Chapter 5
Integration
5.1
Antidifferentiation; the
Indefinite Integral
1.
I=
Z
=
Z
3.
I
x5 dx =
x6
+ C.
6
1
dx =
x2
Z
7.
I
√
(3t2 − 5t + 2)dt
Z
Z
√ Z 1/2
2
= 3 t dt − 5 t dt + 2 dt
3
√ t3/2 t
= 3
− 5
+ 2t + C
3
3/2
√
2 5t3/2
3
= t −
+ 2t + C.
3
9.
I
=
=
=
=
=
2
1
√
(3 y − 3 + )dy
y
y
Z
Z
Z
3 y 1/2 dy − 2 y −3 dy + y −1 dy
−2 3/2 y
y
3
−2
+ ln |y| + C
−2
3/2
=
=
=
√
ex
( + x x)dx
2
Z
Z
1
ex dx + x3/2 dx
2
1 x x5/2
e +
+C
2
5/2
Z
ex
2x5/2
+
+ C.
2
5
√ Z 3
u
1
=
du
− 2 + e2 +
3u 2u
2
Z
Z
1
3
=
u−1 du −
u−2 du
3
2
Z
Z
1
2
+e
du +
u1/2 du
2
1
3 u−1
1 u3/2
2
=
ln |u| −
+e u+
+C
3
2 −1
2 3/2
=
13.
Z
=
I
x−2 dx
1
= −x−1 + C = − + C.
x
Z
I = 5dx = 5x + C.
5.
11.
I
=
15.
Z
17.
2y 3/2 + y −2 + ln |y| + C
1
2y 3/2 + 2 + ln |y| + C.
y
1
3
u3/2
ln |u| +
+ e2 u +
+ C.
3
2u
3
Z 2
x + 2x + 1
I =
dx
x2
Z 2
1
=
1 + + 2 dx
x x
1
= x + 2 ln |x| − + C.
x
Z
1
3
2
I =
(x − 2x )
− 5 dx
x
Z
=
(x2 − 2x − 5x3 + 10x2 )dx
=
143
11x3
5x4
− x2 −
+ C.
3
4
144
19.
CHAPTER 5. INTEGRATION
I
Z √
Since the current population is 10,000, it
follows that
t(t2 − 1)dt
Z
Z
5/2
=
t dt − t1/2 dt
=
10, 000 = P (0) = 4(0) + 3(0)5/3 + C
2t7/2
2t3/2
−
+ C.
7
3
=
or C = 10, 000.
P (t) = 4t + 3t5/3 + 10, 000
21. The slope of the tangent is the derivative. Thus
and
f 0 (x) = 4x + 1
P (8) =
=
=
=
and so the function f (x) is
Z
f (x) = (4x + 1)dx = 2x2 + x + C.
Since the graph contains (1, 2),
2 = f (1) = 2 × 12 + 1 + C or C = −1. Thus
2
f (x) = 2x + x − 1.
23. The slope of the tangent is the derivative.
2
Thus f (x) = x − 2 + 2
x
Z
2
and f (x) =
(x3 − 2 + 2)dx
x
Z
=
(x3 − 2x−2 + 2)dx
0
dy
= 3 + 2t + 6t2 .
dt
The distance is
y(t) = 3t + t2 + 2t3 + S0 .
3
x4
2
+ + 2x + C.
4
x
Since the graph contains (1, 3),
=
3 = f (1) =
27.
4(8) + 3(85/3 ) + 10, 000
32 + 3(25 ) + 10, 000
32 + 96 + 10, 000
10, 128 people.
1
5
+ 2 + 2 + C or C = −
4
4
Thus f (x) =
x4
2
5
+ + 2x − .
4
x
4
25. Let P (t) denote the population of the town t
months from now. Since
dP
= 4 + 5t2/3 ,
dt
y(2) = 6 + 4 + 16 + S0 ,
y(1) = 3 + 1 + 2 + S0 ,
so the distance traveled during the second
minute is
|y(2) − y(1)| = 26 − 6 = 20 meters.
29. The rate is N 0 (t) people per hour. In a time
period dt the number of people is
Z
N 0 (t)dt = N (t) + C
The number of people entering the fair from
11:00 a.m. to 1:00 p.m. is
N (4) + C − N (2) − C = N (4) − N (2).
31. The rate at which population changes is
2/3
P (t) is an antiderivative of 4 + 5t . Thus
Z
P (t) =
(4 + 5t2/3 )dt
5/3 3t
= 4t + 5
+C
5
=
4t + 3t5/3 + C.
r(t) = 0.6t2 + 0.2t + 0.5
thousand people per year and the rate of
pollution 5 units per thousand people.
In a time period dt the pollution increases by
5(0.6t2 + 0.2t + 0.5)dt
5.1. ANTIDIFFERENTIATION; THE INDEFINITE INTEGRAL
and in t years it will be
t2
+ 2.5t + C units
2
The change in two years will be
145
37. Let v(r) denote the velocity of blood through
the artery. Since
t3 +
23 +
v 0 (r) = −ar
cm per sec, v(r) is an antiderivative of −ar.
Z
ar2
+ C.
v(r) = (−ar)dr = −
2
22
+ 2.5(2) + C − C = 15 units.
2
33. Let C(q) denote the total cost of producing q
units.
Since marginal cost is
Since v(R) = 0,
aR2
aR2
+ C or C =
2
2
a
v(R) = (R2 − r2 ).
2
dC
= 6q + 1
dq
dollars per unit, C(q) is an antiderivative of
6q + 1.
Z
C(q) = (6q + 1)dq = 3q 2 + q + C1 .
Since the cost of producing the first unit is
$130,
0 = v(R) = −
39. (a) Let P (t) denote the population of the
endangered species. Since the species is
growing at
P 0 (t) = 0.51e−0.03t
130 = C(1) = 3 + 1 + C1 or C1 = 126
per year, P (t) is an antiderivative of
0.51e−0.03t .
Z
P (t) = (0.51e−0.03t dt = −17e−0.03t + C.
C(q) = 3q 2 + q + 126 and C(10) = $436
35. Let P (q) denote the profit from the production
and sale of q units. Since
dP
= 100 − 2q
dq
Since P (0) = 500, 500 = −17 + C or
C = 517.
P (10) = 517 − 17e−0.3 = 504.41 so the
species will be 504 strong 10 years from
now.
dollars per unit, P (q) is an antiderivative of
100 − 2q.
Z
P (q) = (100 − 2q)dq = 100q − q 2 + C.
Since the profit is $700 when 10 units are
produced,
700 = P (10) = 100(10) − (10)2 + C
or C = −200.
P (q) = 100q − q 2 − 200
which attains its maximum value when its
derivative,
dP
= 100 − 2q = 0
dq
or q = 50. Hence the maximum profit is
2
P (50) = 100(50) − (50) − 200 = $2, 300
(b) Writing Exercise —
Answers will vary.
41.
f 00 (x) = −0.12(x − 10)
which vanishes when x = 10.
