p =21
p =32
p =43
p =54
12 4
3p =—1 2
p =—27p =—23
5p =—2
Solutions to the Exercises of Chapter 13
13A. Riemann Sums
1. a = 0, b = 5, m = 4. The partition is 0 < 2 < 3 < 4 < 5. So ∆x1 = 2−0 = 2, ∆x2 = 3−2 =
1, ∆x2 = 4 − 3 = 1,∆x4 = 5 − 4 = 1, and p1 = x1 = 2, p2 = x2 = 3, p2 = x3 = 4, and
p4 = x4 = 5. So
4∑i=1
f(pi)∆xi = f(p1) · 2 + f(p2) · 1 + f(p3) · 1 + f(p4) · 1
= 2(25 − 22) + (25 − 32) + (25 − 42) + (25 − 52)
= 42 + 16 + 9 + 0 = 67.
2. a = 0, b = 4, n = 4. The partition is 0 < 1 < 2 < 3 < 4. ∆x1 = 1 − 0 = 1, ∆x2 = 2 − 1 =
1, ∆x3 = 3−2 = 1, ∆x4 = 4−3 = 1, and p1 = midpoint between 0 and 1 = 12, p2 = midpoint
between 1 and 2 = 32, p3 = midpoint between 2 and 3 = 5
2, p4 = midpoint between 3 and 4
1p =-1
2p =-0.5
3p =0
4p =0.5
5p =1
6p =1.5
= 72. So
n∑i=1
f(pi)∆xi =4∑
i=1
f(pi) · 1 = f(p1) + f(p2) + f(p3) + f(p4)
= 16 −(
1
2
)2
+ 16 −(
3
2
)2
+ 16 −(
5
2
)2
+ 16 − (7
2)2
= 4 · 16 − 1
4− 9
4− 25
4− 49
4= 64 − 1 + 9 + 25 + 49
4
= 64 − 84
4= 64 − 21 = 43.
3. a = −1, b = 2, m = 6, ∆x1 = −0.5 − (−1) = 0.5, ∆x2 = 0 − (−0.5) = 0.5, ∆x3 =
0.5 − 0 = 0.5, ∆x4 = 1.0 − 0.5 = 0.5, ∆x5 = 1.5 − 1.0 = 0.5, ∆x6 = 2 − 1.5 = 0.5, and
p1 = −1, p2 = −0.5, p3 = 0, p4 = 0.5, p5 = 1.0, p6 = 1.5. So
n∑i=1
f(pi)∆xi =6∑
i=1
f(pi)(0.5) = (0.5) [f(p1) + · · · + f(p6)]
= (0.5)[(
(−1)3 + 2)
+((−0.5)3 + 2
)+ (03 + 2) +
((0.5)3 + 2
)+
((1.0)3 + 2
)+
((1.5)3 + 2
)]= (0.5) [(−1 + 2) + (−0.125 + 2) + 2 + (0.125 + 2) + (1 + 2) + (3.375 + 2)]
= (0.5) [2 + 2 + 2 + 2 + 2 + 2 − 1 − 0.125 + 0.125 + 1 + 3.375]
= (0.5) [12 + 3.375]
= (0.5)(15.375) = 7.6875
4. a = 0, b = 2, m = 4, ∆x1 = 0.5 − 0 = 0.5, ∆x2 = 1.0 − 0.5 = 0.5, ∆x3 = 1.5 − 1.0 =
0.5, ∆x4 = 2 − 1.5 = 0.5. So
2
421p =0.25 p =1
3p =1.25 p =2
n∑i=1
f(pi)∆xi =4∑
i=1
f(pi)(0.5)
= (0.5) [f(p1) + f(p2) + f(p3) + f(p4)]
= (0.5)
[1
0.25 + 1+
1
1 + 1+
1
1.25 + 1+
1
2 + 1
]
= (0.5)
[1
1.25+
1
2+
1
2.25+
1
3
]= (0.5) [0.8 + 0.5 + 0.44 + 0.33]
= 1.035
13B. Pushing a Riemann Sum to the Limit
5. i. The partition is 0 < 37< 6
7< 9
7< 12
7< 15
7< 18
7< 21
7= 3. It follows that the points,
p1, . . . , p7 are as listed.
ii.
7∑i=1
f(pi)∆xi =7∑
i=1
f(pi) ·3
7
=[f(p1) + f(p2) + f(p3) + f(p4) + f(p5) + f(p6) + f(p7)
](3
7
)
=[16 −
(1 · 3
7
)2
+ 16 −(
2 · 3
7
)2
+ 16 −(
3 · 3
7
)2
+ 16 −(
4 · 3
7
)2
+ 16 −(
5 · 3
7
)2
+ 16 −(
6 · 3
7
)2
+ 16 −(
7 · 3
7
)2 ](3
7
)
=[16 · 7 −
(12 ·
(3
7
)2
+ 22 ·(
3
7
)2
+ 32 ·(
3
7
)2
+ 42 ·(
3
7
)2
+ 52 ·(
3
7
)2
+ 62 ·(
3
7
)2
+ 72 ·(
3
7
)2
+ 72 ·(
3
7
)2 )] (3
7
)
=
[16 · 7 −
(3
7
)2 (12 + 22 + 32 + 42 + 52 + 62 + 72
)] (3
7
)
3
iii. The use of the sum of squares formula shows that this equals
[16 · 7 −
(3
7
)2 (7(7 + 1)(14 + 1)
6
) ](3
7
)
= 16 · 3 − 9
2
(7 + 1)(14 + 1)
72
= 48 − 9
2
(120
72
)
= 48 − 540
72= 48 − 11.02 = 36.98
6. i. p1 = 1 · 3n, p2 = 2 · 3
n, p3 = 3 · 3
n, p4 = 4 · 3
n, · · · ,
pi = i · 3n, · · · , pn = n · 3
n= 3.
ii. f(pi) = 16 − (pi)2 = 16 −
(i 3n
)2= 16 − i2 ·
(3n
)2
iii.n∑
i=1
f(pi)∆xi =
[n∑
i=1
f(pi)
] (3
n
)=
[n∑
i=1
(16 − i2
(3
n
)2)] (
3
n
)
=
[16 · n−
n∑i=1
i2(
3
n
)2] (
3
n
)= 48 −
(n∑
i=1
i2
)(3
n
)2 (3
n
)
= 48 −[n(n + 1)(2n + 1)
6
] (3
n
) (3
n
)2
= 48 − (n + 1)(2n + 1)
2· 9
n2
= 48 − 9
2
(n + 1)(2n + 1)
n2= 48 − 9
2
(2n2 + 3n + 1
n2
)
= 48 − 9
2
(2 +
3
n+
1
n2
)
7. i. When n is pushed to infinity, ‖Pn‖ = 3n
goes to zero.
ii. Review Steps (1) - (3) of the limit process
lim‖P‖→0
(n∑
i=1
f(pi)∆xi
).
Notice that in Step (1) you were free to take any partition P and that in Step (2) you
were free to take any points p1, · · · , pn with p1 in the first subinterval, p2 in the second,
and so on. The assertion in Step (3) was that the numbersn∑
i=1
f(pi)∆xi
produced by the repetition of Steps (1) and (2) with partitions P whose norms go to
zero, will close in on some number
lim‖P‖→0
n∑i=1
f(pi)∆xi .
4
This number is the same regardless of how the partitions P and the points pi were selected
along the way. This fact was not proved in the text. Rather, it was only made plausible
by making the connection with areas.
