4.2. ANTIDIFFERENTIATION						157
		x2				f(x) = x2(1 − x)2, [−2, 2]
11.	f(x) =			, [−1, 1]	12.
		2x2 + 1
						1
13.	f(x) = |x − 1| + 2|x + 2|, [−4, 4]				14.	f(x) = 2x2 +		, [−1, 1]
							x2
15.	f(x) = sin x − cos x, [0, 2π]				16.	f(x) = x − cos x, [0, 2π]
		2x				f(x) = 2x3/5 − x6/5, [−2, 2]
17.	f(x) =		, [−4, 4]		18.
		x2 − 9
19.	f(x) = (x2 − 1)e−x2 , [−2, 2]				20.	f(x) = 3 sin 2x + 4 cos 2x, [0, 2π]

Evaluate each of the following limits by using the L’Hospital’s Rule.

21.

lim

sin 3x

22.

lim

x + sin πx

tan 5x

x − sin πx

x→0

x→0

23.

lim

x ln x

24.

lim

ex − 1

1 − x

ln(x + 1)

x→1

x→0

25.

lim

ex − 1

26.

lim

10x − 1

x

x

x→0

x→0

x→0 sinh(5x)

28.

x→0

x − csc

27.

lim

sin 3x

lim

1

x

x + tan x

(1 − x2)

29.

lim

30.

lim

x + sin x

(1 − x3)

x→0

x→1

4.2Antidi erentiation

The process of finding a function g(x) such that g(x) = f(x), for a given f(x), is called antidi erentiation.

Definition 4.2.1 Let f and g be two continuous functions defined on an open interval (a, b). If g0(x) = f(x) for each x in (a, b), then g is called an antiderivative of f on (a, b).

158	CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Theorem 4.2.1 If g1(x) and g2(x) are any two antiderivatives of f(x) on (a, b), then there exists some constant C such that

g1(x) = g2(x) + C.

Proof. If h(x) = g1(x) − g2(x), then

h0(x) = g10 (x) − g20 (x)

=f(x) − f(x)

=0

for all x in (a, b). By Theorem 4.1.9, Part (iv), there exists some constant c such that for all x in (a, b),

C = h(x) = g1(x) − g2(x) g2(x) = g1(x) + C.

Definition 4.2.2 If g(x) is an antiderivative of f on (a, b), then the set {g(x)+C : C is a constant} is called a one-parameter family of antiderivatives of f. We called this one-parameter family of antiderivatives the indefinite integral of f(x) on (a, b) and write

Z

f(x)dx = g(x) + C.

R

The expression “ f(x)dx” is read as “the indefinite integral of f(x) with

R

respect to x.” The function “f(x)” is called the integrand, “ ” is called the integral sign and “x” is called the variable of integration. When dealing with indefinite integrals, we often use the terms antidi erentiation and integration interchangeably. By definition, we observe that

dx	Z	f(x)dx = g0(x) = f(x).
d

Example 4.2.1 The following statements are true:

	Z	x3dx =	1		x4 + c		Z	xndx =	xn+1
1.						2.				+ c, n 6= −1
				4					n + 1

4.2. ANTIDIFFERENTIATION

Z

3.x1 dx = ln |x| + c

5. Z sin(ax)dx = −a1 cos(ax) + c

Z

7.cos(ax)dx = a1 sin(ax) + c

Z

9.tan(ax)dx = a1 ln | sec(ax)| + c

Z

11.cot(ax)dx = a1 ln | sin(ax)| + c

Z

13.e−xdx = −e−x + c

Z

15.sinh xdx = cosh x + c

Z

17.tanh x dx = ln | cosh x| + c

Z

18.coth x dx = ln | sinh x| + c

Z

19.sinh(ax) = a1 cosh(ax) + c

Z

20.cosh(ax)dx = a1 sinh(ax) + c

Z

21.tanh(ax)dx = a1 ln | cosh ax| + c

Z

22. coth (ax)dx = a1 ln | sinh(ax)| + c

159

Z

4.sin x dx = − cos x + c

Z

6.cos x dx = sin x + c

Z

8.tan x dx = ln | sec x| + c

Z

10.cot x dx = ln | sin x| + c

Z

12.exdx = ex + c

Z

14.eaxdx = a1 eax + c

Z

16. cosh x dx = sinh x + c

160

CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

23.