(a) f 0 (10) = 7 items per minute.
Z
(b)
f (x) =
f 0 (x)dx
Z
=
(−0.12)(x − 10)dx
= x + 0.6x2 − 0.02x3
(c) f 0 (x) = 0 when x = 20.8.
f (20.8) ≈ 100.
146
CHAPTER 5. INTEGRATION
43. The distance covered during the reaction time
of 0.7 seconds is
Since the velocity was 67 ft/sec when the
brakes were applied, v0 = 67 and
60miles 5, 280ft
hr
× 0.7sec. = 61.6
hr
mile 3600 sec.
v(t) = −23t + 67.
ft. The speed is
Since v =
dD
= −28t + 88 = 0 ft/sec
dt
s(t) =
when the car comes to rest, so t = 22
7 sec.
The distance traveled in this time will be
D
22
7
= −14×
22
7
2
+
88 × 22
+61.6 = 199.89
7
s(t) = −
x
b dx =
=
=
=
Z
e(ln b)x dx
Z
1
e(ln b)x ln bdx
ln b
x
eln b
+C
ln b
bx
+C
ln b
23 2
t + 67t.
2
(c) The car will stop when v(t1 ) = 0 or
67
t1 =
= 2.91 sec.
23
The distance traveled is s(2.91) = 97.5869
ft.
According to the graphing utility (after
zooming)
s(0.775) = 45.0 and v(0.775) = 49.2.
5.2
Integration by Substitution
1. Let u = 2x + 6. Then du = 2dx or dx =
Hence
Z
=
and
v = −23t + v0 .
3. Let u = 4x − 1. Then du = 4dx or dx =
Hence
Z
√
4x − 1dx =
du
.
2
Z
1
u5 du
2
(2x + 6)6
+C
12
(2x + 6)5 dx =
49. (a) Let v(t) be the velocity of the car. Then
dv
= −23
dt
23 2
t + 67t + s0 .
2
(b)
45. H(x) = C because the tangent line is horizontal
at every point. Two horizontal lines differ by a
constant C but are parallel.
If F 0 (x) = G0 (x) the tangent lines to F (x) and
G(x) are parallel at every point. One curve can
be translated a distance C from the other.
Z
(−23t + 67)dt = −
For convenience s0 = 0.Then
ft. The camel may or may not get hit.
If the camel sits on the road in the car’s path,
it will make contact with the car.
If the camel is standing and its legs are in the
car’s path, the car will hit it.
If, on the other hand, the car is positioned
between the camel’s front and rear legs, and if
the hood of the car is more than 0.89 ft. (10.7
inches) in length, the camel will escape
undamaged.
47.
Z
dS
the distance traveled is
dt
1
4
Z
u1/2 du
du
.
4
5.2. INTEGRATION BY SUBSTITUTION
147
1 2u3/2
+C
4 3
(4x − 1)3/2
+C
6
=
=
5. Let u = 1 − x. Then du = −dx or dx = −du.
Z
Z
1−x
Hence
e
dx = − eu du = −e1−x + C
7. Let u = x2 . Then du = 2xdx or xdx =
Hence
Z
2
xex dx =
1
2
Z
eu du =
1
du.
2
Hence
Z
t(t2 + 1)5 dt
=
=
=
=
17. Let u = x5 + 5x4 + 10x + 12. Then
du = (5x4 + 20x3 + 10)dx or
1
(x4 + 4x3 + 2)dx = du. Hence,
5
Z
3x4 + 12x3 + 6
dx
x5 + 5x4 + 10x + 12
1 x2
e +C
2
9. Let u = t2 + 1. Then du = 2tdt or tdt =
11. Let u = x3 + 1. Then du = 3x2 dx or
1
x2 dx = du. Hence,
3
Z
Z
1
2 3
3/4
x (x + 1) dx =
u3/4 du
3
1 4u7/4
=
+C
3 7
4(x3 + 1)7/4
=
+ C.
21
13. Let u = y 5 + 1. Then du = 5y 4 dy or
1
y 4 dy = du. Hence,
5
Z
Z
2y 4
2
du
dy
=
y5 + 1
5
u
2
=
ln |u| + C
5
2
=
ln |y 5 + 1| + C.
5
15. Let u = x2 + 2x + 5. Then du = (2x + 2)dx or
1
(x + 1)dx = du. Hence,
2
Z
(x + 1)(x2 + 2x + 5)12 dx
x4 + 4x3 + 2
dx
x5 + 5x4 + 10x + 12
Z
3
1
3
du = ln |u| + C
5
u
5
3
ln |x5 + 5x4 + 10x + 12| + C.
5
= 3
1
du.
2
Z
1
u5 du
2
(t2 + 1)6
+ C.
12
Z
u13
1
u12 du =
+C
2
26
(x2 + 2x + 5)13
+ C.
26
=
=
Z
19. Let t = u2 − 2u + 6. Then dt = (2u − 2)du or
3
(3u − 3)du = dt. Hence,
2
Z
Z
3u − 3
3
1
du =
dt
(u2 − 2u + 6)2
2
t2
Z
3
=
t−2 dt
2
3
1
=
−
+C
2
t
3
= −
+ C.
2(u2 − 2u + 6)
5
1
dx = dx.
5x
x
Z
Z
ln 5x
Hence
dx =
udu
x
(ln 5x)2
=
+ C.
2
21. Let u = ln 5x. Then du =
23. Let u = ln x. Then du =
Z
1
dx =
x(ln x)2
Z
1
dx. Hence,
x
1
du
u2
1
1
= − +C =−
+ C.
u
ln x
148
CHAPTER 5. INTEGRATION
25. Let u = ln(x2 + 1). Then du =
2x
dx.
x2 + 1
2x ln(x2 + 1)
dx
x2 + 1
Z
u2
=
udu =
+C
2
[ln(x2 + 1)]2
=
+ C.
2
p
dy
27.
= x x2 + 5
dx
and (2, 10) is on the curve.
Z
y =
x(x2 + 5)1/2 dx
Z
1
=
(x2 + 5)1/2 d(x2 + 5)
2
1 (x2 + 5)3/2
(x2 + 5)3/2
=
+C =
+ C.
2
3/2
3
Hence
Z
¿From the given point we obtain
1
10 = (4 + 5)3/2 + C = 9 + C or C = 1.
3
1
y = (x2 + 5)3/2 + 1.
3
29. (a)
x0 (t)
= −2(3t + 1)1/2
Z
2
x(t) = −
(3t + 1)1/2 (3dt)
3
4p
= −
(3t + 1)3 + C
9
Since x(0) = 4, C =
40
and
9
4p
40
(3t + 1)3 +
9
9
4 √ 3 40
x(4) = −
13 +
≈ −16.4
9
9
x(t) = −
(b)
(c)
3=−
Thus t ≈ 0.4
31. (a)
4p
40
(3t + 1)3 +
9
9
1
1
t
=
−
(t + 1)2
t + 1 (t + 1)2
1
= ln(t + 1) +
+C
t+1
x0 (t) =
x(t)
Since x(0) = 0, C = −1 and
x(t) = − ln(t + 1) +
1
−1
t+1
5
≈ −2.4
6
1
(c)
3 = ln(t + 1) +
−1
t+1
Thus t ≈ 53.