Now turn to the procedure outlined in Exercise 6. Notice that it is a specific instance of
the process described in Steps (1) and (2). It follows that the limiting number
limn→∞
[48 − 9
2
(2 +
3
n+
1
n2
)]
must be equal to
lim‖P‖→0
(n∑
i=1
f(pi)∆xi
).
iii. Because ∫ b
a
f(x) dx = lim‖P‖→0
(n∑
i=1
f(pi)∆xi
)
by definition, we find that∫ 3
0
(16 − x2)dx = limn→∞
[48 − 9
2
(2 +
3
n+
1
n2
)]
= 48 − 9
2· 2 = 48 − 9 = 39.
Correction: Note that the answer 37 (as stated in the text) is incorrect.
13C. Applying the Fundamental Theorem of Calculus
8. From the fact that ddx
cosx = − sinx, we see that F (x) = − cosx is an anti-derivative of
f(x) = sinx. Therefore,
∫ π2
0
sin x = (− cosx)∣∣π
2
0= − cos
π
2− (− cos 0).
A look at the graph of cosx (see Figure 10.29) shows that cos π2
= 0 and cos 0 = 1. So
∫ π2
0
sinx dx = 1.
9. We need an anti-derivative of tanx = sin xcos x
. Let g(x) = cosx. So g′(x) = − sin x, and tanx =
−g′(x)g(x)
. A review of Section 10.3 tells us that ddx
[− ln g(x)] = −g′(x)g(x)
. So ddx
[− ln(cosx)] = tanx.
Therefore,
5
∫ π4
0
tanxdx = − ln (cosx)∣∣π
4
0= − ln cos
π
4− (− ln cos 0)
= − ln
√2
2+ ln 1 = − ln 2
12 + ln 2 + 0
= −1
2ln 2 + ln 2 =
1
2ln 2 ≈ 0.347.
10. Recall from Section 8.6 that ddx
tanx = sec2 x. So
∫ π4
0
sec2 dx = tanx∣∣π
4
0= tan
π
4− tan 0 = 1 − 0 = 1.
The answers to the next two exercises depend on formulas from Sections 10.1 and 10.3.
11.
∫ ln 5
ln 2
ex dx = ex∣∣ln 5
ln 2= eln 5 − eln 2 = 5 − 2 = 3.
12.
∫ 7
3
1
xdx = lnx
∣∣73
= ln 7 − ln 3 = ln7
3≈ 0.847.
13. This integral can be solved by use of the formula
∫ √a2 − x2
xdx =
√a2 − x2 − a ln
(a + (a2 − x2)
12
x
)+ C
as follows: ∫ 4
1
√25 − x2
xdx =
[√25 − x2 − 5 ln
(5 +
√25 − x2
x
)] ∣∣∣∣4
1
=√
9 − 5 ln
(5 +
√9
4
)−
√24 + 5 ln
(5 +
√24
1
)
≈ 3 − 5 ln 2 −√
24 + 5 ln(5 +√
24)
≈ 3 − 3.466 − 4.899 + 11.462 ≈ 6.097.
14. Making use of the discussion preceding this exercise, we get∫ 6
20
(2x− 7)dx = −∫ 20
6
(2x− 7)dx = −(x2 − 7x)∣∣206
= − [(400 − 140) − (36 − 42)] = −(260 + 6) = −266.
6
15. By the discussion preceding Exercise 14,∫ 1
8
(x
32 + 4x
12 − π
)dx = −
∫ 8
1
(x
32 + 4x
12 − π
)dx
= −[(
2
5x
52 + 4 · 2
3x
32 − πx
) ∣∣∣∣8
1
]
= −[(
2
58
52 +
8
38
32 − 8π
)−
(2
5+
8
3− π
)]≈ −[72.408 + 60.340 − 25.133 − (0.4 + 2.667 − 3.142)]
≈ −107.69.
16. Once again by the discussion preceding Exercise 14,∫ 0
π
(cosx− 8x2
)dx = −
∫ π
0
(cosx− 8x2
)dx
= −[(
sinx− 8
3x3
) ∣∣∣∣π
0
]= −
[(0 − 8
3π3
)− 0
]
=8
3π3.
The fact to use in the next three exercises is:
d
dx
(∫ x
a
f(t)dt
)= f(x).
17. F ′(x) = x(1 + x3)9.
18. F ′(x) = sinx (cosx− 4x2) .
19. Let F (x) =
∫ x
0
(sin t + 2t−3
)dt. By the formula already referred to, F ′(x) = sinx + 2x−3.
Notice that G(x) = F (3x2). An application of the chain rule shows us that
G′(x) = F ′(3x2) · 6x =[sin 3x2 + 2
(3x2
)−3]6x
= 6x sin 3x2 + 12x1
(3x2)3= 6x sin 3x2 +
12
27
1
x5
= 6x sin 3x2 +4
9x5.
7
13D. The Substitution Method
20. With u = 4x− 5, we get dudx
= 4. So du = 4dx anddx = du4. So
∫(4x− 5)
12dx =
∫u
12 · du
4=
1
4
∫u
12 du =
1
4
[2
3u
32 + C ′
]
=2
12(4x− 5)
32 +
C ′
4=
1
6(4x− 5)
32 + C.
21. With u = 1 − 5x2, we get dudx
= −10x. So du = −10x dx. Therefore,
∫10x(1 − 5x2)
23 dx =
∫u
23 (−du) = −
∫u
23 du = −
[3
5u
53 + C ′
]
= −3
5
(1 − 5x2
) 53 − C ′
= −3
5
(1 − 5x2
) 53 + C.
22. Because du = 2x dx, we get∫x cosx2 dx =
∫(cosu)(
1
2du) =
1
2
∫cosu du
=1
2[sinu + C ′]
=1
2sinx2 + C.
23. Since dudt
= cos t, we get
∫sin3 t cos t dt =
∫u3 du =
u4
4+ C
=1
4sin4 t + C.
24. Note that du = dx and x = u− 1. So∫(x− 1)(x + 1)
12 dx =
∫(u− 2)u
12 du =
∫ (u
32 − 2u
12
)du
=2
5u
52 − 2 · 2
3u
32 + C
=2
5(x + 1)
52 − 4
3(x + 1)
32 + C.
8
25. Because dx = du and x = u− 3, we get∫x2(x + 3)
12 dx =
∫(u− 3)2u
12 du =
∫(u2 − 6u + 9)u
12 du
=
∫ (u
52 − 6u
32 + 9u
12
)du
=2
7u
72 − 6 · 2
5u
52 + 9 · 2
3u
32 + C
=2
7(x + 3)
72 − 12
5(x + 3)
52 + 6(x + 3)
32 + C.
26. Because dx = du and x = u + 2, we see that∫x2
(x− 2)3dx =
∫(u + 2)2
u3du =
∫u2 + 2u + 4
u3du
=
∫ (u−1 + 2u−2 + 4u−3
)du
= lnu− 2u−1 − 2u−2 + C
= ln(x− 2) − 2(x− 2)−1 − 2(x− 2)−2 + C.
27. Note that dudϕ
= sec2 ϕ, so du = sec2 ϕdϕ. Therefore,∫sec2 ϕ
tan2 ϕ + 1dϕ =
∫du
u2 + 1.
By a formula from Section 10.5,
∫du
u2 + 1= tan−1 u + C. So
∫sec2 ϕ
tan2 ϕ + 1dϕ = tan−1(tanϕ) + C
= ϕ + C.
There is a simpler approach to this problem. Dividing the identity
sin2 ϕ + cos2 ϕ = 1
by cos2 ϕ, gives us the identity tan2 ϕ + 1 = sec2 ϕ. So∫sec2 ϕ
tan2 ϕ + 1dϕ =
∫dϕ = ϕ + C.