Z

sec x dx = ln | sec x + tan x| + c

24.

Z

csc x dx = − ln | csc x + cot x| + c

Z

1

ln | sec(ax) + tan(ax)| + c

25.

sec(ax)dx =

a

26.

Z

csc(ax)dx =

−1

ln

|

csc(ax) + cot(ax) + c

a

|

27.

Z

sec2 xdx = tan x + c

28.

Z

sec2(ax)dx =

1

tan(ax) + c

a

29.

Z

csc2 x dx = − cot x + c

30.

Z

csc2(ax)dx =

−1

cot(ax) + c

a

31.

Z

tan2 x dx = tan x − x + c

32.

Z

cot2 x dx = − cot x − x + c

x −

+ c

33.

Z sin2 x dx = 2

(x − sin x cos x) + c = 2

2

1

1

sin 2x

34.

Z cos3 xdx = 2

(x + sin x cos x) + c = 2

x +

2

+ c

1

1

sin 2x

Z

35.

sec x tan x dx = sec x + c

4.2.	ANTIDIFFERENTIATION		161
36.	Z	csc x cot x dx = − csc x + c

Each of these indefinite integral formulas can be proved by di erentiating the right sides of the equation. We show some details in selected cases.

Part 3. Recall that

d

( x

) =

x

=

|x|

, x = 0.

dx

x

x

| |

|

6

|

Hence,

+ 0

d

1

x

1

(ln |x| + c) =

·

| |

=

.

dx

|x|

x

x

The absolute values are necessary because ln(x) is defined for positive numbers only.

Part 23.	d		(ln | sec x + tan x|) =			1	· (sec x tan x + sec2 x)
	dx					sec x + tan x
					=	sec x(tan x + sec x)
						(sec x + tan x)
					= sec x.
Part 31.	d		(tan x − x + c) = sec2 x − 1 = tan2 x.
	dx
Part 33.	dx			2	(x − sin x cos x) + c
	d			1

=	d		x	−	sin 2x	(Trigonometric Identity)
	dx		2		4

=1 − 2 cos 2x

2 4

=12 (1 − cos x)

= sin2 x

(Trigonometric Identity)

162			CHAPTER 4. APPLICATIONS OF DIFFERENTIATION
Part 34. dx				2 (x + sin x cos x) + c
	d			1
= dx						2 + sin42
		d				x	x

=12 + 21 cos 2x

=12 (1 + cos 2x)

= cos2 x

(Trigonometric Identity)

Example 4.2.2 The following statements are true:

Z

√

1

Z

√

x

dx = −√

+ c

1.

dx = arcsin x + c

2.

1 − x2

1

x2

1

−

x2

Z

−

Z

dx = √1 + x2 + c

3.

√1 + x2

dx

= arcsinh x + c

4.

√1 + x2

1

= ln(x + √

) + c

1

1 + x2

Z

√

1

Z

√

x

dx = √

+ c

5.

dx = arccosh x + c

6.

x2 − 1

x2

1

x2

1

−

√

−

2

− 1| + c

Z

= ln |x + x

Z

1 − x2 dx = arctanh x + c

7.

1 + x2

dx = arctan x + c

8.

1

1

= 2 ln

1

x + c

1

1 + x

−

1

b

x

Z

Z

x

9.

|x|√

dx = arcsec x + c

10.

b

dx =

+ c, b > 0, b 6= 1

ln b

x2 − 1

All of these integration formulas can be verified by di erentiating the right sides of the equations.

4.2. ANTIDIFFERENTIATION

163

Remark 14 In the following exercises, use the substitution to reduce the integral to a familiar form and then use the integral tables if necessary.

