Solutions to Suggested Problems
Section 5.1
1. P(x) = MP(x)dx = (−0.8x + 2000)dx = −0.4x2 + 2000x + c. Setting
x = 30 and P(30) = 15000 gives −0.4(30)2 + 2000(30) + c = 15000, from
which we get c = −44, 640. Therefore, P(x) = −0.4x2 + 2000x − 44, 640.
R
R√
xdx = x1/2 dx = 32 x3/2 + c, an antiderivative of f (x) is 23 x3/2 .
3. Since
R
R
5. Since (s − 1s )ds = sds − 1s ds = 12 s2 − ln|s| + c, an antiderivative of
h(s) is 12 s2 − ln|s|.
R
R
R
R
√
7. Since (x5 −3 x+x−4 )dx = x5 dx− x1/3 dx+ x−4 dx = 16 x6 − 43 x4 3−
1 −3
+ c, an antiderivative of f (x) is 61 x6 − 34 x4 3 − 13 x−3 .
3x
R
R
d
x2
dx (1 + e )
11. Since F 0 (x) =
f (x).
R
2
= 2xex = f (x), F(x) is an antiderivative of
13. Since F 0 (x) =
d
x−1
dx (ln| x+1 |)
(x+1)−(x−1)
(x−1)(x+1)
= f (x), F(x) is and antiderivative of f (x).
=
2
x2 −1
=
d
dx (ln|x − 1| − ln|x + 1|)
=
1
x−1
1
− x+1
=
15. We want F 0 (x) = f (x). Since F 0 (x) =
5A = 1. So A = 51 .
5A
5x+1 ,
17. Since F 0 (x) =
or 2A = 1. So, A = 12 .
23. Since
2Ax
,
x2 +1
we get
2Ax
x2 +1
=
x
,
x2 +1
d
A
dx (Aln|x − 2| + Bln|x + 3| + c) = x−2
we get
5A
5x+1
=
1
5x+1 ,
or
B
+ x+3
, we get
A
B
(A + B)x + (3A − 2B)
1
=
+
=
x2 + x − 6 x − 2 x + 3
x2 + x − 6
Therefore, we must have A + B = 0 and 3A − 2B = 1. The solutions are
A = 1/5 and B = −1/5.
1
27. Since 12 xdx = 14 x2 + c, we can take F1 (x) =
2
2
2
1, F3 (x) = x4 , F4 (x) = x4 + 1, and F5 (x) = x4 + 2.
R
x2
4
− 2, F2 (x) =
x2
4
−
29. Since 2e−x dx = −2e−x + c, we can take F1 (x) = −2e−x , F2 (x) =
−2e−x + 1, F3 (x) = −2e−x + 2, F4 (x) = −2e−x + 3, and F5 (x) = −2e−x + 4.
R
31. From the graph of f , we get
(
−2
f (x) =
x−4
if 0 ≤ x ≤ 2,
if 2 < x ≤ 6.
Taking antiderivatives gives
(
−2x + c
F(x) = 1 2
2 x − 4x + b
if 0 ≤ x ≤ 2,
if 2 < x ≤ 6.
for some constants c and b. In order for F to be continuous, the two branches
must coincide at x = 2, which means that −4 + c = −6 + b, or b = 2 + c.
Thus, we get
(
−2x + c
if 0 ≤ x ≤ 2,
F(x) = 1 2
2 x − 4x + 2 + c if 2 < x ≤ 6.
The initial condition F1 (0) = 2 gives c = 2, and
(
−2x + 2
if 0 ≤ x ≤ 2,
F1 (x) = 1 2
2 x − 4x + 4 if 2 < x ≤ 6.
The initial condition F2 (0) = 1 gives c = 1, and
(
−2x + 1
if 0 ≤ x ≤ 2,
F2 (x) = 1 2
2 x − 4x + 3 if 2 < x ≤ 6.
Therefore, F1 (1) = 0, F2 (1) = −1, and F1 (3) − F2 (3) = 1.