28. Try u = 12x7 + 19. So dudx
= 84x6. Hence du = 84x6 dx and (looking ahead), x6dx = 184du. So∫
5x6
12x7 + 19dx =
∫5( 1
84du)
u=
∫5
84u−1du =
5
84lnu + C
=5
84ln(12x7 + 19) + C.
9
29. Try u = 1 + 2x + 4x2. So dudx
= 2 + 8x = 2(1 + 4x). Hence du = 2(1 + 4x)dx. Therefore,∫(1 + 4x)(1 + 2x + 4x2)
12dx =
∫u
12du
2=
∫1
2u
12du
=1
2· 2
3u
32 + C
=1
3(1 + 2x + 4x2)
32 + C.
30. With u = sin t, we get du = cos t dt, and hence∫sin6 t cos t dt =
∫u6du =
u7
7+ C
=1
7sin7 t + C.
31. Try u = ez + 1. So du = ezdz and ez = u− 1. We now get∫(ez + 1)
12 e2zdz =
∫(ez + 1)
12 ez · ezdz
=
∫u
12 (u− 1)du =
∫(u
32 − u
12 )du
=2
5u
52 − 2
3u
32 + C
=2
5(ez + 1)
52 − 2
3(ez + 1)
32 + C.
32. Let u = 1 + 4x13 . So du
dx= 4
3x− 2
3 and hence x− 23dx = 3
4du. Also, when x = 1 and x = 8, u = 5
and u = 9, respectively. Therefore,∫ 8
1
x− 23
√1 + 4x
13dx =
∫ 9
5
u12 · 3
4du =
∫ 9
5
3
4u
12du
=
[3
4· 2
3u
32
∣∣∣∣9
5
]=
1
29
32 − 1
25
32
=1
2· 33 − 1
2
(√5)3
=1
2
(27 − 5
√5).
33. Let u = 1 + 2x. So dudx
= 2 and dx = 12du. Also, when x = 0 and 3 respectively, u = 1 and 7.
Therefore, ∫ 3
0
dx3√
(1 + 2x)2=
∫ 7
1
1
2
du
(u2)13
=
∫ 7
1
1
2u− 2
3 du
=
[1
2(3)u
13
∣∣∣∣7
1
]=
3
2u
13
∣∣∣∣7
1
=3
23√
7 − 3
2.
10
34. Let u = t10. So dudt
= 10t9 and (looking ahead) t9dt = du10. For t = 0 and 1, u = 0 and 1
respectively. Therefore, ∫ 1
0
t9 tan(t10)dt =
∫ 1
0
tanu · du10
=1
10
∫ 1
0
tanu du.
It remains to find an antiderivative of tanu. But this was already done in the solution of
Exercise 9 above, ddu
[− ln(cosu)] = tanu. So
∫ 1
0
t9 tan(t10)dt =1
10
[− ln(cosu)
∣∣∣∣1
0
]=
1
10[− ln(cos 1) + ln 1]
≈ 1
10(− ln(0.540) + 0) ≈ 0.616
10≈ 0.062.
35. Let u = lnx. So dudx
= 1x
and du = dxx
. For x = 1 and 3, respectively u = 0 and ln 3. Therefore,∫ 1
3
(lnx)2
xdx = −
∫ 3
1
(lnx)2
xdx
= −∫ ln 3
0
u2 du = −[
1
3u3
∣∣∣∣ln 3
0
]
= −(1
3(ln 3)3 − 0) = − ln 3.
13E. Computations of Areas
36. Because f(x) ≥ 0 over the interval in question, this area is equal to∫ 1
−1
x4 dx =1
5x5
∣∣1−1
=1
5− 1
5(−1) =
2
5.
37. Because f(x) ≥ 0 over [−2,−1], the area is∫ −1
−2
1
x2dx =
∫ −1
−2
x−2 dx = −x−1∣∣−1
−2
= −(−1)−1 −(−(−2)−1
)= 1 − 1
2=
1
2.
38. The area is equal to ∫ 4
1
x12dx =
2
3x
32
∣∣41
=2
34
32 − 2
3
=2
3(23 − 1) =
14
3.
11
39. Refer back to Figure 5.25 and note that the graph of f(x) = x13 lies below the x-axis for
−3 ≤ x ≤ 0 and above the x-axis for 0 ≤ x ≤ 8. Hence
∫ 0
−3
x13 dx = −(area below the x-axis and above the graph from x = −3 to x = 0)
and the area that is to be computed is equal to
−∫ 0
−3
x13 dx +
∫ 8
0
x13 dx = −
[3
4x
43
∣∣0−3
]+
[3
4x
43
∣∣80
]
= −(
0 − 3
4(−3)
43
)+
(3
48
43 − 0
)
=3
4(− 3
√3)4 +
3
424 =
3
4(3
3√
3) + 12
=9
43√
3 + 12.
40. Because cos x ≥ 0 for 0 ≤ x ≤ π4, the same is true for f(x) = secx. So by the first part of
Section 13.3, the area is
∫ π4
0
sec x dx = ln(secx + tanx)∣∣π
4
0
= ln(sec
π
4+ tan
π
4
)− ln (sec 0 + tan 0)
= ln
(2√2
+ 1
)− ln (1 + 0) = ln
(1 +
√2)− 0
≈ 0.881.
41. The area is
∫ 3
0
√x + 1 dx. This integral is easily solved by the method of substitution: u =
x + 1; du = dx; u = 1 and 4, respectively, when x = 0 and 3. So∫ 3
0
√x + 1 dx =
∫ 4
1
u12du =
2
3u
32
∣∣41
=2
3(8 − 1) =
14
3.
Can you think of a reason why this area is the same as that of Exercise 38? [Hint: Refer to
Section 10.8.]
42. Because the graph of f(x) = x(x2+1)2
lies below the x-axis for −1 ≤ x ≤ 0 and above the x-axis
for 0 ≤ x ≤ 2, the area in question is
−∫ 0
−1
x
(x2 + 1)2dx +
∫ 2
0
x
(x2 + 1)2dx.
These integrals are solved by the substitution u = x2 + 1 as follows: dudx
= 2x, so xdx = 12du.
12
1 3
2 5
x = 1
0
Therefore, the sum of the two integrals is equal to
−∫ 1
2
1
2
du
u2+
∫ 5
1
1
2
du
u2=
1
2
∫ 2
1
u−2du +1
2
∫ 5
1
u−2du
=1
2
(−u−1
∣∣21
)+
1
2
(−u−1
∣∣51
)=
1
2
(−1
2+ 1
)+
1
2
(−1
5+ 1
)
=1
4+
2
5=
13
20.
43.
∫ 3
1
1
xdx = lnx
∣∣31
= ln 3 − ln 1 = ln 3 ≈ 1.099.
44. The area is given by
∫ 5
2
1
x− 1dx. This integral is solved by the substitution u = x − 1 as
follows: ∫ 5
2
1
x− 1dx =
∫ 4
1
1
udu = lnu
∣∣41
= ln 4 = 2 ln 2 ≈ 2(0.693) ≈ 1.386.
13
- 2- 4
2 5
A 2
A 1
45. Let f(x) = xx2+4
. Because
f ′(x) =(x2 + 4) − x(2x)
(x2 + 4)2= − x2 − 4
(x2 + 4)2,
observe that f ′(x) ≥ 0 for −2 ≤ x ≤ 2 and f ′(x) ≤ 0 for all other x. So the graph of f(x)
is decreasing over x = −4 ≤ x ≤ −2, increasing over −2 ≤ x ≤ 2, and then decreasing over
2 ≤ x ≤ 5. A rough sketch of the situation is
So the area to be computed is equal to
−∫ 0
−4
x
x2 + 4dx +
∫ 5
0
x
x2 + 4dx.