(b)
x(4) = − ln 5 −
33. The rate of growth is
G0 = 1 +
1
(x + 1)2
and (2, 5) satisfies the function.
Z
G =
[1 + (x + 1)−2 ]dx
Z
= x + (x + 1)−2 d(x + 1)
(x + 1)−1
+C
−1
1
= x−
+ C.
x+1
= x+
1
¿From the given point we obtain 5 = 2 − + C
3
10
or C =
.
3
1
10
7
y = 0−
+
= meters.
0+1
3
3
35. The rate of growth is P 0 (t) = e0.02t and (0, 50)
satisfies the function.
Z
P (t) =
e0.02t dt
Z
1
=
e0.02t d(0.02t)
0.02
e0.02t
=
+ C.
0.02
¿From the given point we obtain 50 = 50 + C or
C = 0.
P (10) = 50e0.2 = 61.07 million people.
5.2. INTEGRATION BY SUBSTITUTION
149
(b) According to the graphing utility
L(4) = 0.345 and L(12) = 0.345, so the
ozone level is the same at 11:00 a.m. as it
is at 7:00 p.m.
37. (a) Let L(t) denote the ozone level t hours
after 7:00 a.m. Since
dL
dt
=
=
0.24 − 0.03t
√
36 + 16t − t2
(0.24 − 0.03t)(36 + 16t − t2 )−1/2
39. (a) Let V (t) denote the value of the machine
after t years. Since
parts per million per hour, then by
substituting u = 36 + 16t − t2 ,
Z
L(t) = (0.24 − 0.03t)(36 + 16t − t2 )−1/2 dt
Z
= 0.03 (8 − t)(36 + 16t − t2 )−1/2 dt
dV
= −960e−t/5
dt
dollars per year,
V (t) =
= 0.03(36 + 16t − t2 )1/2 + C.
If V0
C
Since the ozone level was 0.25 at 7:00 a.m.,
√
0.25 = L(0) = 0.03 36 + C = 0.18 + C
= 0.03(36 + 16(8) − 82 )1/2 + 0.07
= 0.03(10) + 0.07 = 0.37
parts per million, which occurs at 3:00
p.m. (8 hours after 7:00 a.m.).
= 4, 800e−t/5 + C.
= V (0) = 4, 800e0 + C,
= V0 − 4, 800.
V (t) = 4, 800e−t/5 + V0 − 4, 800.
L(t) = 0.03(36 + 16t − t2 )1/2 + 0.07.
L(8)
(−960e−t/5 )dt
Thus,
or C = 0.07. Hence,
dL
The peak ozone level occurs when
= 0,
dt
that is, when 0.24 − 0.03t = 0 or t = 8.
Note that L(t) has its absolute maximum
dL
at this critical point since
> 0 (L is
dt
increasing) for 0 < t < 8,
dL
and
< 0 (L is decreasing) for 8 < t.
dt
Thus the peak ozone level is
Z
(b) If V0 = $5, 200, then
V (10) = 4, 800e−2 + 5, 200 − 4, 800
≈ $1, 049.61.
√
dP
= 3 x + 1.
dx
41.
P0 = 300 cents.
P (x) =
3(x + 1)3/2
+ P0 .
3/2
P (8) = 2(8 + 1)3/2 + 300 or $3.54.
43. (a)
30x
(3 + x)2
3+x
3
= 30
−
(3 + x)2
(3 + x)2
3
p(x) = 30 ln |3 + x| +
+C
3+x
p0 (x) =
With p(0) = 2.25, C = 27.75 − 30 ln 3 and
3
p(x) = 30 ln |3 + x| +
−27.75−30 ln 3
3+x
150
CHAPTER 5. INTEGRATION
(b)
p(4)
and integrate to get
Z
Z
1
dy =
3dx,
y
ln |y| = 3x + C1 ,
|y| = e3x+C1 = eC1 e3x or y = Ce3x
3
= 30 ln 7 +
− 27.75 − 30 ln 3)
7
≈ $10.50 per Weenie
(c) p(0.785)approx3 Weenies will be supplied.
11 − x
R0 (x) = √
, C 0 (x) = 2 + x + x2
45.
14 − x
where C is the constant ±eC1 .
5. Separate the variables of
dy
= ey
dx
Let 14 − x = u, du = −dx. Then
Z
u−3
√ du
R(x) = −
u
Z
= − (u1/2 − 3u−1/2 )du
2 3/2
u − 6u1/2
= −
3
√
2
= 6 14 − x − (14 − x)3/2 + C1
3
We also have
Z
x3
x2
C(x) = (2 + x + x2 )dx = 2x +
+
+ C2
2
3
and integrate to get
Z
Z
e−y dy =
dx,
−e−y
where C is the constant −C1 . Hence,
ln e−y
−y
Introduction to
Differential Equations
then y
=
Z
(3x2 + 5x − 6)dx
and integrate to get
Z
Z
ydy =
xdx,
y2
2
=
x2
+ C1 or y 2 = x2 + C
2
√
y = ± x2 + C, where C is the constant 2C1 .
9.
dy
dx
y −1/2 dy
=
√
xy or
= x1/2 dx
2 3/2
2y 1/2 =
x +C
3
dy
y
=
dx
x−1
dy
dx
=
y
x−1
ln |y| = ln |x − 1| + C1
11.
5x2
= x3 +
− 6x + C.
2
3. Separate the variables of
dy
= 3y
dx
ln(C − x),
ln(C − x), or y = − ln(C − x)
dy
x
=
dx
y
dy
= 3x2 + 5x − 6,
dx
1.
=
=
7. Separate the variables of
P (9) − P (5)
= R(9) − C(9) − [R(5) − C(5)]
= −295.54 − (−64.167) = −231.373
5.3
= x + C1 or e−y = C − x
|y|
y
= C1 ,
= eC1 = C
|x − 1|
x−1
where C is any real number.
ln
5.3. INTRODUCTION TO DIFFERENTIAL EQUATIONS
151
y+3
dy
=
dx
(2x − 5)6
13.
y3
3
dy
= (2x − 5)−6 dx
y+3
1
(2x − 5)−5 + C1
10
1
y = −3 + C exp −
10(2x − 5)5
dx
xt
=
dt
2t + 1
dx
x
=
ln |x| =
ln |x| + ln
√
4
2t + 1
=
√
x 4 2t + 1
=
x =
1
1
1−
dt
2
2t + 1
t
1
− ln |2t + 1| + C1
2 4
t
+ C1
2
eC1 et/2 = Cet/2
Cet/2
√
4
2t + 1
2y 3 = 3x2 + C.
Since y = 3 when x = 2, 54 = 12 + C or C = 42.
