dx is called the differential. For now, we will just treat it as part of the integral symbol. It tells us the independent variable of the function (usually, but not always, x) and, in a sense, is where the increase in the exponent comes from. It does have meaning on its own, but we will explore that later. Looking at the integral as an anti-derivative, that is, as an operation that reverses the derivative, we should be able to figure out the basic process. Remember:
d
dx [x n] nx n1
and Dx [constant] is always 0
(or, multiply the power in front and subtract one from the power). If we are starting with the derivative and want to reverse the process, the power must increase by one and we should divide by the new power. Also, we do not know, from the derivative, if the original function had a constant that became zero, let alone what the constant was.
The Anti-Power Rule
1
1
nn x
x dx Cn
for 1n
The "+ C" is to account for any constant that might have been there before the derivative was taken.
113
NB. This Rule will not work if n = -1, because it would require that we divide by
zero. But we know from the Derivative Rules what yields x 1 or 1x as the
derivative--Ln x. So we can complete the anti-Power Rule as:
The Anti-Power Rule
1
1
nn x
x dx Cn
if n 1
1 ln *dx x C
x
*NB. The absolute values are necessary to maintain the domain, since negative numbers cannot be logged.
Since x x xD f x g x D f x D g x
and x xD c f x c D f x ,
then
f x g x
dx f x dx g x dx
c f x dx c f x dx
These allows us to integrate a polynomial by integrating each term separately.
OBJECTIVES Find the Indefinite Integral of a polynomial. Use Integration to solve rectilinear motion problems.
114
Ex 1 3x2 4x 5
dx
2 1 1 1 0 1
2
3 2 1
3 4 5 3 4 52 1 1 1 0 1
3 4 5
3 2 1
x x xx x dx C
x x xC
3 22 5x x x C
Ex 2 x4 4x2 51
x
dx
4 1 2 1 0 1
4 2 1 4 54 5 ln
4 1 2 1 0 1x
x x xx x dx C
x
5 31 45 ln
5 3x x x x C
Ex 3 2 3 4 x x dx
x
1
2 23 3
1 12 1 3
4 4
4ln 12 1 13
x x dx x x dxx x
x xx C
4
3 31 34ln
3 4x x x C
115
Integrals of products and quotients can be done easily IF they can be turned into a polynomial.
Ex 4 x2 x3 2x 1
dx
4 1
2 3 23 3 3
7 44 33 3
2 1 2 2
2 27 44 3
3 3
x x x dx x x x x dx
x x x xC
7 4
4 33 31 6 1 3
2 7 3 4x x x x C
Example 5 is called an initial value problem. It has an ordered pair (or initial value pair) that allows us to solve for C.
Ex 5 f ' x 4x3 6x 3 . Find f x if 0 13f
3
4 2
( ) 4 6 3
3 3
f x x x dx
x x x C
24(0) 0 3 0 3 0 13
13
f C
C
f (x) x4 3x2 3x13
116
Ex 6 The acceleration of a particle is described bya t 3t2 8t 1. Find the
distance equation for x t if 0 3v and 0 1x .
2
3 21
3 2
1
1
3 2
( ) 3 8 1
4
3 0 4 0 0
3
( ) 4 3
v t a t dt t t dt
t t t C
C
C
v t t t t
3 2
4 3 22
4 3 2
2
2
( ) 4 3
1 4 13
4 3 21 4 1
1 0 0 0 3 04 3 2
1
x t v t dt t t t dt
t t t t C
C
C
x t 1
4t4
4
3t3
1
2t2 3t 1
It is important to note that the value of the C is not always equal to the value given at the beginning of the problem. In the problems above, t = 0, and we had polynomial functions – if this were not the case, C could have very different values.
117
Ex 7 The acceleration of a particle is described bya t 12t2 6t 4 . Find the
distance equation for x t if 1 0v and 1 3x .
2
3 21
3 2
1
1
3 2
( ) 12 6 4
4 3 4
0 4 1 3 1 4 1
5
( ) 4 3 4 5
v t a t dt t t dt
t t t C
C
C
v t t t t
3 2
4 3 22
4 3 2
2
2
( ) 4 3 4 5
2 5
3 1 1 2 1 5 1
6
x t v t dt t t t dt
t t t t C
C
C
x t t4 t3 2t2 5t 6
118
3.1 Homework Set A Perform the Anti-differentiation.
1. 6x2 2x 3
dx 2. x3 3x2 2x 4
dx
3. 2
x3
dx 4. 8x4 4x3 9x2 2x 1
dx
5. x3 4x2 5 dx 6. 4x 1 3x 8
dx
7. x 6
x
dx 8. x2 x 3
x
dx
119
9. x 1 3 dx 10. 4x 3
2
dx
11. x 3 x32 6
x
dx 12. 4x3 x 3
x2
dx
Solve the initial value problems.
13. f ' x 3x2 6x 3 . Find f(x) if f(0) = 2.
14. f ' x x3 x2 x 3. Find f(x) if f(1) = 0.
15. f ' x x 2 3 x 1 . Find f(x) if f(4) = 1.
120
16. The acceleration of a particle is described by a t 36t2 12t 8 . Find the
distance equation for x(t) if v(1) = 1 and x(1) = 3.
17. The acceleration of a particle is described by a t t2 2t 4 . Find the
distance equation for x(t) if v(0) = 2 and x(0) = 4.
121
3.1 Homework Set B Perform the Anti-differentiation.
1.
x2 5x 6 dx 2.
x2 4x 7
x
dx
3.
x5 7x3 2x 9
2xdx
4.
x3 3x2 3x 1
x 1dx
5.
y2 5 2
dy 6.
t2 7t 9
t 1dt
7.