These integrals can be solved by the substitution u = x2 + 4. Because dudx
= 2x and hence
xdx = 12du, we get that the area is
−∫ 4
20
1
2
du
u+
∫ 29
4
1
2
du
u= −1
2
[lnu
∣∣420
]+
1
2
[lnu
∣∣294
]= −1
2(ln 4 − ln 20) +
1
2(ln 29 − ln 4) =
1
2[ln 29 + ln 20 − 2 ln 4]
=1
2ln
(29)(20)
16=
1
2ln
145
4≈ 1.795.
13F. Integration by Parts
46. Starting with u = x and dv = sinx dx, we get du = dx and v = − cos x. So∫x sinx dx =
∫u dv = uv −
∫v du
= −x cosx−∫
− cosx dx
= −x cosx +
∫cosx dx
= −x cosx + sinx + C.
14
47. ∫ π2
0
(x + x cosx) dx =
∫ π2
0
x dx +
∫ π2
0
x cosx dx
=1
2x2
∣∣π2
0+
∫ π2
0
x cosx dx
=1
8π2 +
∫ π2
0
x cosx dx.
To compute
∫x cosx dx, we let u = x and dv = cosx dx and proceed by integration by parts:
du = dx, v = sinx, and ∫x cosx dx =
∫u dv = uv −
∫v du
= x sinx−∫
sinx dx
= x sinx + cosx + C.
So ∫ π2
0
x cosx dx = (x sinx + cosx)∣∣π
2
0
=(π
2· 1 + 0
)− (0 + 1) =
π
2− 1.
Therefore, ∫ π2
0
(x + x cosx)dx =1
8π2 +
π
2− 1.
48. Proceeding as suggested, we have u = lnx, dv = x dx; du = 1xdx and v = x2
2. So∫
x lnx dx =
∫u dv = uv −
∫v du
=x2
2lnx−
∫x2
2· 1
xdx =
x2
2lnx− 1
2
∫x dx
=x2
2lnx− 1
4x2 + C.
49. The suggestion is to let u = lnx and dv = x2dx. So du = 1xdx, v = x3
3, and we get∫
x2 lnx dx =
∫u dv = uv −
∫v du
=x3
3lnx−
∫x3
3· 1
xdx
=x3
3lnx− 1
3
∫x2 dx
=x3
3lnx− 1
9x3 + C.
15
50. With u = x and dv = e5xdx, we get du = dx and v = 15e5x. So∫
xe5xdx =
∫u dv = uv −
∫v du
= x · 1
5e5x −
∫1
5e5xdx
=x
5e5x − 1
5· 1
5e5x + C
=x
5e5x − 1
25e5x + C.
Therefore, ∫ 1
0
xe5xdx =
(x
5e5x − 1
25e5x
) ∣∣10
=1
5e5 − 1
25e5 −
(0 − 1
25
)
=4
25e5 +
1
25.
51. The integral should be
∫x2e5xdx. Does u = x2 and dv = exdx accomplish anything? Let’s
see. Since du = 2xdx and v = 15e5x, we get∫
x2e5xdx =
∫u dv = uv −
∫v du
= x2 · 1
5e5x −
∫2x · 1
5e5xdx
=x2
5e5x − 1
2
∫xe5xdx.
Notice that the integral has been reduced to the one already solved in Exercise 50. So∫x2e5x dx =
x2
5e5x − 1
2
[x
5e5x − 1
25e5x + C ′
]
=x2
5e5x − x
10e5x +
1
50e5x + C.
52. Given the context, we will try to solve
∫ex sinx dx by integration by parts. Let u = sinx
and dv = exdx. So du = cosx dx and v = ex. Hence∫ex sin x dx = ex sinx−
∫ex cosx dx
as required.
16
53. A quick answer to this question can be obtained by solving the equation derived in Exercise
52 for∫ex cosx dx. But this would not help us to solve Exercise 54. It is therefore more
useful to try to solve∫ex cosx dx by parts. Let u = cosx and dv = exdx. So du = − sinx dx
and v = ex, and ∫ex cosx dx = ex cosx +
∫ex sinx dx.
Derive this formula by solving the integral of Exercise 52 in a different way (but still by parts).
54. By substituting the equation derived in Exercise 52 into that derived in Exercise 53, we get∫ex cosx dx = ex cosx + ex sinx−
∫ex cosx dx.
So 2∫ex cosx dx = ex cosx + ex sin x + C ′ and therefore,∫
ex cosx dx =1
2ex cosx +
1
2ex sinx + C.
Check that similarly, ∫ex sinx dx =
1
2ex sinx− 1
2ex cosx + C.
55. We begin by trying a substitution on
∫ln(x + 1)dx. To avoid confusion (with the v, du
notation of the integration by parts method that will follow shortly) we let z = x + 1. So
dz = dx and ∫ln(x + 1)dx =
∫ln z dz.
Now let u = ln z and dv = dz. So du = 1zdz and v = z, and∫
ln z dz =
∫u dv = uv −
∫v du
= z ln z −∫
z · 1
zdz = z ln z −
∫dz
= z ln z − z + C.
So
∫ln(x + 1)dx = (x + 1) ln(x + 1) − (x + 1) + C. Therefore,
∫ 1
0
ln(x + 1)dx = [(x + 1) ln(x + 1) − (x + 1)]∣∣10
= (2 ln 2 − 2) − (ln 1 − 1)
= 2 ln 2 − 1.
56. Let z = t12 . So dz = 1
2t−
12 dt and
∫cos t
12 dt =
∫(cos z) · 2z dz = 2
∫z cos z dz. Now let u = z
and dv = cos z dz. So du = dz and v = sin z. Therefore,∫z cos z dz =
∫u dv = uv −
∫v du
= z sin z −∫
sin z dz = z sin z + cos z + C ′.
17
It follows that ∫cos t
12 dt = 2 [z sin z + cos z + C ′]
= 2t12 sin t
12 + 2 cos t
12 + C.
13G. Trigonometric Substitution
57. Let x = 12sin θ with 0 ≤ θ ≤ π
2. Note that x = 0 when θ = 0 and x = 1
2when θ = π
2.
Also, dxdθ
= 12cos θ, so dx = 1
2cos θ dθ. With these substitutions, the integral is transformed as
follows:
∫ 12
0
√1 − 4x2 dx =
∫ π2
0
√1 − sin2 θ · 1
2cos θ dθ.
Because 1 − sin2 θ = cos2 θ and cos θ ≥ 0 for 0 ≤ θ ≤ π2,
∫ π2
0
√1 − sin2 θ · 1
2cos θ dθ =
∫ π2
0
1
2cos2 θ dθ =
∫ π2
0
1
4(1 + cos 2θ)dθ
=1
4(θ +
1
2sin 2θ)
∣∣π2
0=
1
4
(π
2+
1
2sin π − 0
)
=π
8.
The formula cos2 θ = 1+cos 2θ2
is an easy consequence of the addition formula for the cosine
(see Exercises 2C) and the equality sin2 θ + cos2 θ = 1. Compute
∫ 12
0
√1 − 4x2 dx
in a totally different way by making use of the circle x2 + y2 = 14
of radius 12.
58. Let x = tan θ with −π4≤ θ ≤ π
4. Note that x = −1 when θ = −π
4and x = 1 when θ = π
4.
Also, dx = sec2 θ dθ. Therefore
∫ 1
−1
1
(x2 + 1)12
dx =
∫ π4
−π4
sec2 θ dθ
(tan2 θ + 1)12
.