Thus 2y 3 = 3x2 + 42 or
y=
or C =
4
. Hence,
5
e5x
4
+ .
5
5
dy
x
= 2,
dx
y
Z
Z
2
y dy = xdx,
y=
19.
e0
+C
5
3x2 + 42
2
1/3
.
y −2 dy = (4 − x)1/2 dx
−
1
2
C
= − (4 − x)3/2 −
y
3
3
C
Note: − is just for convenience, but make
3
sure you introduce a constant of integration
immediately after your last integration.
y=
is correct also, since
Ce(2t+1)/4 = Cet/2 e1/4 = C2 et/2 .
dy
17.
If
= e5x ,
dx
Z
e5x
then y = e5x dx =
+ C.
5
1=
√
dy
= y2 4 − x
dx
21.
Ce(2t+1)/4
Note: x = √
4
2t + 1
Since y = 1 when x = 0,
x2
C
+ ,
2
6
(remember the form of the constant of
integration is arbitrary)
ln |y + 3| = −
15.
=
3
C + 2(4 − x)3/2
Since y = 2 when x = 4, C =
y=
23.
3
and
2
6
3 + 4(4 − x)3/2
dy
y+1
=
dt
t(y − 1)
y−1
dt
dy =
y+1
t
2
dt
1−
dy =
y+1
t
y − 2 ln |y + 1| = ln |t| + C
2 − 2 ln 3 = C
y − 2 ln |y + 1| − ln |t| = 2 − 2 ln 3
y − 2 = ln(y + 1)2 + ln t − ln 9
y = 2 + ln
t(y + 1)2
9
152
25.
CHAPTER 5. INTEGRATION
Q = B − Ce−kt , so Ce−t = B − Q.
dQ(t)
dt
(b) Solving the differential equation leads to
dt
50
t
ln Q = − + C1
50
Q = Ce−t/50
dQ
Q
= −Ce−kt (−k)
= kCe−kt = k(B − Q).
27. Let Q denote the number of bacteria. Then,
dQ
is the rate of change of Q, and since this
dt
rate of change is proportional to Q, it follows
that
dQ
= kQ
dt
where k is a positive constant of
proportionality.
dQ
29. Let Q denote the investment. Then
is the
dt
rate of change of Q, and since this rate of
change is equal to 7 % of the size of Q, it
follows that
dQ
= 0.07Q
dt
dP
31. Let P denote the population. Then
is the
dt
rate of change of P , and since this rate of
change is the constant 500, it follows that
dP
= 500
dt
33. Let N be the total population and Q(t) the
number of people who have caught the disease.
For positive proportionality constant k,
= −
Since Q(0) = 1, 600, the equation becomes
Q(t) = 1, 600e−t/50
D(p) = a − bp and S(p) = r + sp
37.
dp
dt
= k(a − bp − r − sp)
= k[a − r − (b + s)p]
Z
Z
dp
= kdt
a − r − (b + s)p
−
1
ln |a − r − (b + s)p| = kt + C1
b+s
a − r − (b + s)p = Ce−(b+s)kt
a − r + Ce−(b+s)kt
b+s
a−r
lim P (t) =
t→∞
b+s
p=
dQ(t)
= kQ(t)[N − Q(t)]
dt
35. (a) Let Q(t) be the amount of fluoride in the
reservoir at time t. Since there are 200
million gallons of solution in the reservoir,
Q(t)
the concentration of fluoride is
. 4
200
million gallons of solution flow out of the
39. Writing Exercise —
tank per day. Thus
Answers will vary.
Q(t) pounds
4 million gallons
41. (a)
dP
day
200 million gallons
dt
flows out of the reservoir per day, or
dP
P (k − mP )
dQ
Q
=−
dt
50
= P (k − mP )
= dt =
A
B
+
P
k − mP
5.4. INTEGRATION BY PARTS
153
Multiplying by the Least Common
Denominator leads to
1 = A(k − mP ) + BP .
1
With P = 0, A = , and
k
k
m
with P = , B =
m
k
Z
Z
(b)
dp 1
m dP
1
+
t =
k
P
k
k − mP
1
=
(ln |P | − ln |k − mP |) + C1
k
(c)
P
ln
k − mP
P
k − mP
(mC3 ekt + 1)P
P
I
I
7.
1.
I=
.
f (x) = x
f 0 (x) = 1
I
3.
I
f (x) = 1 − x g(x) = ex
f 0 (x) = −1
G(x) = ex
G(t) =
t2
2
Z
t2
1 t2
ln 2t −
dt
2
t 2
Z
t2
1
=
ln 2t −
tdt
2
2
t2
1
=
ln 2t −
+ C.
2
2
Z
I = ve−v/5 dv
g(v) = e−v/5
G(v) = −5e−v/5
−v/5
= −5ve
f 0 (x) = 1
I
=
=
=
11.
g(t) = t
=
f (x) = x
e−x dx
= −(x + 1)e−x + C.
Z
I = (1 − x)ex dx.
(2 − x)ex + C.
Z
I = t ln 2tdt.
+5
Z
e−v/5 dv
= −5(v + 5)e−v/5 + C.
Z
Z
√
I = x x − 6dx = x(x − 6)1/2 dx.
9.
g(x) = e−x
G(x) = −e−x
= −xe−x +
ex dx
f (v) = v
f 0 (v) = 1
xe−x dx
Z
Z
.
k
1
and D =
, with C3 an
m
mC3
arbitrary constant.
Z
= (1 − x)ex +
f (t) = ln 2t
1
f 0 (t) =
t
= C3 ekt
Integration by Parts
(−ex )dx
5.
where C =
5.4
Z
=
= kt − C2
= kC3 ekt
k
m
=
1 −kt
1+
e
mC3
C
=
1 + De−kt
= (1 − x)ex −
g(x) = (x − 6)1/2
2
G(x) = (x − 6)3/2
3
Z
2
2
3/2
x(x − 6) −
(x − 6)3/2 dx
3
3
2
4
x(x − 6)3/2 − (x − 6)5/2 + C
3
15
2
3/2
(x − 6) (x + 4) + C
5
Z
I = x(x + 1)8 dx.
f (x) = x
f 0 (x) = 1
g(x) = (x + 1)8
(x + 1)9
G(x) =
9
154
CHAPTER 5. INTEGRATION
Z
1
1
I =
x(x + 1)9 −
(x + 1)9 dx
9
9
1
(x + 1)10
=
x(x + 1)9 −
+ C.
9
90
Z
x
√
I=
dx.
x+2
1
f (x) = x g(x) = √
= (x + 2)−1/2
x+2
f 0 (x) = 1 G(x) = 2(x + 2)1/2
13.
I
2x(x + 2)1/2 −
Z
2(x + 2)1/2 dx
2
1/2
(x + 2)3/2 + C.
2x(x + 2) − 2
3
Z
I = x2 e−x dx.
=
=
15.
19.