4t2 1 3t3 7 dt 8.
x5 3x3 x 7
2xdx
122
Answers: 3.1 Homework Set A
1. 6x2 2x 3
dx 2. x3 3x2 2x 4
dx
3 22 3x x x C 4 3 214
4x x x x C
3. 2
x3
dx 4. 8x4 4x3 9x2 2x 1
dx
232 x C 5 4 3 283
5x x x x x C
5. x3 4x2 5 dx 6. 4x 1 3x 8
dx
6 42 5
3 4x x C 3 229
4 82
x x x C
7. x 6
x
dx 8. x2 x 3
x
dx
3 1
2 2212
3x x C
12 21
2 3ln2
x x x C
9. x 1 3 dx 10. 4x 3
2
dx
4 3 21 3
4 2x x x x C 3 216
12 93
x x x C
11. x 3 x32 6
x
dx 12. 4x3 x 3
x2
dx
3 5 1
2 2 22 612
3 5x x x C
12 122 2 3x x x C
13. f ' x 3x2 6x 3 . Find f (x) if f (0) = 2.
3 23 3 2f x x x x
14. f ' x x3 x2 x 3. Find f (x) if f (1) = 0.
4 3 21 1 1 373
4 3 2 12f x x x x x
123
15. f ' x x 2 3 x 1 . Find f (x) if f (4) = 1.
3
2 23 10 352
2 3 3f x x x x
16. The acceleration of a particle is described by a t 36t2 12t 8 . Find the
distance equation for x(t) if v(1) = 1 and x(1) = 3.
4 3 23 2 4 13 11x t t t t t
17. The acceleration of a particle is described by a t t2 2t 4 . Find the
distance equation for x(t) if v(0) = 2 and x(0) = 4.
4 3 21 12 2 4
12 3x t t t t t
3.1 Homework Set B
1.
x2 5x 6 dx 2.
x2 4x 7
x
dx
3 21 5
3 26x x x C 21
4 7ln2
x xx C
3.
x5 7x3 2x 9
2xdx
4.
x3 3x2 3x 1
x 1dx
5 31 7
10 69ln xx x x C 3 21
3x x x C
5.
y2 5 2
dy 6.
t2 7t 9
t 1dt
5 31 10
5 325y y y C 21
26 3ln 1t t t C
7.
4t2 1 3t3 7 dt 8.
x5 3x3 x 7
2xdx
7 4 312 3 287
7 4 3t t t t C 5 31 1 1 7
ln10 2 2 2
x x x x C
124
3.2: Integration by Substitution--the Chain Rule The other three basic derivative rules--The Product, Quotient and Chain Rules--are a little more complicated to reverse than the Power Rule. This is because they yield a more complicated function as a derivative, one which usually has several algebraic simplifications. The Integral of a Rational Function is particularly difficult to unravel because, as we saw, a Rational derivative can be obtained by differentiating a composite function with a Log or a radical, or by differentiating another rational function. Reversing the Product Rule is as complicated, though for other reasons. We will leave both these subjects for a more advanced Calculus Class. The Chain Rule is another matter. Composite functions are among the most pervasive situations in math. Though not as simple at reverse as the Power Rule, the overwhelming importance of this rule makes it imperative that we address it here. Remember:
The Chain Rule: d
dx f g(x)
f ' g(x) g '(x)
The derivative of a composite turns into a product of a composite and a non-composite. So if we have a product to integrate, it might be that the product came from the Chain Rule. The integration is not done by a formula so much as a process that might or might not work. We make an educated guess and hope it works out. You will learn other processes in Calculus for when it does not work.
Integration by Substitution (The Unchain Rule) 0) Notice that you are trying to integrate a product. 1) Identify the inside function of the composite and call it u. 2) Find du from u. 3) If necessary, multiply a constant inside the integral to create du, and balance it by multiplying the reciprocal of that constant outside the integral. (See EX 2) 4) Substitute u and du into the equation. 5) Perform the integration by Anti-Power (or Transcendental Rules, in next section.) 6) Substitute your initial expression back in for the u.
125
This is one of those mathematical processes that makes little sense when first seen.
But after seeing several examples, the meaning suddenly becomes clear. Be
patient.
OBJECTIVE Use the Unchain Rule to integrate composite, product expressions.
Ex 1 3x2 x3 5 10
dx
x3 5 10
is the composite function. u x3 5
du 3x2 dx
10
2 3 10
11
3 5
11
x x dx u du
uC
11
315
11x C
Ex 2 x x2 5 3
dx
x2 5 3
is the composite function. So u x2 5
du 2x dx
3 32 2
3
4
15 5 2
2
1
2
1
2 4
x x dx x x dx
u du
uC
4
215
8x C
Notice that our du had a constant in it that was not in our initial problem. We simply introduced the constant inside our integral, and multiplied by the reciprocal outside the integral. The problem hasn’t changed because we have just multiplied by 1 (the product of a number and its reciprocal is 1).
126
We can only employ that trick with constants. This is very important. If you feel like you need to introduce a variable, either you have picked your u incorrectly, or this technique simply won’t work, and we need more advanced techniques to perform the integration.
Ex 3 x3 x x4 2x2 54
dx
x4 2x2 54 is the composite function. So
u x4 2x2 5
du 4x3 4x dx 4 x3 x dx
4 43 4 2 4 2 3
4
14
54
12 5 2 5 4
4
1
4
1
4
154
4
x x x x dx x x x x dx
u du
u du
uC
5
44 212 5
5x x C
127
Ex 4 3x2 4x 5
x3 2x2 5x 2 3
dx u x3 2x2 5x 2
du 3x2 4x 5 dx
2 -33 2 2
33 2
-3
2
3 4 5 2 5 2 3 4 5
2 5 2
2
x xdx x x x x x dx
x x x
u du
uC
2
3 2
1
2 2 5 2C
x x x
Sometimes, the other factor is not the du, or there is an extra x that must be replaced with some form of u.