Recall that tan2 θ + 1 = sec2 θ. Because cos θ ≥ 0 for −π4≤ θ ≤ π
4, sec θ ≥ 0 for such θ, and
hence ∫ π4
−π4
sec2 θ dθ
(tan2 θ + 1)12
=
∫ π4
−π4
sec θ dθ = ln(sec θ + tan θ)∣∣π
4
−π4
= ln
(2√2
+ 1
)− ln
(2√2− 1
)
18
θa
x
= ln(√
2 + 1)− ln
(√2 − 1
)
= ln
(√2 + 1√2 − 1
)≈ 1.763.
59. Taking x = 3 sec θ, we get x2 − 9 = 9 sec2 θ − 9 = 9(sec2θ − 1) = 9 tan2 θ. Also, dx =
3 sec θ tan θdθ. So∫x2
√x2 − 9
dx =
∫9 sec2 θ
3 tan θ3 sec θ tan θ dθ = 27
∫sec3 θ dθ.
60. Exercise 57 suggests that sin θ might play a role. After experimenting a little, let x = a sin θ.
So a2−x2 = a2(1− sin2 θ) = a2 cos2 θ and dx = a cos θ dθ. Assuming that a ≥ 0 and cos θ ≥ 0,
we get ∫ √a2 − x2
xdx =
∫a cos θ
a sin θ· a cos θ dθ =
∫a cos2 θ
sin θdθ
= a
∫1 − sin2 θ
sin θdθ
= a
∫1
sin θdθ − a
∫sin θ dθ.
= a
∫1
sin θdθ + a cos θ.
The integral
∫1
sin θdθ is solved in a way analogous to the solution of
∫1
cos θdθ =
∫sec θ dθ
in Section 13.3. The function 1sin x
is the cosecant csc x. Try solving
∫ √a2 − x2
xdx completely
in terms of x by starting with the substitution x = a cos θ. Make use of Section 10.5 and the
right triangle
that depicts the equation x = a cos θ. You will be able to confirm that you did it correctly by
checking your answer against the conclusions of Section 10.4.
19
θ
u
1
13H. Surface Area
61. The surface area is equal to
A =
∫ 5
0
2πx3√
1 + (3x2)2 dx = 2π
∫ 5
0
x3(1 + 9x4
) 12 dx.
This integral can be solved by the substitution u = 1 + 9x4. Because du = 36x3dx, we get∫ 5
0
x3(1 + 9x4
) 12 dx =
∫ 5626
1
1
36u
12du =
1
36
[2
3u
32
∣∣56261
]
=1
54
(56261.5 − 1
)≈ 7814.565.
62. The surface area is equal to
A = 2π
∫ 1
0
e−x√
1 + (−e−x)2 dx = 2π
∫ 1
0
e−x√
1 + e−2x dx.
The substitution u = e−x shows promise (after some trial and error). Since du = −e−xdx, we
get
A = 2π
∫ e−1
1
−√
1 + u2 du = 2π
∫ 1
e−1
√1 + u2 du .
At this point (see the discussion that concludes Section 13.3B), the substitution u = tan θ
suggests itself. Since du = sec2 θdθ,∫ √1 + u2 du =
∫ √1 + tan2 θ sec2 θ dθ =
∫sec3 θ dθ.
This integral has already been solved in Section 13.3C. Using this solution, we get∫ √1 + u2 du =
1
2(sec θ)(tan θ) +
1
2ln(sec θ + tan θ) + C.
To get from θ back to u consider the “picture” of the equation tan θ = u = u1. This is
From this right triangle we see that cos θ = 1√1+u2 and hence that sec θ =
√1 + u2. Therefore,∫ √
1 + u2 du =1
2(sec θ)(tan θ) +
1
2ln(sec θ + tan θ) + C
=1
2u√
1 + u2 +1
2ln
(u +
√1 + u2
)+ C.
20
It follows that
A = 2π
[(1
2u√
1 + u2 +1
2ln
(u +
√1 + u2
)) ∣∣1e−1
]
= π[1 ·
√2 + ln
(1 +
√2)− e−1
√1 + e−2 − ln
(e−1 +
√1 + e−2
)]
Check that e−1√
1 + e−2 = 1e2
√e2 + 1 and e−1 +
√1 + e−2 = 1
e
(1 +
√e2 + 1
). Therefore,
A = π
[√
2 + ln(1 +√
2) −√e2 + 1
e2− ln
(1 +
√e2 + 1
e
)]
A = π
[√
2 + ln(1 +√
2) −√e2 + 1
e2− ln
1
e− ln(1 +
√e2 + 1)
]
≈ π[1.414 + 0.881 − 0.392 + 1 − 1.360] ≈ 4.848.
63. Because f ′(x) = 12(4 − x2)
− 12 (−2x) = −x
(4−x2)12,
A =
∫ 32
− 12
2π√
4 − x2
√1 +
x2
4 − x2dx =
∫ 32
− 12
2π√
4 − x2
√4 − x2 + x2
4 − x2dx
=
∫ 32
− 12
2π√
4 dx =
∫ 32
− 12
4πdx = 4πx∣∣ 3
2
− 12
= 4π
(3
2−
(−1
2
))= 8π.
13I. Up to the Gills
64. The equation of the circle (a quarter of which is shown in Figure 13.45) is
x2 + (y − 5.5)2 = 22.
Solving for y, we get (y − 5.5)2 = 4 − x2, hence y − 5.5 = ±√
4 − x2, and therefore, y =
5.5 ±√
4 − x2. Because the arc of Figure 13.45 lies below the line y = 5.5, the − applies. So
the arc is the graph of the function
f(x) = 5.5 −√
4 − x2, 0 ≤ x ≤ 2.
65. Notice that
f ′(x) = −1
2
(4 − x2
)− 12 (−2x) =
x
(4 − x2)12
.
21
Therefore, the surface area is
A =
∫ b
a
2πf(x)
√1 + (f ′(x))2 dx
=
∫ 2
0
2π(5.5 −
(4 − x2
) 12
) √1 +
x2
4 − x2dx
=
∫ 2
0
2π(5.5 −
(4 − x2
) 12
) √4 − x2 + x2
4 − x2dx
=
∫ 2
0
2π(5.5 −
(4 − x2
) 12
) 2
(4 − x2)12
dx
=
∫ 2
0
4π
[5.5
(4 − x2)12
− 1
]dx.
66. The integral of Exercise 65 is equal to
A =
∫ 2
0
[22π√4 − x2
− 4π
]dx = 22π
∫ 2
0
1√4 − x2
dx−∫ 2
0
4πdx.
= 22π
∫ 2
0
1√4 − x2
dx− (4πx)∣∣20
= 22π
∫ 2
0
1√4 − x2
dx− 8π.
The remaining integral is solved by the substitution
x = 2 sin θ
with 0 ≤ θ ≤ π2. Note that x = 0 when θ = 0 and x = 2 when θ = π
2. Because dx = 2 cos θdθ,∫ 2
0
1√4 − x2
dx =
∫ π2
0
1√4 − 4 sin2 θ
2 cos θdθ
= 2
∫ π2
0
cos θdθ√4 cos2 θ
=
∫ π2
0
dθ =π
2.
Therefore,
A = (22π)(π
2
)− 8π = 11π2 − 8π = 83.433 µm2.
13J. Points in the Polar Plane
67. We will make use of the transformation equations
x = r cos θ and y = r sin θ
and elementary facts about the sine and cosine. (See Sections 1.4 and 4.4.)