I=
= −x2 e−x −
x3
ln x −
3
x3
=
ln x −
3
Z
Z
ln x
dx
=
I=
x2
I
21.
=
Z
1
1
I = − ln x +
dx
x
x2
1
= − (ln x + 1) + C.
x
Z
Z
2
I = x3 ex dx = x2 (xex )dx.
23.
f (x) = x2
I
= −x2 e−x + 2(−xe−x +
17.
f (x) = x3
f 0 (x) = 3x2
I
= x3 ex − 3
25.
g(x) = ex
G(x) = ex
(x3 − 3x2 + 6x − 6)ex + C
Z
1 2 x2
1 x2
=
x e − 2x
e
dx
2
2
x2 x2 1 x2
=
e − e + C.
2
2
Z
I = x7 (x4 + 5)8 dx
f (x) = x4
f 0 (x) = 4x3
Z
x2 ex dx
Z
3 x
2 x
= x e − 3(x e − 2 xex dx)
Z
= x3 ex − 3x2 ex + 6(xex − ex dx)
=
I
e−x dx)
= −e−x (x2 + 2x + 2) + C
Z
I = x3 ex dx.
2
g(x) = xex
1 2
f 0 (x) = 2x G(x) = ex
2
f2 (x) = x g2 (x) = e−x
f20 (x) = 1 G2 (x) = −e−x
Z
Z
1
x2 dx
3
x3
+C
9
1
ln xdx
x2
1
f (x) = ln x g(x) = 2
x
1
1
f 0 (x) =
G(x) = −
x
x
Z
2x(−e−x )dx
Z
= −x2 e−x + 2 xe−x dx.
x2 ln xdx.
f (x) = ln x g(x) = x2
1
x3
f 0 (x) =
G(x) =
x
3
f1 (x) = x2 g1 (x) = e−x
f10 (x) = 2x G1 (x) = −e−x
I
Z
I
=
=
=
g(x) = x3 (x4 + 5)8
1 4
G(x) =
(x + 5)9
36
Z
1 4 4
1
9
x (x + 5) −
x3 (x4 + 5)9 dx
36
9
x4 4
1
(x + 9)9 −
(x4 + 5)10 + C
36
360
1
(x4 + 5)9 (9x4 − 5) + C
360
5.4. INTEGRATION BY PARTS
155
27. Let f (x) be
√ the function whose tangent has
slope x ln x.
√
Then f 0 (x) = x ln x for x > 0 and
Z
Z
√
1
f (x) = x ln xdx =
x ln xdx
2
To integrate by parts,
f (x) = ln x g(x) = x
1
x2
f 0 (x) =
G(x) =
x
2
f (x)
=
=
=
=
Z
1
x ln xdx
2
2
Z 2 1
1
x
x
ln x −
dx
2
x
2
2
Z
1 x2 ln x 1
−
xdx
2
2
2
2
2
x ln x x
−
+C
4
8
Since (2, f (2)) = (2, −3), that is, when x = 2,
f = −3,
22 ln 2 22
−3 =
−
+C
4
8
1
5
−3 = ln 2 − + C, or C = − − ln 2. Thus
2
2
f (x) =
x2 ln x x2
5
−
− − ln 2
4
8
2
Since no units are produced when t = 0,
Q(0) = 0 = −200(2) + C or C = 400. Hence,
Q(t) = −200(t + 2)e−0.5t + 400
and the number of units produced during the
first three hours is Q(3) = 176.87.
31. Let q denote the number of units produced and
C(q) the cost of producing the first q units.
dC
= (0.1q + 1)e0.03q and
Then,
dq
Z
f (q) = 0.1q + 1 g(q) = e0.03q
1 0.03q
f 0 (q) = 0.1
G(q) =
e
0.03
Thus
C(q)
=
=
200 =
dQ
= 100te−0.5t and
dt
Z
Q(t) = 100 te−0.5t dt
and C(20) = 274.6
33. (a)
I=
=
100[−2te
−
xn eax dx
g(x) = eax
1
f 0 (x) = nxn−1 G(q) = eax
a
Z
xn ax n
I=
e −
xn−1 eax dx
a
a
(b)
Z
Z
f (x) = xn
Thus
Q(t)
1
10 0.03q
(0.1q + 1)e0.03q −
e
+ 177.4
0.03
0.09
g(t) = e−0.5t
e−0.5t
G(t) = −
= −2e−0.5t
0.5
−0.5t
1
10 .3
(2)e.3 −
e + C1
0.03
0.03
or C1 = 177.4. Thus
29. Let t denote time and Q(t) the number of units
produced. Then,
f 0 (t) = 1
Z
1
10
[(0.1q + 1)e0.03q −
e0.03q dq]
0.03
3
1
10 0.03q
(0.1q + 1)e0.03q −
e
+ C1
0.03
0.09
When q = 10, C = 200 and
C(q) =
f (t) = t
(0.1q + 1)e0.03q dq.
C(q) =
−0.5t
(−2)e
= −200te−0.5t − 400e−0.5t + C
= −200(t + 2)e−0.5t + C
Z
dt]
=
x3 e5x dx
1 3 5x 3
x e −
5
5
Z
x2 e5x dx
156
CHAPTER 5. INTEGRATION
Z
1 3 5x 3 1 2 5x 2
5x
=
x e −
x e −
xe dx
5
5 5
5
1 3 5x
3
=
x e − x2 e5x
5
25
Z
6 1 5x 1
5x
xe −
e dx
+
25 5
5
e5x
=
(125x3 − 75x2 + 30x − 6) + C
625
5.5
Review Problems
Review Problems
Z 1.
2.
Z
=
x5 − 3x2 +
1
x2
dx
=
Z
=
x6
1
− x3 − + C.
6
x
(x5 − 3x2 + x−2 )dx
√
1
+ 5 + x)dx
x
2x3/2
− ln |x| + 5x +
+ C.
3
(x2/3 −
3x5/3
5
1
3. Let u = 3x + 1. Then du = 3dx or dx = du.
3
Z
Z
√
1
3x + 1dx =
u1/2 du
3
1 2u3/2
=
+C
3 3
2(3x + 1)3/2
=
+ C.
9
4. Let u = 3x2 + 2x + 5. Then du = (6x + 2)dx or
1
(3x + 1)dx = du. Hence
2
Z
p
I =
(3x + 1) 3x2 + 2x + 5dx
Z
1
1 2u3/2
=
u1/2 du =
+C
2
2 3
(3x2 + 2x + 5)3/2
=
+ C.
3
5. Let u = x2 + 4x + 2. Then du = (2x + 4)dx or
1
(x + 2)dx = du. Hence
2
Z
I =
(x + 2)(x2 + 4x + 2)5 dx
Z
1
(x2 + 4x + 2)6
=
u5 du =
+ C.
2
12
6. Let u = x2 + 4x + 2. Then du = (2x + 4)dx or
1
(x + 2)dx = du. Hence
2
Z
x+2
I =
dx
x2 + 4x + 2
Z
1
1
1
=
du = ln |u| + C
2
u
2
1
2
=
ln |x + 4x + 2| + C.