Ex 5 x 1 x 1
dx
u x 1
x u 1
du dx
1
2
3 12 2
5 32 2
1 1 1 1
2
2
25 3
2 2
x x dx u u du
u u du
u u du
u uC
5 3
2 22 41 1
5 3x x C
128
Ex 6 x3 x2 4 3
2
dx
u x2 4
x2 u 4
du 2x dx
3 32 23 2 2 2
32
5 32 2
7 52 2
14 4 2
2
14
2
14
2
1 47 52
2 2
x x dx x x x dx
u u du
u u du
u uC
7 5
2 22 21 44 4
7 5x x C
129
3.2 Homework Set A Perform the Anti-differentiation.
1. 5x 3 3
dx 2. x3 x4 5
24
dx
3. 1 x3 2
dx 4. 2 x
23
dx
5. x 2x2 3
dx 6. dx
5x 2 3
130
7. x3
1 x4
dx 8. x 1
x2 2x 33
dx
9. x5 x2 4 2
dx 10. x 3 x 1 2
dx
3.2 Homework Set B
1.
2x 5 x2 5x 6 6
dx 2.
3t2 t3 1 5
dt
3.
10m15
m2 3m14Êdm
4.
3x2
1 x3 5
dx
131
5.
4s1 5ds 6.
5t
t2 1dt
7.
3m2
m3 8dm
8.
181x 1 5dx
9.
v2
5 v3dv
132
Answers: 3.2 Homework Set A
1. 5x 3 3
dx 2. x3 x4 5
24
dx
41
5 320
x C 25
415
100x C
3. 1 x3 2
dx 4. 2 x
23
dx
7 41 1
7 2x x x C
533
25
x C
5. x 2x2 3
dx 6. dx
5x 2 3
3
2 212 3
6x C
2
1
10 5 2C
x
7. x3
1 x4
dx 8. x 1
x2 2x 33
dx
4112
x C 2
3232 3
4x x C
9. x5 x2 4 2
dx 10. x 3 x 1 2
dx
5 4 3
2 2 21 84 4 4
10 3x x x C
7 5 32 2 2
2 8 8
7 5 33 3 3 Cx x x
3.2 Homework Set B
1.
2x 5 x2 5x 6 6
dx 2.
3t2 t3 1 5
dt
7
215 6
7x x C
631
16
t C
3. 4 2
10 15
3 1
mdm
m m
4.
3x2
1 x3 5
dx
3
42203 1
3m m C
431
14
x C
133
5.
4s1 5ds 6.
5t
t2 1dt
61
4 124
s C 25ln 1
2t C
7.
3m2
m3 8dm
8.
181x 1 5dx
3ln 8m C 61
181 11086
x C
9.
v2
5 v3dv
31ln 5
3v C
134
3.3: Anti-derivatives--the Transcendental Rules
The proof of the Transcendental Integral Rules can be left to a more formal Calculus course. But since the integral is the inverse of the derivative, the discovery of the rules should be obvious from looking at the comparable derivative rules.
Derivative Rules
2
sin cos
cos sin
tan sec
ln
u u
u u
d duu u
dx dx
d duu u
dx dx
d duu u
dx dx
d due e
dx dx
d dua a a
dx dx
2
csc csc cot
sec sec tan
cot csc
1
1
lna
d duu u u
dx dx
d duu u u
dx dx
d duu u
dx dx
d duLn u
dx u dx
d duLog u
dx u a dx
d
dxsin-1 u
1
1 u2Du
d
dxcos-1 u
1
1 u2Du
d
dxtan-1 u
1
u2 1Du
d
dxcsc-1 u
1
u u2 1Du
d
dxsec-1 u
1
u u2 1Du
d
dxcot-1 u
1
u2 1Du
135
Transcendental Integral Rules
2
cos sin
sin cos
sec tan
ln
u u
uu
u du u C
u du u C
u du u C
e du e C
aa du C
a
2
csc cot csc
sec tan sec
csc cot
1 ln
u u du u C
u u du u C
u du u C
du u Cu
du
1 u2
sin1u C
du
1u2
tan1u C
du
u u2 1
sec1u C
Note that there are only three integrals that yield inverse trig functions where there were six inverse trig derivatives. This is because the other three rules derivative rules are just the negatives of the first three. As we will see later, these three rules are simplified versions of more general rules, but for now we will stick with the three.
OBJECTIVE Integrate functions involving Transcendental operations.
Ex 1 sin x 3cos x
dx
sin x 3cos x
dx sin x
dx 3 cos x
dx
cos 3sin x x C
136
Ex 2 ex 4 3csc2 x
dx
ex 4 3csc2 x
dx ex
dx 4 dx 3 csc2 x
dx
4 3cot xe x x C
Ex 3 sec x sec x tan x
dx
sec x sec x tan x
dx sec2 x
dx sec x tan x
dx
tan sec x x C Of course, the Unchain Rule will apply to the transcendental functions quite well.
Ex 4 sin 5x
dx
u 5x
du 5dx
1sin 5 sin 5 5
51
sin 51
cos 5
x dx x dx
u du
u C
1cos 5
5x C
Ex 5 sin6 x cos x
dx
u sin x
du cos x dx
6 6
7
sin cos
1
7
x x dx u du
u C
71sin
7x C
137
Ex 6 x5 sin x6
dx
u x6
du 6x5dx
5 6 6 51sin sin 6
61
sin 6
1cos
6
x x dx x x dx
u du
u C
61cos
6x C
Ex 7 cot3 x csc2 x
dx
2
cot
csc
u x
du x dx
3 2 3 2
3
4
cot csc cot csc
1
4
x x dx x x dx
u du
u C
41cot
4x C
Ex 8 cos x
x
dx u x x
12
du 1
2x
-12 dx
1
2x1
2dx
cos 1 2 cos
2
2 cos
2sin
xdx x dx
x x
u du
u C
2sin x C
138
Ex 9 xex2 1
dx
u x2 1
du 2x dx
2 21 11 2
2
1
21
2
x x
u
u
xe d x e x dx
e du
e C
2 11
2xe C
Ex 10 x
1 x4
dx u x2
du 2x dx
4 22
2
1
1 1 2
21 1
1 1
2 1
1sin
2
xdx x dx
x x
duu
u C
1 21sin
2x C
139
3.3 Homework Set A Perform the Anti-differentiation.
1. x4 cos x5
dx 2. sin 7x 1
dx
3. sec2 3x 1
dx 4.
sin x
x
dx
5. tan4 x sec2 x
dx 6.