22
i. x = 3 cos π4
= 3 ·√
22
= 32
√2 and y = 3 sin π
4= 3 ·
√2
2= 3
2
√2. So (x, y) =
(32
√2, 3
2
√2).
ii. x = −2 cos(−π
6
)= −2 cos π
6= −2
√3
2= −
√3 y = −2 sin
(−π
6
)= (−2)
(− sin π
6
)=
(−2)(−1
2
)= 1. So (x, y) =
(−√
3, 1).
iii. x = 3 cos 7π3
= 3 cos(2π + π
3
)= 3 cos π
3= 3 · 1
2= 3
2and y = 3 sin 7π
3= 3 sin
(2π + π
3
)=
3 sin π3
= 3 ·√
32
= 32
√3. So (x, y) =
(32, 3
2
√3).
iv. x = 5 cos 0 = 5 and y = 5 sin 0 = 0. So (x, y) = (5, 0).
v. x = −2 cos π2
= (−2)0 = 0 and y = −2 sin π2
= (−2)1 = −2. So (x, y) = (0,−2).
vi. x = −2 cos 3π2
= −2 cos(
π2
+ π)
= (−2)(− cos π
2
)= 0 and y = −2 sin 3π
2= −2 sin
(π2
+ π)
=
2 sin π2
= 2. So (x, y) = (0, 2).
vii. Since this problem is virtually identical to (vi), we do(4,−5π
4
)instead.
x = 4 cos(−5π
4
)= 4 cos 5π
4= 4 cos
(π + π
4
)= −4 cos π
4= −2
√2 and y = 4 sin
(−5π
4
)=
−4 sin 5π4
= −4 sin(π + π
4
)= 4 sin π
4= 2
√2. So (x, y) =
(−2
√2, 2
√2)
viii. x = 0 cos 6π7
= 0 and y = 0 sin 6π7
= 0. So (x, y) = (0, 0).
ix. x = − cos(−23π
3
)= − cos
(23π3
)= − cos
(8π − π
3
)= − cos
(−π
3
)= − cos π
3= −1
2and
y = − sin(−23π
3
)= sin 23π
3= sin
(8π − π
3
)= sin
(−π
3
)= − sin π
3= −
√3
2. So (x, y) =(
−12,−
√3
2
).
x. In view of (ix) we will change this to(5,−15
4π).
x = 5 cos(−15
4π)
= 5 cos(4π − 15
4π)
= 5 cos π4
= 5√
22
= 52
√2 and y = 5 sin
(−154π)
=
5 sin(4π − 15
4π)
= 5 sin π4
= 52
√2. So (x, y) =
(52
√2, 5
2
√2).
xi. In view of (vi), do(1, 3π
2
)to
(−3, 14
6π)
instead.
x = −3 cos(
146π)
= −3 cos(2π + 2
6π)
= −3 cos π3
= −32
and y = −3 sin(
146π)
=
−3 sin(2π + 2
6π)
= −3 sin π3
= −3 ·√
32. So (x, y) =
(−3
2,−3
2
√3).
xii. x = 3 cos(−5π
6
)= 3 cos
(π − π
6
)= −3 cos
(−π
6
)= −3 cos π
6= −3 ·
√3
2= −3
2
√3 and
y = 3 sin(−5π
6
)= 3 sin
(π − π
6
)= −3 sin
(−π
6
)= 3 sin π
6= 3
2. So (x, y) =
(−3
2
√3, 3
2
).
68. Suppose a point P (other than the origin 0) has polar coordinates (r, θ). Then
(r, θ + 2π), (r, θ − 2π), (r, θ + 4π), (r, θ − 4π),
and more generally, (r, θ + 2kπ), where k can be any integer (positive or negative), are polar
coordinates of P . Observe that any set of polar coordinates of P with first coordinate r has
the form (r, θ + 2kπ). Note that (−r, θ + π) also represents P and hence that any set polar
coordinates of P with first coordinate −r has the form (−r, θ+π+2kπ) = (−r, θ+(2k+1)π).
It follows that if a single set of polar coordinates can be determined for P , then all others are
given by the “recipe” above. So in the problems that follow it suffices to specify a single set
of polar coordinates.
i. The ray θ = π4
goes through the Cartesian point (3,3). Taking r =√x2 + y2 =
√9 + 9 =√
18 = 3√
3 provides the rest. So the polar coordinates are(3√
3, π4
).
23
ii. The ray θ = −π4
goes through the Cartesian point (4,−4). Because r =√
42 + (−4)2 =√32 = 4
√2, the coordinates are
(4√
2,−π4
).
iii. Since the ray θ = π2
goes through (0, 5), the polar coordinates are(5, π
2
).
iv. The ray θ = π goes through (−4, 0) so that the polar coordinates are (4, π).
Up to now we were able to determine a set of polar coordinates simply by “inspection”. From
now on we will be more systematic and use the transformation equations
tan θ =y
xand r = ±
√x2 + y2.
v. Because x = 3 and y = 3√
3, yx
= 3√
33
=√
3. By Table 1.2 of Section 1.4 and Figure 26 of
Section 4.4, θ = π3
is the only possible θ with −π2≤ θ ≤ π
2. Note that r = ±
√9 + 9 · 3 =
±√
36 = ±6. Because the point (3, 3√
3) is in the first quadrant,(6, π
3
)is one answer.
vi. Note that tan θ = yx
=
√3
3
− 13
= −3√
33
= −√
3. So tan(−θ) =√
3. Hence −θ = π3, or
θ = −π3
, is the only possible θ with −π2≤ θ ≤ π
2(refer to Table 1.2 of Section 1.4 and
Figure 26 of Section 4.4). Because r = ±√
19
+ 39
= ±√
49
= ±23
and the point(−1
3,√
33
)is in the second quadrant, it follows that
(−2
3,−π
3
)are polar coordinates of the point.
vii. Because tan θ =√
3−3
= −√
3√3√
3= − 1√
3, tan(−θ) = 1√
3. Again by Table 1.2 of Section 1.4
and Figure 26 of Section 4.4, −θ = π6, or θ = −π
6, is the only possible θ with −π
2≤ θ ≤ π
2.
Note that r = ±√
9 + 3 = ±√
12 = ±2√
3. Since (−3,√
3) is in the second quadrant, it
follows that(−2
√3,−π
6
)are polar coordinates of the point.
Correction: Change the point in vii. to (3,−√
3). This puts it into the fourth quadrant.
viii. As in the earlier problem, tan θ = 2−2
√3
= − 1√3
implies that θ = −π6
is the only possible
θ with −π2≤ θ ≤ π
2. Since r = ±
√4 · 3 + 4 = ±
√16 = ±4, and (−2
√3, 2) is in the
second quadrant,(−4,−π
6
)are polar coordinates of the point.
ix. To represent the origin 0 in polar coordinates we need r = 0. But θ can be arbitrary. So
(0, θ) with any θ is a set of polar coordinates for 0.
x. Because tan θ = yx
= −5−5
√3
= 1√3, we get (as before) that θ = π
6is the only possible
θ with −π2≤ θ ≤ π
2. Note that r = ±
√25 · 3 + 25 = ±
√100 = ±10. Because the
point (−5√
3,−5) is in the third quadrant,(−10, π
6
)is a representation of it in polar
coordinates.
13K. Equations in Polar Coordinates
Exercises 69-75 are applications of the transformation equations
x = r cos θ, y = r sin θ, r = ±√x2 + y2, and tan θ =
y
x.
69. The equation 2x + 3y = 4 transforms to
2r cos θ + 3r sin θ = 4 or r(2 cos θ + 3 sin θ) = 4,
24
and hence (since 2 cos θ + 3 sin θ cannot be zero) to
r =4
2 cos θ + 3 sin θ.