2
7. Let u = 2x2 + 8x + 3. Then du = (4x + 8)dx or
3
(3x + 6)dx = du. Hence
4
Z
3x + 6
I =
dx
(2x2 + 8x + 3)2
Z
Z
3
1
3
=
du
=
u−2 du
4
u2
4
3
= − u−1 + C
4
3
= −
+ C.
4(2x2 + 8x + 3)
8. Let u = x − 5. Then du = dx. Hence
Z
I =
(x − 5)12 dx
Z
(x − 5)13
=
u12 du =
+ C.
13
9. Method 1:
Let u = x − 5. Then du = dx and x = u + 5.
Z
Hence I =
x(x − 5)12 dx
Z
=
(u13 + 5u12 )du
=
(x − 5)14
5(x − 5)13
+
+ C.
14
13
5.5. REVIEW PROBLEMS
157
Method 2:
13.
f (x) = x
f 0 (x) = 1
So I
=
=
=
=
12
g(x) = (x − 5)
(x − 5)13
G(x) =
13
g(x) = e3x
e3x
G(x) =
3
f 0 (x) = 1
Hence I
Hence I
Hence I
−x/2
Z
(−2e−x/2 )dx)
Z
= −2xe−x/2 + 2 e−x/2 dx
= −2xe
−
= −2xe−x/2 − 4e−x/2 + C
= −2(x + 2)e−x/2 + C.
3
x3 (x2 ex )dx
3
x3 ex
−
3
Z
3
x2 ex dx
3
=
14.
I=
=
Z
x3 ex
1 3
− ex + C
3
3
3
1 3
(x − 1)ex + C.
3
(2x + 1)e0.1x dx.
f (x) = 2x + 1 g(x) = e0.1x
f 0 (x) = 2
G(x) = 10e0.1x
Hence I
= 10e0.1x (2x + 1) − 20
=
=
15.
Z
e0.1x dx
10(2x + 1)e0.1x − 200e0.1x + C
10(2x − 19)e0.1x + C.
Z
I = x ln 3xdx.
f (x) = ln 3x g(x) = x
1
x2
f 0 (x) =
G(x) =
x
2
3x Z 3x xe
e
5
−
dx
3
3
5
5 3x
=
x−
e + C.
3
9
Z
I = xe−x/2 dx.
g(x) = e−x/2
G(x) = −2e−x/2
=
Z
g(x) = x2 ex
3
ex
G(x) =
3
3
=
f (x) = x
f 0 (x) = 1
3
x5 ex dx =
f 0 (x) = 3x2
x(x − 5)12 dx
Z
x(x − 5)13
(x − 5)13
−
dx
13
13
x(x − 5)13
1 (x − 5)14
−
+C
13
13
14
x(x − 5)13
(x − 5)14
−
+ C.
13
182
f (x) = x
Z
f (x) = x3
Z
1
10. Let u = 3x. Then du = 3dx or dx = du.
3
Z
Hence I = 5 e3x dx
Z
5
=
eu du
3
5e3x
=
+ C.
3
Z
11. I = 5xe3x dx.
12.
Rewrite I =
16.
x2 ln 3x
Hence I =
−
2
x2 ln 3x
=
−
2
Z
I = ln 3xdx.
Z
1
xdx
2
x2
+ C.
4
f (x) = ln 3x g(x) = 1
1
f 0 (x) =
G(x) = x
x
Hence I
= x ln 3x −
Z
dx
= x ln 3x − x + C.
158
CHAPTER 5. INTEGRATION
1
17. Let u = ln 3x. Then du = dx and so
x
Z
Z
ln 3x
I =
dx = udu
x
(ln 3x)2
=
+ C.
Z2
ln 3x
18.
I=
dx.
x2
1
f (x) = ln 3x g(x) = 2
x
1
1
0
f (x) =
G(x) = −
x
x
Z
ln 3x
1
Hence I = −
+
dx
x
x2
ln 3x 1
= −
− +C
x
x
1
= − (ln 3x + 1) + C.
x
19. Rewrite
Z
Z
I = x3 (x2 + 1)8 dx = x2 [x(x2 + 1)8 ]dx.
f (x) = x2
g(x) = x(x2 + 1)8
(x2 + 1)9
f 0 (x) = 2x G(x) =
18
So I
=
=
x2 (x2 + 1)9
1
−
x(x2 + 1)9 dx
18
9
x2 (x2 + 1)9
(x2 + 1)10
−
+ C.
18
180
Z
20. Let u = x2 + 1. Then du = 2xdx and so
Z
Z
I = 2x ln(x2 + 1)dx = (1)(ln u)du
f (u) = ln u
1
f 0 (u) =
u
Hence I
g(u) = 1
G(u) = u
= u ln u −
Z
u
= u ln u −
Z
du
1
du
u
= u ln u − u + C
= u(ln u − 1) + C
= (x2 + 1)[ln(x2 + 1) − 1] + C.
21. The slope of the tangent is the derivative.
Hence f 0 (x) = x(x2 + 1)3
and so f is an antiderivative of x(x2 + 1)3 .
Z
That is f (x) =
x(x2 + 1)3 dx
=
(x2 + 1)4
+ C.
8
Since the graph of f passes through the point
24
(1, 5), 5 = f (1) =
+ C = 2 + C or C = 3.
8
Hence f (x) =
(x2 + 1)4
+3
8
22. Let Q(x) denote the number of commuters
using the new subway line x weeks from now. It
is given that
dQ
= 18x2 + 500
dx
commuters per week. Hence, Q(x) is an
antiderivative
of 18x2 + 500. That is,
Z
Q(x) = (18x2 + 500)dx = 6x3 + 500x + C.
Since 8,000 commuters currently use the
subway, 8, 000 = Q(0) = C. Hence,
Q(x) = 6x3 + 500x + 8, 000, and the number of
commuters who will be using the subway in 5
weeks is Q(5) = 11, 250.
23. Let Q(x) denote the number of inmates in
county prisons x years from now.
It is given that
dQ
= 280e0.2x
dx
inmates per year. Hence, Q(x) is an
antiderivative of 280e0.2x . That is,
Z
Q(x) = 280e0.2x dx = 1, 400e0.2x + C
Since the prisons currently house 2,000 inmates,
2, 000 = Q(0) = 1, 400 + C or C = 600. Hence,
Q(x) = 1, 400e0.2x + 600, and the number of
inmates 10 years from now will be
Q(10) = 1, 400e2 + 600 = 10, 945.
5.5. REVIEW PROBLEMS
24.
y
159
dy
= x3 − 3x2 + 5,
dx
Z
=
(x3 − 3x2 + 5)dx
=
28.
y=
x4
− x3 + 5x + C.
4
dy
= 0.02xy
dx
+C1
dy
ln x
=
dx
y
ydy = ln xdx
2
y
= x ln |x| − x + C
2
Since y = 100 when x = 1, 5, 000 = −1 + C or
C = 5, 001.
p
y = 2(x ln |x| − x + 5, 001).