Ln x
x
dx
7. e6x
dx 8.
cos 2x
sin3 2x
dx
140
9. 2
2
ln 1
1
x xdx
x
10.
e x
x
dx
11. cot x csc2 x
dx 12.
1
x2sin
1
x
cos
1
x
dx
13. x
1 x4
dx 14. cosx
1 sin2 x
dx
141
3.3 Homework Set B
1.
x5 sin 3x xex2 dx 2.
cosx
1 sin xdx
3.
sec2 2x dx 4.
csc2 ex ex
dx
5.
sec ln x tan ln x 3x
dx
6.
x5 7
x2 e2x sec2 x
dx
7. ex cscex cotex dx 8.
ex 2 ex 1 dx
142
9.
x2 sec2 x3 2xex2
dx
10.
sec2 y tan5 y dy . Verify that your integration is correct by taking the
derivative of your answer.
11.
cosesin
2 1
d
. Verify that your integration is correct by taking
the derivative of your answer.
12.
tsec2 4t2 tan 4t2 dt . Verify that your integration is correct by taking
the derivative of your answer.
143
13.
x2 sinx3 dx 14.
te5t21 dt
15.
ey 1 2
dy 16.
xsec2 x2 tanx2 dx
17.
sin 3t cos5 3t dt 18.
xcosx2
esin x2
dx
19.
tan ln sec d 20.
e4y 2y2 7cos3y dy
144
21.
sin x 4 cos7 x 4
dx
22.
2x
x2 5 sec2 3x xex2
dx
23.
e2t sec2 e2t
dt 24.
18lnm
mdm
25.
2ycos y2 sin4 y2
dy
145
3.3 Homework Set A
1. x4 cos x5
dx 2. sin 7x 1
dx
51sin
5x C
1cos 7 1
7x C
3. sec2 3x 1
dx 4.
sin x
x
dx
1
tan 3 13
x C 2cos x C
5. tan4 x sec2 x
dx 6.
ln
xdx
x
51tan
5x C 21
ln 2
x C
7. e6x
dx 8.
cos 2x
sin3 2x
dx
61
6xe C 21
csc 24
x C
9. 2
2
ln 1
1
x xdx
x
10.
e x
x
dx
2 21ln 1
4x C 2 xe C
11. cot x csc2 x
dx 12.
1
x2sin
1
x
cos
1
x
dx
3
22cot
3x C 21 1
sin2
Cx
13. x
1 x4
dx 14. cosx
1 sin2 x
dx
1 21tan
2x C x C , 2
2x n
146
3.3 Homework Set B
1.
x5 sin 3x xex2 dx 2.
cosx
1 sin xdx
261 1 1
cos 36 3 2
xx x e C ln 1 sin x C
3.
sec2 2x dx 4.
csc2 ex ex
dx
1
tan 22
x C cot x Ce
5.
sec ln x tan ln x 3x
dx
6.
x5 7
x2 e2x sec2 x
dx
1
sec ln3
x C
6 1 21 17 tan
6 2xx x e x C
7. ex cscex cotex dx 8.
ex 2 ex 1 dx
csc xe C 213 2
2x xe e x C
9.
x2 sec2 x3 2xex2
dx
231
tan3
xx e C
10.
sec2 y tan5 y dy . Verify that your integration is correct by taking the
derivative of your answer.
61tan
6y C 61
tan6
dy C
dy
2 5sec tany y
11.
cosesin
2 1
d
. Verify that your integration is correct by taking
the derivative of your answer.
sin 21ln 1
2e C sin 21
ln 12
de C
d
sin2
cos1
e
147
12.
tsec2 4t2 tan 4t2 dt . Verify that your integration is correct by taking
the derivative of your answer.
3
221tan 4
12t C
3221
tan 412
dt C
dt
2 2 2sec 4 tan 4t t t
13.
x2 sinx3 dx 14. 25 1tte dt
31cos
3x C
25 11
10te C
15.
ey 1 2
dy 16.
xsec2 x2 tanx2 dx
212
2y ye e y C
3221
tan3
x C
17.
sin 3t cos5 3t dt 18.
xcosx2
esin x2
dx
61cos 3
18t C
2sin1
2xe C
19.
tan ln sec d 20.
e4y 2y2 7cos3y dy
21ln sec
2C 4 31 2 7
sin34 3 3
ye y y C
21.
sin x 4 cos7 x 4
dx
22.
2x
x2 5 sec2 3x xex2
dx
61
6sec 4x C
22 1 1ln 5 tan 3
3 2xx x e x C
23.
e2t sec2 e2t
dt 24.
18lnm
mdm
21tan
2te C 29ln m C
148
25.
2ycos y2 sin4 y2
dy
3 21csc
3y C
149
3.4: Separable Differential Equations
Vocabulary: Differential Equation (differential equation) – an equation that contains an
unknown function and one or more of its derivatives. General Solution – The solution obtained from solving a differential equation. It
still has the +C in it. Initial Condition – Constraint placed on a differential equation; sometimes called
an initial value. Particular Solution – Solution obtained from solving a differential equation when
an initial condition allows you to solve for C. Separable Differential Equation – A differential equation in which all terms with
y’s can be moved to the left side of an equals sign ( = ), and in which all terms with x’s can be moved to the right side of an equals sign ( = ), by multiplication and division only.
OBJECTIVES Given a separable differential equation, find the general solution. Given a separable differential equation and an initial condition, find a particular solution.
Ex 1 Find the general solution to the differential equation
dy
dx
x
y.
dy
dx
x
y Start here.
ydy xdx Separate all the y terms to the left side of the equation
and all of the x terms to the right side of the equation.
ydy xdx Integrate both sides.