70. This equation transforms to
i. 9r2 cos2 θ + r2 sin2 θ = 4r sin θ. Assume (for the moment that r = 0). So
9r cos2 θ + r sin2 θ = 4 sin θ or r(9 cos2 θ + sin2 θ) = 4 sin θ.
Now 9 cos2 θ + sin2 θ cannot be zero because if it were, then sin θ = 0, hence 9 cos2 θ
= 0 and thus cos θ = 0. But sin θ and cos θ cannot both be zero. Why not? So
ii. r = 4 sin θ9 cos2 θ+sin2 θ
. Note that equations (i) and (ii) are not algebraically the same. For
instance(0, π
2
)satisfies (i) but not (ii). But for any (r, θ) with r = 0, they imply the
same relationship between r and θ. Also, the graphs of the two equations are the same
because the origin lies on both.
71. This equation transforms to
i. r2 = r cos θ(r2 cos2 θ − 3r2 sin2 θ). If r = 0, then
1 = r cos θ(cos2θ − 3 sin2 θ),
and hence, because cos θ(cos2 θ − 3 sin2 θ) cannot be zero,
ii. r = 1cos θ(cos2 θ−3 sin2 θ)
. Note that this time the graphs of (i) and (ii) are the same, except
at the origin 0 which is on the graph of (i) but not on the graph of (ii).
72. The equation r = 5 transforms to ±√x2 + y2 = 5, or x2 + y2 = 25.
73. Notice that the origin 0 =(0, π
2
)is on the graph of r = 3 cos θ. It follows that r2 = 3r cos θ
has the same graph. This last equation transforms to x2 + y2 = 3x.
74. tan θ = 6 becomes yx
= 6.
75. Because the origin 0 = (0, 0) is on the graph of r = 2 sin θ tan θ, the graph of this equation is
the same as that of r2 = 2r sin θ tan θ. This transforms to x2 + y2 = 2y · yx. To include the
origin on the graph, we rewrite this as x3 + xy2 = 2y2.
76. The graph is the set of all (r, θ) with r = 6. This is the circle with center 0 and radius 6.
77. The graph is the set of all points of the form(r,−8π
6
). Consider the ray θ = −8π
6. Since r can
be any number (positive, negative, or zero) the graph is the (straight) line through the origin
that this ray determines. Because tan(−8π
6
)= − tan
(4π3
)= − tan
(π + π
3
)= − tan π
3= −
√3,
the Cartesian equation of this line is y = −√
3x.
78. The graph of r = sin θ is the same as that of r2 = r sin θ. Its Cartesian equivalent is x2 + y2 =
y. This equation can be analyzed by completing the square on x2 + y2 − y = 0. Because,
25
x2 + y2 − y + (12)2 = (1
2)2, we get
x2 +
(y − 1
2
)2
=
(1
2
)2
.
It follows that the graph of r = sin θ is the circle with center the Cartesian point(0, 1
2
)and
radius 12.
79. This equation transforms to y + x = 1, or y = −x + 1. This is the line with slope −1 and
y-intercept 1.
80. The graph of r = 2 cos 2θ is virtually identical to that of r = cos 2θ which is studied in
Example 13.23. The only difference is that the petals of the “rose”, see Figure 13.35, have
length 1 in the example and length 2 in this exercise.
-2 -1 1 2
-2
-1
1
2
81. Lets start with the graph of r = −4 sin 3θ in Cartesian coordinates. Since
sin 3θ = sin(3θ + 2π) = sin 3
(θ +
2
3π
)
this graph has period 23π. So it is the graph of sin θ (which has period 2π) compressed by
-6 -4 -2 2 4 6Q
-1
-0.5
0.5
1
r
26
a factor of 3. The Cartesian graph of r = sin 3θ is sketched above. It follows quickly from
this that the Cartesian graph of r = −4 sin 3θ is as drawn below.
-6 -4 -2 2 4 6Q
-4
-2
2
4
r
Now let sketch r = −4 sin 3θ in “polar”. As the ray θ rotates from 0 to π6, r slides from 0 to
−4 and as θ goes from π6
to π3, r goes from −4 back to 0. The loop on the lower left of the
figure below is traced out in the process. Similarly, as the ray θ rotates from π3
to 2π3, r slides
from 0 to 4 (at θ = π2) back to 0. So the upper loop is traced out. Next, as the ray θ rotates
from 2π3
to π, r slides from 0 to −4 (at θ = 5π6
) back to 0. So the loopon the lower right is
-3 -2 -1 1 2 3
-2
-1
1
2
3
4
traced out. The rest of the graph is just a repetition of what we already have. For example,
as θ moves from 0 to −π3, the, loop on the lower right is traced out. As θ moves from π to 4π
3,
it is the loop on the lower left.
82. The considerations of Example 13.23 show that the Cartesian graph of sin 2θ is as sketched
below
27
-6 -4 -2 2 4 6Q
-1
-0.5
0.5
1
r
So the Cartesian graph of 9 sin 2θ is
-6 -4 -2 2 4 6Q
-7.5
-5
-2.5
2.5
5
7.5
r
Now to the polar graph of r2 = 9 sin 2θ. Observe that 9 sin 2θ ≥ 0 so θ must fall into[0, π
2
],[π, 3π
2
],[2π, 5π
2
], . . . or
[−π
2,−π
],[−3π
2,−5π
2
], . . . . As the ray θ rotates from 0 to
π2, r = +3
√sin 2θ slides from 0 to 3 (at θ = π
4) and back to 0. In the process, the loop on the
upper right in the graph below is traced out. But r = −3√
sin 2θ is also possible. This time
28
-2 -1 1 2
-2
-1
1
2
the loop on the lower left is traced out as θ varies from 0 to π2. Similar considerations
show that as θ varies from π to 3π2, r = +3
√sin 2θ traces out the loop on the lower left and
r = −3√
sin 2θ traces out the loop on the lower left. As θ varies from −π2
to −π, r = +3√
sin 2θ
and r = −3√
sin 2θ traces these loops again. Repeating these considerations shows us that
the graph of r2 = 9 sin 2θ is complete as sketched.
83. The Cartesian graph of r = 2 − cos θ is shown below
-7.5 -5 -2.5 2.5 5 7.5Q
0.5
1
1.5
2
2.5
3r
Turning to the polar graph of r = 2− cos θ, we see that r: slides from 1 to 2 as θ moves from
0 to π2; slides from 2 to 3 as θ moves from π
2to π; slides from 3 to 2 as θ moves from π to 3π
2;
and slides from 2 to 1 as θ moves from 3π2
to 2π. This four step process traces out the loop
sketched in the graph below.
29
x = d
F = (a, 0)
—L4
2- —L
2—L
D
-3 -2 -1 1
-2
-1
1
2
Because the Cartesian graph r = 2 − cos θ is symmetric about the vertical axis, the pattern
is exactly the same as θ moves from 0 to −3π2. The periodicity of the cosine tells us that the
polar graph of r = 2− cos θ is complete as shown. (Incidentally, it is not transparent why this
graph is a “snail”.)
13L. Parabolas and Hyperbolas in Polar Coordinates
84. Equation (∗) becomes 4Lx+4y2 = L2. Dividing through by 4L we get x+ 1Ly2 = L
4and hence
x = − 1Ly2 + L
4. By remarks in Section 4.3, this is a parabola which opens up to the left. The
x-intercept is L4. The y-intercepts are gotten by solving 1
Ly2 = L
4for y; so they are y = ±L
2.
Thus the graph of x = − 1Ly2 + L
4looks as shown below.
