chapter 3 overview: indefinite integrals...111 chapter 3 overview: indefinite integrals as noted in...

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


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


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


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


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


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


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


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


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


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.


y ' cos3x

sin2y; y

3

2, and graph the

solution.

159


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


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


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


dy

dx y2 sin

2x


dy

dx at all points


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


dy

dx x2 y


dy

dx at all points


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


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


dy

dx 2x y


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


dy

dx y2 sin

2x


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


dy

dx x2 y


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



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



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



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.

chapter 3 overview: indefinite integrals...111 chapter 3 overview: indefinite integrals as noted in...

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