1
2y2
1
2x2 C You only need C on one side of the equation and we put
it on the side containing the x.
150
y2 x2 C Multiply both sides by 2. Note: 2C is still a constant, so
we’ll continue to note it just by C.
x2 y2 C This equation should seem familiar. It’s the family of
circles centered at the origin with radius C . Usually, if it’s possible, we will solve our equation for y – so our solution can be
written as y C x2 .
Note that we could check our solution by taking the derivative of our solution.
x2 y2 C
2x 2y
dy
dx 0
dy
dx
x
y.
This process is called implicit differentiation because we are not explicitly solved for y. You may remember it from last year, and we will discuss it at length in a
later chapter. The key idea is that the derivative of y is dy
dx by the Chain Rule.
Steps to Solving a Differential Equation: 1. Separate the variables. Note: Leave constants on the right side of the
equation. 2. Integrate both sides of the equation. Note: only write the +C on the right
side of the equation. Here’s why – you will have a constant on either of your equation when you integrate both sides of your differential Equation, but you would wind up subtracting one from the other eventually, and two constants subtracted from one another is still a constant, so we only write +C on one side of the equation.
3. Solve for y, if possible. If you integrate and get a natural log in your result,
solve for y. If there is no natural log, solve for C. Note: eln y
y because e
raised to any power is automatically positive, so the absolute values are not necessary.
4. Plug in the initial condition, if you are given one, and solve for C.
151
Ex 2 Find the particular solution to y 2xy3y , given y 3 2 .
ln
dyx y
dx
dyx dx
y
y x x C
2
2 3
2 3
3
x x Cy e 2 3
x x Cy e e2 3 Remember that a b a bx x x
x xy Ke 2 3 Since K could be positive or negative, we can
get rid of the absolute values here.
y Ke 03 2 2
K = 2
y 2ex23x
What if
dy
dx y x ? Since this differential equation can’t be separated, we can’t
use this technique. If we multiplied both sides by dx, we would end up with a y with the dx. We couldn’t just subtract the y because we have only dealt with dy or dx multiplied by the function we are integrating. This requires a more advanced technique that is usually introduced in a class specifically on differential equations in college.
How about
dy
dx ex2
? While we can separate the variables, we get an integral that
we cannot solve with the Unchain Rule. For xdy e dx 2
we cannot choose a u
that will make this integral work with the Unchain Rule.
152
Ex 3 Find the general solution to the differential equation
dm
dt mt .
dm
dt mt Start here.
dm
m tdt Separate all the m terms to the left side of the equation
and all of the t terms to the right side of the equation.
dmtdt
m
Integrate both sides.
ln m
1
2t2 C You only need C on one side of the equation and we put
it on the side containing the x.
eln m
e12
t2C e both sides of the equation to solve for y.
t Cm e e21
2
m Ke12
t2
eC is still a constant, so we will just note it as K.
Again, we could check our solution by taking the derivative of our solution. Also notice that the variables we used did not matter – what is important is making sure that the dm is in the numerator on the m side, and the dt is on the numerator on the t side.
Ex 5 Solve the differential equation
dy
dx
9x2 2x 5
2y ey
y
y
y
y
y
dy x x
dx y e
y e dy x x dx
y e dy x x dx
y e x x x C
y e x x x C
2
2
2
2 3 2
2 3 2
9 2 5
2
2 9 2 5
2 9 2 5
3 5
3 5
This is the general solution. Since it is not possible to explicitly solve this for y, we have to leave it in this form.
153
Ex 6 Find the particular solution to
dr
dt
3t2 sint
4r given that
r 0 3 .
sin
sin
sin
cos
cos
cos
cos
cos
dr t t
dt r
rdr t t dt
rdr t t dt
r t t C
C C
r t t
tr t
tr t
2
2
2
2 3
2 3
2 3
3
3
3
4
4 3
4 3
2
2 3 0 0 17
2 17
1 17
3 2 2
1 17
3 2 2
Since our initial value, r 0 3 , was positive, we only need the positive portion of
the radical. It is implied in these processes that we are dealing with functions for our final equation (by the function notation on the intial value), so we have to pick a single function, either the + or the –, for our final answer. What would have happened if you had solved for r before plugging in the initial condition?
r2 t3
2
1
2cost C
r t3
3
1
2cost C
3 03
3
1
2cos0C C
17
2
r t3
3
1
2cost
17
2
Again, you can check your solution by taking the derivative of your solution.
154
3.4 Homework Set A Solve the differential equation.
1.
dy
dx
y
x
2. x2 1 y ' xy
3. 1 sec y tan y y ' x2 1
4.
dy
dx
e2x
4 y3
155
5.
dy
dx
x2 x3 3
y2
Find the solution of the differential equation that satisfies the given initial condition.
6.
dy
dx
2x3
3y2 ; y 2 0
7.
dy
dx y2 1 ; y(1) 0
8.
du
dt
2t sec2 t
2u; u(0) -5
156
9. Solve the initial-value problem
y ' sin x
sin y; y(0)
2, and graph the
solution.
10. Solve the equation e-y y ' cosx 0 and graph several members of the family
of solutions. How does the solution curve change as the constant C varies?
157
3.4 Homework – Set B Solve the differential equation.
1.
dy
dx 4xy3
2. y ' y2 sinx
3.
dv
dt 2 2v t tv
4.
dy
dt
t
y y2 1
158
5.
d
dr
1 r
Find the solution of the differential equation that satisfies the given initial condition.
6.
dy
dx
ycosx
1 y2 ; y 0 1
7. Find an equation of the curve that satisfies
dy
dx 4x3 y and whose
y-intercept is 7.
8. Solve the initial-value problem
y ' cos3x
sin2y; y
3
2, and graph the
solution.