37.
R 3t
e dt = 13 e3t + c. Check: dtd ( 31 e3t + c) = e3t .
R 1 x
R −x
1 R x
1
−x
x
−x
39.
2 (e + e )dx = 2 ( e dx + e dx) = 2 (e − e ) + c. Check:
d 1 x
1 x
−x
−x
dx ( 2 (e − e ) + c) = 2 (e + e ).
2
41. y2 (5y3 + 1y )dy = (5y5 + y)dy = 56 y6 + 12 y2 + c. Check:
1 2
1
5
2
3
2 y + c) = 5y + y = y (5y + y ).
R
R
d 5 6
dy ( 6 y
+
5/3 + c.
47. y = (x4 − 4x2/3 )dx = x4 − 4 x2/3 = 15 x5 − 12
5x
R
R
R
dy
= 1 + 2x, we have y = (1 + 2x)dx = x + x2 + c. The initial
51. Since dx
condition y(0) = −1 gives 0 + 02 + c = −1, so c = −1. Therefore, y =
x + x2 − 1.
R
dy
= x + e−x , we have y = (x + e−x )dx = 12 x2 − e−x + c. The
53. Since dx
initial condition y(0) = 1 gives 12 (0)2 − e0 + c = 1, so c = 2. Therefore,
y = 21 x2 − e−x + 2.
R
0.08t , we have Q = (e0.08t )dt = 1 e0.08t +c = 12.5e0.08t +
55. Since dQ
dt = e
0.08
c. The initial condition Q(0) = 80, 000 gives 12.5 + c = 80, 000, so c =
79, 987.5. Therefore, Q(t) = 12.5e0.08t + 79, 987.5.
R
61. (a) Since P0 (x) = 15 − 0.01x, we have P(x) = 15x − 0.005x2 + c. the
initial condition P(0) = −1000 gives c = −1000. Therefore, P(x) = 15x −
0.005x2 − 1000.
(b) P(1, 000) = 15 · (1, 000) − 0.005 · (1, 000)2 − 1, 000 = 9, 000.
(c) Solving P0 (x) = 15−0.01x = 0 gives x = 1, 500. Since P00 (x) = −0.01 <
0, the graph of P(x) is concave down. The maximum profit that the company
can earn in a single week is P(1, 500) = 15 · (1, 500) = 0.005 · (1, 500)2 −
1, 000 = 10, 250 dollars.
63. If S(t) isRthe savings generated at the end of t years, S0 (t) = 300, 000e−0.03t ,
and S(t) = 300, 000e−0.03t dt = −10, 000, 000e−0.03t + c. The initial condition S(0) = 0 gives c = 10, 000, 000. So, S(t) = 10, 000, 000(1 − e−0.03t ).
Setting 10, 000, 000(1−e−0.03t ) = 2, 000, 000 gives e−0.03t = 45 , or −0.03t =
1
ln( 54 ), or t = − 0.03
ln( 54 ) ≈ 7.44. The system will pay for itself after 7.44
years.
65. Since y0 = 2x − 3, we have y = (2x − 3)dx = x2 − 3x + c. The graph
passes through (0, 4), so we ahve c = y(0) = 4. Therefore, y = x2 − 3x + 4.
Solving y0 = 2x − 3 = 0 gives x = 3/2, at which there is a minimum. The
minimum value is y(3/2) = 7/4, which is the minimum height of the graph.
R
3
71. Write PBN (t) and PA (t) for the profits earned by Book Nook and Adams
0 (t) = 0.92P0 (t), we see P
Books at the time t respectively. Since PBN
BN =
A
0.92PA +c. Now PBN (0) = −2, 000 and PA (0) = −2, 400. Thus, c = −2, 000+
0.92 · 2, 400 = 208 and PBN = 0.92PA + 208 = 0.92 · 3, 100 + 208 = 3, 060.
75. (a) Since the fertilizer is leaving the tank, the volume V (t) remaining in
−0.2t ,
the tank is decreasing and satisfies the differential equation dV
dt = −20e
with the initial condition V (0) = 100.