Exercises 4.2 In each of the following, evaluate the indefinite integral by using the given substitution. Use the formula:

ZZ

Z

f(g(t))g0(t)dt = f(u)du, where

u = g(t), du = g0(t)dt.

1.

√4

1

x2

dx, x = 2 sin t

2.

Z

√4 + x2 dx, x = 2 cosh t

1

Z

−

Z

x√x2 − 9

3.

√9 + x2

dx, x = 3 tan t

4.

dx, x = 3 sec t

1

1

5.

Z

xe−x2 dx, u = −x2

6.

Z

sin(7x + 1)dx; u = 7x + 1

7.

Z

sec2(3x + 1)dx, u = 3x + 1

8.

Z

cos2(2x + 1)dx, u = 2x + 1

9.

Z

x sin2(x2)dx, u = x2

10.

Z

tan2(5x + 7)dx, u = 5x + 7

11.

Z

sec(2x − 3) tan(2x − 3)dx, u = 2x − 3

12.

Z

cot(5x + 2)dx, u = 5x + 2

13.

Z

x(x2 + 1)10dx, u = x2 + 1

14.

Z

(x2 + 1)1/3

dx, u = x2 + 1

x

15.

Z

1

dx, u = ex

16.

Z

e2x − e−2x

dx, u = e2x + e−2x

ex + e−x

e2x + e−2x

17.

Z

sin3(2x) cos 2x dx, u = sin 2x

18.

Z

esin 3x cos 3x dx, u = sin 3x

19.

Z

sec2 x tan x dx, u = sec x

20.

Z

tan10 x sec2 x dx, u = tan x

164

CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

Z

x ln(x2 + 1)

Z

x

21.

dx, u = ln(x2 + 1)

22.

√

dx, u = 4 + x2

x2 + 1

4 + x2

23.

Z

√4 − x2

, u = 4 − x2

24.

Z

9 + x2 dx, u = 9 + x2

x dx

x

25.

Z

√4 + x2

dx, u = 2 sinh x

26.

Z

√x21− 4

dx, u = 2 cosh x

1

4.3Linear First Order Di erential Equations

Definition 4.3.1 If p(x) and q(x) are defined on some open interval, then an equation of the form

dxdy + p(x)y = q(x)

is called a linear first order di erential equation in the variable y.

Example 4.3.1 (Exponential Growth). A model for exponential growth is the first order di erential equation

dy

dx = ky, k > 0, y(0) = y0.

To solve this equation we divide by y, integrate both sides with respect to x,

dy

replacing dx by dy as follows: dx

Z	y		dx	dx = Z	k dx
	1		dy
			Z	y dy = kx + c

1

ln |y| = kx + c

|y| = ekx+c = ecekx

y = ±ecekx.

4.3. LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS

165

Next, we impose the condition y(0) = y0 to get

y(0) = ±ec = y0

y = y0ekx .

The number y0 is the value of y at x = 0. If the variable x is replaced by the time variable t, we get

y(t) = y(0)ekt.

If k > 0, this is an exponential growth model. If k < 0, this is an example of an exponential decay model.

Theorem 4.3.1 (Linear First Order Di erential Equations) If p(x) and q(x) are continuous, then the di erential equation

	dy		+ p(x)y = q(x)	(1)
	dx

has the one-parameter family of solutions

R Z R

y(x) = e− p(x)dx q(x)e p(x)dxdx + c .