159
Answers: 3.4 Homework Set A
1.
dy
dx
y
x
y kx
2. x2 1 y ' xy
y C x 2 1
3. 1 sec y tan y y ' x2 1
secx
y x C
3
1
3
4.
dy
dx
e2x
4 y3
xy e C
14
21
2
5.
dy
dx
x2 x3 3
y2
y x C
133
2323
3
6.
dy
dx
2x3
3y2 ; y 2 0
y x
13
412
2
7.
dy
dx y2 1 ; y(1) 0
tany x 1
8.
du
dt
2t sec2 t
2u; u(0) -5
tanu t t 2 25
160
9. Solve the initial-value problem
y ' sin x
sin y; y(0)
2, and graph the
solution.
cos cosy x 1 1
x
y
10. Solve the equation e
-y y ' cosx 0 and graph several members of the family
of solutions. How does the solution curve change as the constant C varies?
lnsin
yx C
1
x
yy = ln(1/(sin(x)+1))y = ln(1/(sin(x)+2))y = ln(1/(sin(x)+3))y = ln(1/(sin(x)))y = ln(1/(sin(x)-.5))
y = ln(1/(sin(x)+4))
161
3.4 Homework – Set B
1.
dy
dx 4xy3
yx C
2
1
4
2. y ' y2 sinx
cos x C
y
1
3.
dv
dt 2 2v t tv
t
tv Ke
22
2 1
4.
dy
dt
t
y y2 1
t
y C
2323
12
5.
d
dr
1 r
r r C
2
3323
2
6.
dy
dx
ycosx
1 y2 ; y 0 1
ln siny
y x 2 1
2 2
7. Find an equation of the curve that satisfies
dy
dx 4x3 y and whose
y-intercept is 7.
xy e4
7
162
8. Solve the initial-value problem
y ' cos3x
sin2y; y
3
2, and graph the
solution.
cos siny x
11 21 3
2 3
x
y
163
3.5 Slope Fields
Vocabulary: Slope field – Given any function or relation, a slope field is drawn by taking
evenly spaced points on the Cartesian coordinate system (usually points having integer coordinates) and, at each point, drawing a small line with the slope of the function.
Objectives: Given a differential equation, sketch its slope field. Given a slope field, sketch a particular solution curve. Given a slope field, determine the family of functions to which the solution curves belong. Given a slope field, determine the differential equation that it represents.
We use slope fields for several reasons, some of which are important mathematically, and others that are more pragmatic.
They provide for a way to visually see an antiderivative as a family of functions. When we find an indefinite integral, there is always the “+C”. A slope field shows us a picture of the variety of solution curves.
Some differential equations do not have an antiderivative that can be
represented as a known function (like
dy
dx ex2
). Slope fields allow us to
represent the antiderivative of functions that are impossible to integrate.
The most important reason, however, is that the AP Calculus test asks these questions. Whether or not there is any practical value to the mathematics of it, because we must answer questions on an AP test, we have to learn it.
Slope fields give us a graphical solution to a differential equation even is we cannot find an analytical solution. This just means that even though we don’t have an actual equation for a solution, we still know what the solution looks like.
Key Idea: Remember that dy
dx is the slope of a tangent line. Slope fields are
represented the way they are because of this fact.
164
Steps to Sketching a Slope Field:
1. Determine the grid of points for which you need to sketch (many times the points are given).
2. Pick your first point. Note its x and y coordinate. Plug these numbers into the differential equation. This is the slope at that point.
3. Find that point on the graph. Make a little line (or dash) at that point whose slope represents the slope that you found in Step 2. Positive slopes point up as you look left to right, and negative slopes point down. The bigger the numeric value, the steeper the slope.
4. Repeat this process for all the points needed.
Ex 1 Sketch the slope field for
dy
dx y x at the points indicated.
2
1
-1
-2
-4 -2 2
165
Ex 2 Sketch the slope field for x
dy
dx 1 at the points indicated.
We first need to solve this equation for
dy
dx, so
dy
dx x
1. Then we find the slope
values for each of the (x, y) points indicated. Notice that there is no y in the equation, so y does not matter in finding the slopes. That should mean that the slope for any individual x is identical regardless of the y.
x
y
Notice that there are no points for the values where x = 0. This is because there are no slope values for these points (we would be dividing by zero). There are two schools of thought on this:
1. Since there are no numeric values for dy
dx at this point, we should not draw
in a slope.
2. Since an undefined value on dy
dx often indicates a vertical tangent line, we
should show the slopes as vertical.
166
The AP Calculus test simply avoids this issue and never asks you to draw a slope field at a point where it would be undefined. We will consider that these undefined values will not be drawn in. Below is a larger version of the slope field you generated on the previous page:
x
y
If we were to solve the differential equation by separation of variables, this is what we would get: dy
dx x
1
ln
dy dxx
dy dxx
y x C
1
1
If you look at the slope field, it does look like our graph of lny x from last year.
Their appears to be a vertical asymptote at x = 0, but the slope field shows a graph on the left side of the axis as well – this is because of the absolute value. The negative x values become positive.
167
Suppose we want the solution curve that passes through the point (1,1). Analytically, we would just do what we did in the section on separation of variables.
ln
ln
y x C
C
C
1
1
1
lny x 1
To sketch the solution curve, however, we don’t need to do that at all. All we need is to find the point (1,1) on our slope field and start sketching our curve. At that point, our sketch must match the slope perfectly (since that is the exact slope of the curve at that point). Then we just bend the curve in the way that the slopes are pointing – we just follow the trend given by our field. Ex. 3: Sketch the solution curve for the slope field below that passes through (1,1).
x
y
168
Below is what your solution curve should have looked like.
x
y
Notice that we needed our solution curve on both sides of the axis, even though there appears to be a vertical asymptote at the line x = 0. If we were just given this picture, however, and not the differential equation that corresponds to it, we could not assume that the original curve was discontinuous at x = 0 (a curve can be continuous and not differentiable, if you recall from last year).
Steps to Sketching a Solution Curve: 1. Find the initial condition point on the graph. 2. Sketch the beginning of your curve so that it matches the slope at that point
exactly.