(b) Integrating gives V (t) = 100e−0.02t + c. Since V (0) = 100, we must
have 100 + c = 100, or c = 0. Thus, V (t) = 100e−0.02t .
77. First, we check the initial condition: c(0) = (c0 −C)e−0 +C = c0 . Next,
−kAt/V . Finally, substituting
taking the derivative gives c0 (t) = − kA
V (c0 −C)e
for c(t) on the right-hand side of the differential equation gives
kA
kA
kA
[C − c(t)] = [c − (c0 −C)e−kAt/V −C] = − (c0 −C)e−kAt/V .
V
V
V
Section 5.2
1. Let u = x2 + 1. Then du = 2xdx, and
2x
dx =
x2 + 1
Z
Z
du
= ln|u| + c = ln|x2 + 1| + c = ln(x2 + 1) + c.
u
3. Let u = −t. Then du = −dt, or dt = −du, and
Z
Z
−t
e dt =
u
e (−du) = −
Z
eu du = −eu + c = −e−t + c.
5. Let u = 3x + 4. Then du = 3dx, or dx = 13 du, and
Z √
Z
3x + 4dx =
1
1
u1/2 · du =
3
3
7. Let u = ln(x2 + 1). Then du =
Z
xln(x2 + 1)
dx =
x2 + 1
Z
Z
2
2
u1/2 du = u3/2 + c = (3x + 4)3/2 + c.
9
9
2x
,
x2 +1
or
x
dx
x2 +1
= 12 du, and
1
1
1
u( du) = u2 + c = (ln(x2 + 1))2 + c.
2
4
4
4
√
√
√
1
9. Let u = 1+ x, so that x = u−1 and 1− x = 2−u. Also, du = 2√
dx,
x
√
so that dx = 2 xdu = 2(u − 1)du. Therefore,
√
Z
Z
Z
1− x
2
(2 − u)(u − 1)
√ dx = 2
du = 2 (−u + 3 − )du
1+ x
u
u
2
= −u + 6u − 4ln|u| + c
√
√
√
= −(1 + x)2 + 6(1 + x) − 4ln|1 + x| + c
√
√
= 4 x − x − 4ln|1 + x| +C
(Here C = c+5).
√
11. (a) Let u = x. Then du =
Z
1
√ dx =
1+ x
1
√
2
√
dx,
or
2
xdu = 2udu. Thus,
x
2u
du
u+1
Z
√
√
2
=
(2 −
)du = 2u − 2ln|u + 1| + c = 2 x − 2ln( x + 1) + c
u+1
Z
(b) The two answers differ only by a constant. That is,
√
√
√
√
2(1 + x) − 2ln(1 + x) + c = 2 x − 2ln(1 + x) + c + 2.
15. Let u = x2 + 4. Then du = 2xdx, and
Z
√
2x
x2 + 4
Z
dx =
1
√ du =
u
Z
u−1/2 du = 2u1/2 + c = 2
p
x2 + 4 + c.
Differentiating gives
d p 2
1
2x
(2 x + 4 + c) = 2 · (x2 + 4)−1/2 · 2x + 0 = √
.
dx
2
x2 + 4
17. Let u = x2 − 3. Then du = 2xdx, or xdx = 21 du and
Z
2
5
(x − 3) xdx =
Z
p
1
1
1
u5 · du = u6 + c = (x2 − 3)6 + c = 2 x2 + 4 + c.
2
12
12
Differentiating gives
d 1 2
1
( (x − 3)6 + c) = · (x2 − 3)5 · 2x + 0 = (x2 − 3)5 x.
dx 12
6
5
19. Let u = x2 + x + 1. Then du = (2x + 1)dx, and
Z
2x + 1
dx =
2
(x + x + 1)3
1
1
du = − u−2 + c
3
u
2
1
1
= − (x2 + x + 1)−2 + c = − 2
+ c.