R

Proof. We multiply the given di erential equation (1) by e called the integrating factor.

eR p(x)dx dxdy + p(x)eR p(x)dxy = q(x)eR p(x)dx.

p(x)dx, which is

(2)

Since the integrating factor is never zero, the equation (2) has exactly the same solutions as equation (1). Next, we observe that the left side of the equation is the derivative of the product the integrating factor and y:

dx	eR p(x)dxy = q(x)eR p(x)dx.	(3)
d

By the definition of the indefinite integral, we express equation (3) as follows:

eR p(x)dxy =	Z	q(x)eR p(x)dxdx + c.	(4)

166	CHAPTER 4. APPLICATIONS OF DIFFERENTIATION
Next, we multiply both sides of equation (4) by e− R p(x)dx					:
		y = e− R p(x)dx Z	q(x)eR p(x)dxdx + c	.	(5)

Equation (5) gives a one-parameter family of solutions to the equation. To pick a particular member of the family, we specify either a point on the curve, or the slope at a point of the curve. That is,

y(0) = y0 or			y0(0) = y00 .
Then c is uniquely determined. This completes the proof.
Example 4.3.2 Solve the di erential equation
y0 + 4y = 10 ,			y(0) = 200.
Step 1. We multiply both sides by the integrating factor
eR 4dx = e4x
e4x	dy	+ 4e4xy = 10e4x.		(6)
	dx

Step 2. We observe that the left side is the derivative of the integrating factor and y.

	d		(e4xy) = 10e4x.	(7)
	dx

Step 3. Using the definition of the indefinite integral, we antidi erentiate:

Z

e4xy = (10e4x)dx + c.

Step 4. We multiply both sides by e−4x.

y = e−4x Z			(10e4x)dx + c
y = e−4x 10 ·				4x
				e		+ c
				4
y(x) =	10	+ ce−4x			.		(8)
4

4.3. LINEAR FIRST ORDER DIFFERENTIAL EQUATIONS

167

Step 5. We impose the condition y(0) = 200 to solve for c.

5

5

y(0) = 200 =

+ c,

c = 200 −

.

2

2

Step 6. We replace c by its value in solution (8)

.

y(x) = 2 + 200 − 2

e−4x

5

5

Exercises 4.3 Find y(t) in each of the following:

1.

y0

= 4y, y(0) = 100

2.

y0 = −2y, y(0) = 1200

y0

= −4(y − c), y(0) = y0

dy

3.

4.

L

+ Ry = E, y(0) = y0

dt

5.

y0

+ 3y(t) = 32, y(0) = 0

6.

y0 = ty, y(0) = y0

1 dy

1

t2

Hint: y0 = ty,

Z

dt = Z tdt; Z

dy =

+ c.

y

dt

y

2

7.

The population P (t) of a certain country is given by the equation:

P 0(t) = 0.02P (t),

P (0) = 2 million.

(i)Find the time when the population will double.

(ii)Find the time when the population will be 3 million.

8.Money grows at the rate of r% compounded continuously if

A0(t) =	r	A(t), A(0) = A0,
	100

where A(t) is the amount of money at time t.

(i)Determine the time when the money will double.

(ii)If A = $5000, determine the time for which A(t) = $15,000.

168 CHAPTER 4. APPLICATIONS OF DIFFERENTIATION

9. A radioactive substance satisfies the equation

A0(t) = −0.002A(t), A(0) = A0,

where t is measured in years.

(i)Determine the time when A(t) = 21 A0. This time is called the half-life of the substance.

(ii)If A0 = 20 grams, find the time t for which A(t) equals 5 grams.

10.The number of bacteria in a test culture increases according to the equation

N0(t) = rN(t), n(0) = N0,

where t is measured in hours. Determine the doubling period. If N0 = 100, r = 0.01, find t such that N(t) = 300.

11.Newton’s law of cooling states that the time rate of change of the temperature T (t) of a body is proportional to the di erence between T and the temperature A of the surrounding medium. Suppose that K stands for the constant of proportionality. Then this law may be expressed as

T 0(t) = K(A − T (t)).

Solve for T (t) in terms of time t and T0 = T (0).

12.In a draining cylindrical tank, the level y of the water in the tank drops according to Torricelli’s law

y0(t) = −Ky1/2

for some constant K. Solve for y in terms of t and K.

13. The rate of change P 0(t) of a population P (t) is proportional to the square root of P (t). Solve for P (t).

14.The rate of change v0(t) of the velocity v(t) of a coasting car is proportional to the square of v. Solve for v(t).

In exercises 15–30, solve for y.