3. Sketch the rest of the curve by using the slopes as guidelines. The curve does not necessarily go through any slopes other than the initial condition point.
169
Below is the slope field for dy x
dx y
2
. You may notice that the slopes trace out a
very unusual looking curve. While we could separate the variables and solve, it is
useful to look at the slope field as a representation of all of the solutions.
x
y
The solution to the differential ends up being 31
3y x C . Below are several
of the solution curves that illustrate various different values of C (from left to right are the values C = 16, 9, 4, 2, 1, 0, –2, –4, and –9.
x
y
170
Ex 5 Recall from the last section we looked at
dy
dx
x
y. Below is its slope field.
x
y
If you remember from the previous section, the analytic solution was the equation of a circle. If you look at the slope field, the field pretty obviously traces out circular solutions.
171
Ex 4 Match the slope fields with their differential equations.
(a)
dy
dx 3y 4x (b)
dy
dx e x2
(c)
dy
dx cos x (d)
dy
dx
x
y
I
x
y
II
x
y
III
x
y
IV
x
y
172
Ex 5 Match the slope fields with their solution curve.
(a) y 3 (b) y x 2 (c) y x2 4 (d) y
1
x2 (e) y x 3
I
x
y
II
x
y
III
x
y
IV
x
y
V.
x
y
173
Ex 6 Match the slope fields with their solution curve.
(a) yx
1
(b) lny x (c) siny x (d) cosy x
I.
x
y
II
x
y
III
x
y
IV
x
y
174
3.5 Homework Set A
1. A slope field for the differential equation
y ' y 11
4y2
is shown.
Sketch the graphs of the solutions that satisfy the given initial conditions.
(a) 0 1y (b) 0 1y (c) 0 3y (d) 0 3y
x
y
175
Match the differential equation with its slope field (labeled I-IV). Give reasons for your answer.
2. y ' y1 3. y ' y x 4. y ' y2 x2 5. y ' y3 x3
I
x
y
II
x
y
III
x
y
IV
x
y
176
6. Use the slope field labeled below to sketch the graphs of the solutions that satisfy the given initial conditions.
(a) 0 1y (b) 0 0y (c) 0 1y
x
y
7. Sketch a slope field for the differential equation on the axes provided. Then use it to sketch three solution curves.
y '1 y
x
y
177
8. Sketch the slope field of the differential equation. Then use it to sketch a solution curve that passes through the given point.
y ' y 2x ; (1,0)
x
y
9. A slope field for the differential equation 'y y y y 2 4 is shown. Sketch
the graphs of the solutions that satisfy the given initial conditions.
(a) 0 1y (b) 2 3y (c) 2 1y
x
y
178
3.5 Homework Set B
Draw the slope fields for the intervals x 2,2
and
y 1,1
1.
dy
dx x 2y 2.
dy
dx xcos
2y
3.
dy
dx y2 4.
dy
dx x2
5.
dy
dx xy2 6.
dy
dx x ln y
179
7.
dy
dx xy 8.
dy
dx
xey
2
9. Given the differential equation,
dy
dx 2x y
a. On the axis system provided, sketch the slope field for the
dy
dxat all points
plotted on the graph.
b. If the solution curve passes through the point (0, 1), sketch the solution
curve on the same set of axes as your slope field.
c. Find the equation for the solution curve of
dy
dx 2xy , given that
y 0 1
x
y
180
10. Given the differential equation,
dy
dx y2 sin
2x
a. On the axis system provided, sketch the slope field for the
dy
dx at all points
plotted on the graph.
b. There is some value of c such that the horizontal line y = c satisfies the
differential equation. Find the value of c.
c. Find the equation for the solution curve, given that y 11
x
y
181
11. Given the differential equation,
dy
dx x2 y
a. On the axis system provided, sketch the slope field for the
dy
dx at all points
plotted on the graph.
b. If the solution curve passes through the point (0, 1), sketch the solution
curve on the same set of axes as your slope field.
c. Find the equation for the solution curve, given that y 0 1
x
y
182
Answers: 3.5 Homework Set A
1. A slope field for the differential equation
y ' y 11
4y2
is shown.
Sketch the graphs of the solutions that satisfy the given initial conditions. (a) y(0) = 1 (b) y(0) = –1 (c) y(0) = –3 (d) y(0) = 3
x
y
183
Match the differential equation with its slope field (labeled I-IV). Give reasons for your answer.
2. y ' y1 3. y ' y x 4. y ' y2 x2 5. y ' y3 x3
I
x
y
II
x
y
3. y ' y x , plug in points, slopes match 5. y ' y3 x3 , slopes very steep,
cubed numbers are large, so slopes should be as well.
III
x
y
IV
x
y
4. y ' y2 x2 , quadrants are identical, 2. y ' y1, no x so rows all the same
which would happen with the squaring.
184
6. Use the slope field labeled below to sketch the graphs of the solutions that satisfy the given initial conditions.
(a) 0 1y (b) 0 0y (c) 0 1y
x
y
7. Sketch a slope field for the differential equation on the axes provided. Then use it to sketch three solution curves.
y '1 y
x
y
185
8. Sketch the slope field of the differential equation. Then use it to sketch a solution curve that passes through the given point.
y ' y 2x ; (1,0)
x
y
9. A slope field for the differential equation 'y y y y 2 4 is shown. Sketch
the graphs of the solutions that satisfy the given initial conditions.