2
2(x + x + 1)2
Z
Differentiating gives
1
1
2x + 1
d
(− (x2 +x+1)−2 +c) = − ·(−2)(x2 +x+1)−3 ·(2x+1) = 2
.
dx 2
2
(x + x + 1)3
21. Let u = −0.1t 4 . Then du = −0.4t 3 dt, or t 3 dt = −2.5du and
Z
3 −0.1t 4
t e
dt = −2.5
Z
4
eu du = −2.5eu + c = −2.5e−0.1t + c.
Differentiating gives
4
4
4
d
(−2.5e−0.1t + c) = −2.5e−0.1t · (−0.1 · 4t 3 ) = t 3 e−0.1t .
dt
23. Let u = 2x + 1. Then du = 2dx, or dx = 12 du, and
Z
3
dx =
2x + 1
Z
3 1
3
3
· du = ln|u| + c = ln|2x + 1| + c.
u 2
2
2
Differentiating gives
d 3
3
1
3
( ln|2x + 1| + c) = ·
·2 =
.
dx 2
2 2x + 1
2x + 1
25. Let u = ln(t). Then du = 1t dt, and
Z
ln(t 2 )
dt =
t
Z
1
2ln(t) · dt = 2
t
Z
udu = u2 + c = (ln(t))2 + c.
Differentiating gives
d
1
2ln(t) ln(t 2 )
((ln(t))2 + c) = 2 · ln(t) · + 0 =
=
.
dt
t
t
t
6
27. Let u = ln(t). Then du = 1t dt, and
Z
dt
=
tln(t)
1 1
· dt =
ln(t) t
Z
Z
1
du = ln|u| + c = ln|ln(t)| + c.
u
Differentiating gives
d
1 1
1
(ln|ln(t)| + c) =
· =
.
dt
lnt t
tln(t)
31. To compute y =
dt = 14 du and
Z
y=
R√
4t + 1dt, we let u = 4t + 1. Then du = 4dt or
1
1
1
u1/2 · du = u3/2 + c = (4t + 1)3/2 + c.
4
6
6
Since 3 = y(0) = 16 + c, we have c =
x
17
6,
so that y = 61 (4t + 1)3/2 + 17
6.
−x
x
−x
x
33. To compute r = eex −e
+e−x dx, we let u = e + e . Then du = (e −
R
e−x )dx, and r = u1 du = ln|u|+c = ln(ex +e−x )+c. The condition r(0) = 0
gives 0 = ln(1 + 1) + c, so c = −ln(2). Thus, r = ln(ex + e−x ) − ln(2) =
x
−x
ln( e +e
2 ).
R
t
2
−1
35. Since dv
dt = a(t) = 5 − t(t + 1) , we have v(t) = (5 − t 2 +1 )dt. Using
the substitution u = t 2 + 1 we obtain v(t) = 5t − 12 ln(t 2 + 1) + c. Since
the boat starts from a position of rest, v(0) = 0, which gives c = 0 and
v(t) = 5t − 12 ln(t 2 + 1). So v(1) = 5 − 12 ln(2) ≈ 4.65 miles per hour.
R
37. Integrating gives y = (70e0.14t −120e0.03t )dt = 70 e0.14t dt −120 e0.03t dt.
Using the substitutions u = 0.14t for the first integral and u = 0.03t for the
second, we obtain
R
y=
R
R
70 0.14t 120 0.03t
e
−
e
+ c = 500e0.14t − 4, 000e0.03t + c.
0.14
0.03
From the initial condition y(0) = 3, 700, we obtain 500−4, 000+c = 3, 700,
so c = 7, 200. Therefore, the fish population after t years is y(t) = 500e0.14t −
4, 000e0.03t + 7, 200 and y(10) = 500e1.4 − 4, 000e0.3 + 7, 200 ≈ 3, 828.
7
R
x
)dx = 1, 600x−1, 600 √ x 2 dx.