2
= eC1 e0.01x ,
2
or y = Ce0.01x where C = ±eC1 .
26. Separate the variables of
dy
xy
=√
dx
1 − x2
30.
dy
= k(80 − y)
dx
and integrate to
Z
1
dy
80 − y
− ln |80 − y|
|80 − y|
y − 80
dy
xdx
=√
y
1 − x2
p
ln |y| = − 1 − x2 + C
get
=
Z
kdx,
= kx + C1 ,
= e−kx−C1 = e−C1 e−kx ,
= Ce−kx ,
Since y = 2 when x = 0 C = 1 + ln 2 and
p
ln |y| = − 1 − x2 + 1 + ln 2
p
|y|
ln | | = 1 − 1 − x2
2
or y = 80 + Ce−kx where C = ±e−C1
√
y = 2e1−
27. Separate the variables of
dy
ex
= e2x−y = y
dx
e
and integrate to get
Z
Z
ey dy = e2x dx.
or
e2x
ey =
+ C1
2
2x
e
y = ln
+C
2
(5x4 − 3x2 − 2)dx = x5 − x3 − 2x + C.
29.
and integrate to get
Z
1
dy = 0.02xdx,
y
ln |y| = 0.01x2 + C1 ,
2
Z
Since y = 4 when x = 1, 4 = 1 − 1 − 2 + C or
C = 6.
Hence
y = x5 − x3 − 2x + 6
25. Separate the variables of
|y| = e0.01x
dy
= 5x4 − 3x2 − 2,
dx
31. If
1−x2
d2 y
= 2,
dx2
then
dy
dx
=
Z
d2 y
dx
dx2
=
Z
2dx
= 2x + C1 .
160
CHAPTER 5. INTEGRATION
dy
Since
= 3 when x = 0, 3 = 2(0) + C1 or
dx
C1 = 3. Hence
dy
= 2x + 3
dx
and y =
Z
(2x + 3)dx = x2 + 3x + C.
Since y = 5 when x = 0, 5 = 02 + 3(0) + C or
C = 5. Hence
y = x2 + 3x + 5.
32. Let V (t) denote the value of the machine after t
years. The rate of change of V is
dV
= k(V − 5, 000),
dt
where k is a positive constant of
proportionality.
Separate the variables and integrate to get
Z
Z
1
dV = kdt,
V − 5, 000
ln(V − 5, 000) = kt + C1 ,
V − 5, 000 = ekt+C1 = eC1 ekt ,
The value of the machine after 8 years is
V (8) = 5, 000 + 35, 000e8k
= 5, 000 + 35, 000(e4k )2
2
5
= 5, 000 + 35, 000
= $22, 857.
7
33. Let P (q) denote the profit, R(q) the revenue,
and C(q) the cost when the level of production
is q units. Since the marginal revenue is
R0 (q) = 200q −1/2 , C 0 (q) = 0.4q,
and profit is revenue minus cost,
dR dC
dP
=
−
= 200q −1/2 − 0.4q
dq
dq
dq
dollars per unit. The profit function P (q) is an
antiderivative of the marginal profit. That is,
Z
P (q) =
(200q −1/2 − 0.4q)dq
=
400q 1/2 − 0.2q 2 + C.
Since profit is $2,000 when the level of
production is 25 units,
2, 000 = P (25) = 400(5) − 0.2(25)2 + C
or
V (t) = 5, 000 + Cekt
where C = eC1 and the absolute values can be
dropped since V − 5, 000 > 0.
Since the machine was originally worth $40,000,
40, 000 = V (0) = 5, 000 + C
or C = 35, 000. Hence,
V (t) = 5, 000 + 35, 000ekt .
Since the machine was worth $30,000 after 4
years,
or C = 125. Hence,
P (q) = 400q 1/2 − 0.2q 2 + 125,
and the profit when 36 units are produced is
P (36) = $2, 265.80.
34. (a) The rate of change of price is
P 0 (x) = 0.2 + 0.003x2 .
P (x) = 0.2x + 0.001x3 + C
= 0.2x + 0.001x3 + 250
30, 000 = V (4) = 5, 000 + 35, 000e4k ,
35, 000e4k = 25, 000 or
e4k =
5
25, 000
= .
35, 000
7
where C = 250 cents ($2.50) since “now”
means x = 0. P (10) = 2 + 1 + 250 = 253
or $2.53.
5.5. REVIEW PROBLEMS
161
so N (t) = 0.1t3 + 0.3t2 + t + C
and the revenue is R(t) = 3N (t).
No revenue is generated initially, so
R(0) = 0 = C.
In 5 days, the revenue will be N (5) = $75.
(b)
If P 0 (x) = 0.3 + 0.003x2
then the price would be 0.1 more cents per
week or P (10) = $2.54.
35. Let V (t) denote the value of the machine t
years from now. Since
dV
= 220(t − 10)
dt
dollars per year, the function V (t) is an
antiderivative of 220(t − 10). Thus,
Z
V (t) =
220(t − 10)dt
=
110t2 − 2, 200t + C.
Since the machine was originally worth $12,000,
it follows that V (0) = 12, 000 = C. Thus, the
value of the machine after t years will be
2
V (t) = 110t − 2, 200t + 12, 000
and the value after 10 years will be
V (10) = $1, 000.
1
h(x) = 0.5 +
36.
(x + 1)2
meters per year. The growth per year is
Z
1
h(x) = f (x)dx = .5x −
+ h0
x+1
During the second year the tree will grow
h(2) − h(1) =
2
meter
3
37. Let N (t) = denote the number of bushels t days
from now. The number of bushels will be an
antiderivative of
dN
= 0.3t2 + 0.6t + 1.
dt
38. Let P (x) denote the population x months from
now. Then
√
dP
= 10 + 2 x,
dx
and the amount by which the population will
increase during a month is
P (x) = 10x +
4x3/2
+C
3
The initial population is P (0) = C The
population at the end of 9 months is
P (9) = 10(9) +
(4)93/2
+C
3
and the population during the next 9 months is
P (9) − P (0) = 126 people.
39. Let N (t) denote the size of the crop (in
bushels) t days from now. Then
dN
= 0.5t2 + 4(t + 1)−1
dt
bushels per day. The increase in size of the crop
over t days is
Z
N (t) = [0.5t2 + 4(t + 1)−1 ]dt
=
0.5 3
t + 4 ln |t + 1| + C
3
N (0) = C. The size of the crop in 6 days is
N (6) and the increase will be
N (6) − N (0) =
0.5 3
(6 ) + 4 ln 7 + C − C = 43.78
3
bushels. Hence, at $2 per bushel, the value of
the crop will increase by 2(43.78) = $87.56.
40. Let Q(t) denote the total consumption (in
billion-barrel units) of oil over the next t years.
162
CHAPTER 5. INTEGRATION
Then the demand (billion barrels per year) is
dQ
the rate of change
of total consumption
dt
with respect to time.