(a) 0 1y (b) 2 3y (c) 2 1y
x
y
186
3.5 Homework Set B
Draw the slope fields for the intervals x 2,2
and
y 1,1
1.
dy
dx x 2y 2.
dy
dx xcos
2y
x
y
x
y
3.
dy
dx y2 4.
dy
dx x2
x
y
x
y
5.
dy
dx xy2 6.
dy
dx x ln y
x
y
x
y
187
7.
dy
dx xy 8.
dy
dx
xey
2
x
y
x
y
9. Given the differential equation,
dy
dx 2x y
a. On the axis system provided, sketch the slope field for the
dy
dxat all points plotted
on the graph. b. If the solution curve passes through the point (0, 1), sketch the solution curve on
the same set of axes as your slope field.
c. Find the equation for the solution curve of
dy
dx 2xy , given that
y 0 1
x
y
c. 2xy e
188
10. Given the differential equation,
dy
dx y2 sin
2x
a. On the axis system provided, sketch the slope field for the
dy
dx at all points plotted
on the graph. b. There is some value of c such that the horizontal line y = c satisfies the differential
equation. Find the value of c.
c. Find the equation for the solution curve, given that y 11
x
y
c. 1
2cos 1
2
y
x
189
11. Given the differential equation,
dy
dx x2 y
a. On the axis system provided, sketch the slope field for the
dy
dx at all points plotted
on the graph. b. If the solution curve passes through the point (0, 1), sketch the solution curve on
the same set of axes as your slope field.
c. Find the equation for the solution curve, given that y 0 1
x
y
c. 3
3x
y e
190
AP Calculus '07-8 Integral Test Directions: Round at 3 decimal places. Show all work.
un du
eu du
au du
sinu du cosu du
sec2
u du csc2
u du
secu tanu du cscu cotu du
sin2
u du cos2
u du
du
a2 u2
du
a2 u2
du
u u2 a2
191
1. x3 3x 2
x x43
5
x7
dx
2. cos sin-1 x dx
1 x2
3. ecos 5x
sin 5x dx
4. 4
2 3 csc cot x x dx
192
5.
dx
x 1 x 3
6. 2 2ln
x
dxx
7. a t 4sin2t describes the acceleration of a particle. Find v t and x t if
v 0 2 and x 0 3.
193
8. x 2x2 8dx
9. f " x 2 3x 3, f ' 0 4 and f 0 3. Find f(x).
194
x
y
10) Shown above is the slope field for which of the following differential equations?
(A) dy
x ydx
(B) dy
y xdx
(C) 1dy
xdx
(D) 2dyx y
dx (E) 2dy
y xdx
11) The slope field for a differential equation ,dy
f x ydx
is shown in the
figure. Which of the following statements must be true? I. All solution curves have the same slope for a given x value. II. As y approaches 1, the solution curve has a vertical tangent. III. The solution curve is undefined at y = 1.
x
y
(A) I. (B) II. (C) I. and II. (D) II. and III. (E) None
195
12) The slope field below corresponds to which of the following differential equations?
x
y
(A) cosdy
xdx
(B) sindy
xdx
(C) 2cscdy
xdx
(D) 2lndy
xdx
(E) 2xdy
edx
13) The slope field below represents solutions to a differential
equation, ,dy
f x ydx
. Which of the following could be a specific solution
to that differential equation?
x
y
(A) siny x (B) xy e (C) lny x (D)2
2
xy (E) siny x
196
14) The slope field below represents solutions to a differential
equation, ,dy
f x ydx
. Which of the following could be a specific solution
to that differential equation?
x
y
(A) 2 tany x (B) 2 coty x (C) 3 2y x (D) 1
2y
x
(E) 2 lny x
15) The slope field for a differential equation ,dy
f x ydx
is shown in the
figure. Which of the following statements are true?
x
y
(A) I. (B) II. (C) II. and III. (D) I. and III. (E) I., II., and III.
I. The curve is symmetrical about the y axis II . At x = 2, the solution curve has a tangent with a slope of –1. III. The solution curve is undefined at y = 0.
197
Integral Test
1. x3 3x 2
x x43
5
x7
dx
7 5
4 2 3 21 3 32
4 2 7x x x x C
2. cos sin-1 x dx
1 x2
x C
3. ecos 5x
sin 5x dx
cos51
5xe C
4. 4
2 3 csc cot x x dx
7
33cot
7x C
5.
dx
x 1 x 3
2
1 Cx
6. 2 2ln
x
dxx
34ln
3x C
7. a t 4sin2t describes the acceleration of a particle. Find v t and x t if
v 0 2 and x 0 3.
2cos2v t t
sin 2 3x t t
8. x 2x2 8dx
198
3
2212 8
6x
9. f " x 2 3x 3, f ' 0 4 and f 0 3. Find f(x).
41 8
' 2 312 3
f x x
51 127
' 2 3180 45
83
f x x x
x
y
10. Shown above is the slope field for which of the following differential equations?
(A) dy
x ydx
(B) dy
y xdx
(C) 1dy
xdx
(D) 2dyx y
dx (E)
2dyy x
dx
11. The slope field for a differential equation ,dy
f x ydx
is shown in the
figure. Which of the following statements must be true? I. All solution curves have the same slope for a given x value. II. As y approaches 1, the solution curve has a vertical tangent. III. The solution curve is undefined at y = 1.
199
x
y
(A) I. (B) II. (C) I. and II. (D) II. and III. (E) None
12. The slope field below corresponds to which of the following differential equations?
x
y
(A) cosdy
xdx
(B) sindy
xdx
(C) 2cscdy
xdx
(D) 2lndy
xdx
(E) 2xdy
edx
13. The slope field below represents solutions to a differential
equation, ,dy
f x ydx
. Which of the following could be a specific solution
to that differential equation?
200
x
y
14. siny x (B) xy e (C) lny x (D)2
2
xy (E) siny x
15. The slope field below represents solutions to a differential
equation, ,dy
f x ydx
. Which of the following could be a specific solution
to that differential equation?
x
y
(A) 2 tany x (B) 2 coty x (C) 3 2y x (D) 1
2y
x
(E) 2 lny x
201
16. The slope field for a differential equation ,dy
f x ydx
is shown in the
figure. Which of the following statements are true?
x
y
(A) I. (B) II. (C) II. and III. (D) I. and III. (E) I., II., and III.
I. The curve is symmetrical about the y axis II . At x = 2, the solution curve has a tangent with a slope of –1. III. The solution curve is undefined at y = 0.