1+x2
√ 1+x
2
By substituting u = 1+x and du = 2xdx, we obtain y = 1, 600(x− 1 + x2 )+
R
38. Integrating gives y = 1, 600(1− √
c. Since y(0) = 0, √
we have 0 − 1, 600 + c = 0, so c = 1, 600.
√ Therefore,
2
y = 1, 600x − 1, 600 x + 1 + 1, 600, and y(6) = 1, 600(7 − 37) ≈ 1, 468.
41. To compute E(y) = −2.52y(0.1y2 + 1)−4 dy, we let u = 0.1y2 + 1.
Then du = 0.2ydy or ydy = 5du. Therefore,
R
E(y) = −12.6
Z
u−4 du = 4.2u−3 + c = 4.2(0.1y2 + 1)−3 + c.
Since E(0) = 3.6, we have 4.2 + c = 3.6, or c = −0.6. Thus, E(y) =
2
−3
2
3
4.2(0.1y
q +√1) − 0.6. Solving E(y) = 0, or (0.1y + 1) = 7, we find
y = 10(3 7 − 1) ≈ 3 years.
Section 5.3
1. Let u = x and dv = e2x dx. Then du = dx and v = 12 e2x . Therefore,
Z
1
xe2x dx = xe2x −
2
Z
1
1
1
1
1 2x
e dx = xe2x − e2x + c = e2x (x − ) + c.
2
2
4
2
2
5. Let u = t 2 and dv = e−t dt. Then du = 2tdt and v = −e−t , and
Z
2 −t
2 −t
t e dt = −t e
Z
+2
te−t dt.
Next, we apply integration by parts again, this time with u = t and dv =
e−t dt. Then du = dt and v = −e−t , and
Z
2 −t
2 −t
t e dt = −t e
2 −t
= −t e
−t
+ 2[−te
−t
− 2te
−
Z
Z
+2
(−e−t )dt]
e−t dt = −t 2 e−t − 2te−t − 2e−t + c.
7. Let u = ln(x) and dv = x2 dx. Then du = 1x dx and v = 13 x3 . Therefore,
Z
1
1
x2 ln(x)dx = x3 ln(x) −
3
3
8
Z
1
1
x2 dx = x3 ln(x) − x3 + c.
3
9
√
9. Let u = x and dv = 3x + 2dx. Then du = dx and v = 29 (3x + 2)3/2 (v is
found by substitution, with t = 3x + 2 and dt = 3dx). Integrating by parts
gives
Z
Z √
2
2
3/2
(3x + 2)3/2 dx,
x 3x + 2dx = x(3x + 2) −
9
9
and, again using the substitution t = 3x + 2, we get
Z √
2
2 1 2
x 3x + 2dx =
x(3x + 2)3/2 − · · (3x + 2)5/2 + c
9
9 3 5
4
2
x(3x + 2)3/2 −
(3x + 2)5/2 + c.
=
9
135
√
R √
11. Let u = x2 and dv = x x2 + 1dx. Then du = 2xdx and v = x x2 + 1dx.
Using the substitution t = x2 + 1 gives v = 13 (x2 + 1)3/2 . Then, integration
by parts gives
Z
Z
p
1
1 2 2
3/2
3
2
x (x + 1) − 2x · (x2 + 1)3/2 dx
x x + 1dx =
3
3
Z
1 2 2
1
=
x (x + 1)3/2 −
(x2 + 1)3/2 (x2 + 1)0 dx
3
3
1 2 2
2
=
x (x + 1)3/2 − (x2 + 1)5/2 + c,
3
15
where we again used the substitution t = x2 + 1 in computing the last integral.
13. Let
1
(x−1)(x+5)
=
A
x−1
B
+ x+5
. By clearing denominators, we get
1 = A(x + 5) + B(x − 1).
Substituting x = 3 gives 1 = A(−5 + 5) + B(−5 − 1), or B = − 16 . Next,
substituting x = 1 gives 1 = A(1 + 5) + B(1 − 1), or A = 16 . Therefore,
1
1
1
1
1
= ·
− ·
.