The fact that this demand is increasing
exponentially at the rate of 10 percent per year
and is currently equal to 40 (billion barrels per
year) implies that the demand is
dQ
= 40e0.1t
dt
billion barrels per year.
Hence, the total yearly consumption will be
Z
Q(t) = 40e0.1t dt = 400e0.1t + C
and Q(0) = 400 + C. At the end of 5 years the
consumption will be
Q(4) = 400e0.5 + C
and during the next 5 years it is
Q(5) − Q(0) = 259.49 billion barrels per year.
41. Let Q(t) denote the number of pounds of salt in
the tank after t minutes.
dQ
Then
is the rate of change of salt with
dt
respect to time (measured in pounds per
minute). Thus,
dQ
dt
Separate the variables and integrate to get
Z
Z
1
1
dQ = −
dt,
Q
50
t
ln |Q| = − + C1 ,
50
Q = eC1 e−t/50 = Ce−t/50 ,
where C = eC1 . Since there are initially 600
pounds of salt in the tank (3 pounds of salt per
gallon times 200 gallons), 600 = Q(0) = C.
Hence,
Q(t) = 600e−t/50 .
The amount of salt in the tank after 100
minutes is Q(100) = 600e−2 = 81.2012 pounds.
42. Let Q(t) denote the population in millions t
years after 1990. The differential equation
describing the population growth is
dQ
= 10k(10 − Q),
dt
where k is a positive constant of
proportionality.
dQ
= 10kdt
10 − Q
− ln |10 − Q| = 10kt + C1
Q = 10 − Ce−10kt
=
(rate at which salt enters)
−(rate at which salt leaves)
pounds entering gallons entering
=
gallon
minute
pounds leaving gallons leaving
−
.
gallon
minute
Now
=
=
Hence
pounds leaving
gallon
pounds of salt in the tank
gallons of brine in the tank
Q
.
200
dQ
Q
Q
=−
(4) = − .
dt
200
50
Since the population was 4 million in 1990,
Q(0) = 4 = 10 − C and so C = 6,
Q = 10 − 6e−10kt
Since the population was 4.74 million in 1995,
Q(5) = 4.74 and
4.74 = 10 − 6e−50k or k = 0.002633
Q(t) = 10 − 6e−0.02633t
43. Let Q(t) denote the amount (in million of
dollars) of new currency in circulation at time
dQ
t. Then
is the rate of change of the new
dt
5.5. REVIEW PROBLEMS
163
currency with respect to time (measured in
miilion dollars per day). Thus
Now to find t so that Q(t) = 0.9(5, 000)
substitute into the last solution
4, 500
= e−18t/5,000 ,
5, 000
dQ
dt
= (rate at which new currency enters)
−(rate at which new currency leaves).
1−
Now, the rate at which new currency enters is
18 million per day. The rate at which new
currency leaves is
thus t =
new currency at time t
times
total currency
(rate at which new currency enters)
Q(t)
=
(18)
5, 000
million per day. Putting it all together,
dQ
18Q
Q
= 18 −
= 18 1 −
dt
5, 000
5, 000
ln
dQ
= 18dt and integrate
1 − Q/5, 000
Q |
|
−5, 000 ln | 1 −
= 18t + C.
5, 000 |
When t = 0, Q(0) = 0 which yields
0 |
|
−5, 000 ln | 1 −
= 18(0) + C
5, 000 |
or C = 0. Therefore, the solution becomes
44. Let P denote the number of people, x the
income. The rate of change of the number of
people
dP
1
= −kP
dt
x
where k is a positive constant of
proportionality.
Separation of variables leads to
dP
−k
=
dt
P
x
Since Q is a part of 5, 000, 1 −
1−
1−
Q
5, 000
Q
>0
5, 000
=−
18t
,
5, 000
Q
= e−18t/5,000 .
5, 000
−k
t + C1
x
ln P =
P = e−kt/x+C1 = e−kt/x eC1 = Ce−kt/x
where C = eC1 . Note that P > 0 and x > 0 in
the context of this problem.
dP
45.
= P (ln P0 )(ln β)β t ,
dt
dP
= (ln P0 )(ln β)β t dt,
P
integrating leads to
Q |
18t
|
ln | 1 −
=−
.
|
5, 000
5, 000
5, 000
ln 10 = 640 days.
18
and integrating yields
Separate variables to obtain
and so ln
18t
1
=−
,
10
5, 000
ln P = (ln P0 )β t + C1 ,
P
= e(ln P0 )β
t
= Celn(P0 )
+C1
βt
= eC1 e(ln P0 )β
= C(P0 )β
t
t
where C = eC1 . Absolute values were dispensed
with because by context P > 0.
164
CHAPTER 5. INTEGRATION
46. Let P be the number of people involved and Q
the number of people implicated.
after substituting for
e3kP =
dQ
= kQ(P − Q)
dt
dQ
= kdt
Q(P − Q)
Before proceeding let’s break up the fraction
(by the method of partial fractions).
1
A
B
=
+
Q(P − Q)
Q P −Q
Now multiply by the least common
denominator. 1 = A(P − Q) + BQ which must
be an identity (that is true for all values), so
when
1
Q = 0, A =
P
1
and when Q = P , B = .
P
Z
Z 1
dQ
dQ
=
+
P
Q
P −Q
Z
= k dt
dQ
Q(P − Q)
1
[ln |Q| − ln |P − Q|] = kt + C1
P
| Q |
ln |
= kP t + C1
P −Q|
Q
= CekP t
P −Q
Since Q = 7 when t = 0, C =
When t = 3 Q = 16,
7
.
P −7
16
7
=
e3P k
P − 16
P −7
When t = 6 Q = 28 and
28
P − 28
=
=
7
e6P k
P −7
7
16(P − 7) 2
P − 7 7(P − 16)
16(P − 7)
7(P − 16)
(28)(7)(P 2 − 32P + 256) = 256(P − 28)(P − 7)
= 256(P 2 − 35P + 196)
196P 2 −6, 272P +50, 176 = 256P 2 −8, 960P +50, 176
(256 − 196)P 2 + (6, 272 − 8, 960)P = 0, and
P = 0 (to be rejected in the context of this
problem) or
P =
2, 688
672
=
= 44.8 ≈ 45 people
60
15
47. Let S be the concentration of the solute inside
the cell, S0 that of the solute outside the cell,
and A the area of the cell wall.
The rate of change of the inside solute is jointly
proportional to the area of the cell surface and
the difference between the solute inside and
outside the wall, so
dS
= kA(S − S0 )
dt
where k is a positive constant of
proportionality, S0 is constant, and so is A.
Separation of variables and integration leads to
Z
Z
1
dS =
kAdt
S − S0
ln |S − S0 | = kAt + C1
S − S0 = ±eC1 ekAt = CekAt
The absolute sign was dropped since C can
conveniently be positive or negative as the need
prescribes. Thus S = S0 + CekAt .