(x − 1)(x + 5) 6 x − 1 6 x + 5
1
1
A
B
15. Let x2 −9x+20
= (x−4)(x−5)
= x−4
+ x−5
. By clearing denominators, we
get 1 = A(x − 5) + B(x − 4), so A = −1 and B = 1. Thus, the given integral
equals
Z
1
dx −
x−5
Z
1
x−5
dx = ln|x − 5| − ln|x − 4| + c = ln|
| + c.
x−4
x−4
9
A
1
B
= x−a
19. Let (x−a)(x−b)
+ x−b
. After clearing denominators, we obtain the
equation 1 = A(x − b) + B(x − a). Substituting x = a gives 1 = A(a − b),
1
1
. Substituting x = b gives 1 = B(b − a) or B = − a−b
. Thus, the
or A = a−b
integral equals
21. Let
Z
1
dx −
x−a
=
A
x−1
1
(
a−b
Z
x
(x−1)(x+2)
Z
1
x−a
1
dx) =
ln|
| + c.
x−b
a−b x−b
B
+ x+2
. After clearing denominators, we obtain the
equation x = A(x + 2) + B(x − 1). Substituting x = 1 gives 1 = 3A or A = 13 ,
and substituting x = −2 gives −2 = −3B or B = 23 . Thus, the integral equals
1
3
Z
1
2
dx +
x−1
3
Z
1
1
2
dx = ln|x − 1| + ln|x + 2| + c.
x+2
3
3
x−3
x−3
23. Let x2 −3x−4
= (x−4)(x+1)
=
obtain the equation
A
x−4
B
+ x+1
. After clearing denominators, we
x − 3 = A(x + 1) + B(x − 4)
Substituting x = 4 gives 1 = 5A so that A = 15 . Substituting x = −1 gives
−4 = −5B, so that B = 45 . Thus, the given integral equals
1
5
Z
4
1
dx +
x−4
5
Z
1
4
1
= ln|x − 4| + ln|x + 1| + c.
x+1 5
5
25. If h(t) denotes the rocket’s height at the time t, then h0 (t) = √t+1
. Inte2t+1
grating gives h(t) = (t + 1)(2t + 1)−1/2 dt. We may compute this integral
by using integration by parts with u = t + 1 and dv = (2t + 1)−1/2 dt. Then
du = dt and v = (2t + 1)1/2 . Thus,
R
1/2
h(t) = (t +1)(2t +1)
−
Z
√
1
(2t +1)1/2 dt = (t +1) 2t + 1− (2t +1)3/2 +c.
3
Since h(0) = 0, we have 1 − 31 + c = 0 or c = − 23 . Thus,
1
2
h(t) = (t + 1)(2t + 1)1/2 − (2t + 1)3/2 − ,
3
3
√
and h(12) = 13 25 − 13 (25)3/2 − 23 = 68
3 ft.
10
10,000
dx. To compute this
29. If t denotes time in days, then t = x(4,000−x)
integral, we use the method of partial fractions, writing
R
10, 000
A
B
= +
.
x(4, 000 − x)
x 4, 000 − x
By multiplying both sides by x(4, 000 − x), we obtain 10, 000 = A(4, 000 −
x) + Bx. Setting x = 0 gives A = 2.5, and setting x = 4, 000 gives B = 2.5.
Therefore,
Z
t =
2.5
dx +
x
Z
2.5
dx
4, 000 − x
= 2.5(ln|x| − ln|4, 000 − x|) + c = 2.5ln|
x
| + c.
4, 000 − x
8
Since x = 8 when t = 0, we must have c = −2.5ln( 3,992
) = 2.5ln(499) ≈
x
15.53. Thus, t = 2.5ln| 4,000−x | + 2.5ln(499). If x = 200, then t ≈ 8.17 days.
31. (a) The range is expanding, since y0 =
9
t 2 +9t
> 0 for t ≥ 1.
9
B
9
(b) Let t 2 +9t
. After clearing denominators, we obtain
= t(t+9)
= At + t+9
9 = A(t + 9) + Bt, whose solution is A = 1, B = −1. Thus,
Z
y=
9
dt =
2
t + 9t
Z
1
1
t
( −
)dt = ln|t| − ln|t + 9| + c = ln|
| + c.
t t +9
t +9
Since 17.7 = y(1) = ln(0.1) + c, we have c = 17.7 − ln(0.1) = 17.7 +
ln(10), and
y = ln|
10t
t
| + 17.7 + ln(10) = ln(
) + 17.7,
t +9
t +9
t ≥ 1.
(c) limt→∞ y(t) = ln(10) + 17.7 ≈ 20 acres.
Section 5.4
2−1
1
1. Use the Riemann sum formula with ∆x = b−a
n = 4 = 4 and f (x) =
1
x . The values of f (x) at the endpoints are as follows: (a) Using left-hand
endpoints:
S4 =
1
1 4 1 2 1 4 1 1 1 1 319
·1+ · + · + · = + + + =
≈ 0.760.
4
4 5 4 3 4 7 4 5 6 7 420
11
x
1
f (x) 1
5/4
4/5
3/2
2/3
7/4
4/7
2
1/2
(a) Using right-hand endpoints:
S4 =
1 4 1 2 1 4 1 1 1 1 1 1 533
· + · + · + · = + + + =
≈ 0.635.
4 5 4 3 4 7 4 2 5 6 7 8 840
(c) The midpoints of the four subintervals and the values of f (x) at these
points are given in the table: Then
x
9/8
f (x) 8/9
S4 =
11/8
8/11
13/8
8/13
15/8
8/15
1 8 1 8 1 8 1 8
2 2
2
2
4448
· + · + · + ·
= + + +
=
≈ 0.691.
4 9 4 11 4 13 4 15 9 11 13 15 6435
3.
S8 = f (0.4) · 0.2 + f (0.6) · 0.2 + ... + f (1.8) · 0.2
= 0.2(15 + 17 + 19 + 16 + 14 + 14 + 12 + 10) = 117(0.2) = 23.4
7. (a) By using midpoints with ∆x = 40, we estimate the area of the upper
region by (60 +86 +94 +108 +100 +44 +41)·40 = 21, 320 square meters,
and the area of the lower region by (40 + 58 + 72 + 66 + 78 + 79 + 68) ·
40 = 18, 440 square meters. Therefore, the area of the lake is approximately
39,760 square meters.
(b) By using left endpoints with ∆x = 20 we estimate the area of the upper
region by (60 + 70 + 86 + 90 + 94 + 98 + 108 + 105 + 100 + 82 + 44 + 40 +
41) · 20 = 20, 360 square meters, and the area of the lower region by (40 +
45 + 58 + 68 + 72 + 67 + 66 + 72 + 78 + 80 + 79 + 80 + 68) · 20 = 17, 460
square meters. Therefore, the arease of the lake is approximately equal to
37,820 square meters.
12
9. The integral is the area of the right-hand triangle, with vertices at (0, 0), (1, 0),
and (1, 2), minus the area of the left-hand triangle, with vertices at (−2, −4), (−2, 0),
and (0, 0). Therefore,
Z 1
2xdx =
−2
1 · 2 (−2)(−4)
−
= −3.
2
2
13. The integral is the areas of the left-hand triangle, with vertices at
(0, 0), (1, 0) and (0, 1), minus the area of the right-hand triangle, with vertices at (1, 0), (3, 0), and (3, −2). Therefore,
Z 3
(1 − x)dx =
0
19.
R2
v(t)dt =
21.
R1
000.05te−0.01t dt
0
0
1
3
−2 = − .
2
2
R2
0 (−32t − 50)dt
25.
Z 3
Z 3
2 f (x)dx = 2
−1
f (x)dx
−1
Z 5
= 2·(
f (x)dx −
Z 5
−1
= 2(−6 − 4) = −20.
13
f (x)dx)
3