Because V =(
L4, 0
)is on the parabola we know that the distance from V to F is equal to the
distance from V to D. So L4− a = d− L
4, and hence L
2= d + a. Because P = (a, b) is on the
parabola, the distance from P to F is equal to the distance from P to D. So b = d−a = L2−2a.
30
Since (a, b) is on the parabola, we see that
a = − 1
L
(L
2− 2a
)2
+L
4.
Therefore, L4− a = 1
L
(L2− 2a
)2and hence, L2
4− aL = L2
4− 2aL + 4a2. It follows that
aL = 4a2. If a = 0, then L = 4a. So a = L4
and hence V = F. But this cannot be so. Why?
Hence a = 0. It follows that the latus rectum is L2
+ L2
= L. Observe that (3) holds because
the axis of the parabola is defined to be the line through F perpendicular to D.
85. Left to the reader.
13M. Areas of Regions Given in Polar Coordinates
86. This is a circle of radius 4, so the area is 16π. Does the area formula
A =
∫ b
a
1
2f(θ)2dθ
provide the same result? Taking r = f(θ) = 4, a = 0, and b = 2π, we get
A =
∫ 2π
0
1
2· 16dθ = 8θ
∣∣2π
0= 16π.
87. The region is a loop above the polar axis traced out by the point
(r, θ) = (3 sin θ, θ)
as θ rotates from 0 to π. (The rotation of θ from π to 2π, or 0 to −π, etc. simply retraces
this loop.) So the area of the region enclosed by the loop is
A =
∫ π
0
1
2(3 sin θ)2dθ =
9
2
∫ π
0
sin2 θdθ.
To evaluate this integral, recall that cos2 θ = 1+cos 2θ2
. So
sin2 θ = 1 − cos2 θ = 1 − 1 + cos 2θ
2=
1 − cos 2θ
2.
It follows that
A =9
2
∫ π
0
1 − cos 2θ
2dθ =
9
4
∫ π
0
(1 − cos 2θ)dθ
=9
4
[(θ − 1
2sin 2θ
) ∣∣π0
]=
9
4π.
By proceeding as in the solution of Exercise 78, one can show that the graph of r = 3 sin θ is
the circle with equation x2 +(y − 3
2
)2=
(32
)2of radius 3
2. So the area is π
(32
)2= 9
4π.
31
θ = —
θ = 0
π3
88. The region in question
has area A =
∫ π3
0
1
2(3 sin θ)2dθ. Using facts from Exercise 87 (and using Examples 4.11 and
4.12), we find that the value of this definite integral is
A =9
4
[(θ − 1
2sin 2θ
) ∣∣π3
0
]=
9
4
(π
3− 1
2sin
2π
3
)
=9
4
(π
3+
1
2sin
(−2π
3
))=
9
4
(π
3− 1
2sin
π
3
)
=9
4
(π
3− 1
2·√
3
2
)=
3
4π − 9
√3
16.
89. The graph of r = 2 + 2 cos θ = 2(1 + cos θ) is in essence the same as that discussed in
Example 13.22. (What is the difference?) So the entire graph is sketched out by letting the
ray determined by θ rotate from θ = 0 to θ = 2π. Therefore the area A of the cardioid is equal
to
A =
∫ 2π
0
1
2f(θ)2dθ =
∫ 2π
0
2(1 + 2 cos θ + cos2 θ)dθ
=
∫ 2π
0
(2 + 4 cos θ + 1 + cos 2θ)dθ
=
(3θ + 4 sin θ +
1
2sin 2θ
) ∣∣2π
0= 6π.
13N. Differential Equations
90. For y = 13e3x, we get dy
dx= 1
3e3x · 3 = e3x. So y = 1
3e3x does satisfy dy
dx= e3x.
91. With y = tanx + secx, dydx
= sec2 x + secx tanx. So
2dy
dx− y2 = 2(sec2 x + secx tanx) − (tan2 x + 2 secx tanx + sec2 x)
= sec2 x− tan2 x = 1.
32
Why the restriction 0 < x < π2? Because cos π
2= 0 means that secx and hence dy
dxis not
defined for x = π2. Note that the less restrictive condition −π
2< x < π
2would have been
sufficient.
92. With y = sin 2x − cos 2x, we get dydx
= 2 cos 2x + 2 sin 2x, and d2ydx2 = −4 sin 2x + 4 cos 2x. So
d2ydx2 + 4y = 0.
93. For y = xe−2x, we get
dy
dx= 1 · e−2x + x(−2e−2x) = e−2x − 2xe−2x, and
d2y
dx2= −2e−2x − (2e−2x + 2x(−2e−2x)) = −4e−2x + 4xe−2x .
Therefore, d2ydx2 + 4 dy
dx+ 4y = (−4e−2x + 4xe−2x) + (4e−2x − 8xe−2x) + 4xe−2x = 0.
13O. The Method of Separation of Variables
94. After separating variables, we get ∫ydy =
∫xdx,
and therefore, y2
2= x2
2+C. So y2 = x2 +2C and hence (after changing C) the general solution
is y = ±√x2 + C. The requirement y(0) = 4, forces 4 = ±
√C. So C = 16 and the − option
is impossible. So the particular solution is
y = (x2 + 16)12 .
95. After separating variables, ∫dy
y=
∫xdx.
Hence ln y = x2
2+ C, and therefore the general solution is
y = eln y = ex2
2+C .
Because y(1) = 3, we get 3 = e12+C . So ln 3 = 1
2+C and C = ln 3− 1
2. The particular solution
is y = ex2
2+ln 3− 1
2 .
96. By separating variables, ∫dy
y=
∫dx
x
and hence ln y = lnx + C. So y = eln y = eln x+C = eln x · eC = Ax, where A = eC . Therefore
the general solution is
33
y = Ax with A > 0.
Note that the initial condition y(2) = 0, cannot be met because y = 0 forces x = 0. So there
is no particular solution that satisfies the given initial condition.
97. Correction: It should be (y − 3)dydt
= 1 instead of (y2 − 3)dydt
= 1. By separating variables,∫(y − 3)dy =
∫dt.
So y2
2− 3y = t + C and (changing C’s),
y2 − 6y − (2t + C) = 0.
By the quadratic formula,
y =6 ±
√36 + 4(1)(2t + C)
2=
6 ± 2√
9 + 2t + C
2.
So y = 3 ±√
2t + 9 + C. To get y = 3 when t = 0, we need 9 + C = 0 or C = −9. So there
are two particular solutions:
y = 3 +√
2t and y = 3 −√
2t.
98. By separating variables, ∫(1 + y)dy =
∫(sinx− cosx)dx
and hence
y +y2
2= − cosx− sin x + C.
So y2 + 2y + (2 sinx + 2 cosx− C) = 0. By the quadratic formula, the general solution is
y =−2 ±
√4 − 4(2 sinx + 2 cosx− C)
2
y = −1 ±√
1 − 2 sinx− 2 cosx + C.
To get a particular solution, we need
0 = −1 ±√
1 − 2 sinπ − 2 cosπ + C
1 = ±√
1 − 0 − 2(−1) + C.
So the + option prevails, and C = −2. Hence the particular solution is
y = −1 +√−1 − sinx− cosx.
Notice that sinx+cosx needs to be negative or 0 for this to make sense. Compare the graphs
in Figures 10.28 and 10.29 and determine an interval between π2
and 2π for which this is the
case.
34
99. After separating variables,
∫(ln y)2
ydy =
∫x2dx
To solve the integral on the left, let u = ln y. So dudy
= 1y
and hence∫(ln y)2
ydy =
∫u2du =
u3
3+ C1
= (ln y)3 + C1.
Therefore (ln y)3+C1 = x3
3+C2 and hence (ln y)3 = x3
3+C. So ln y =
(x3
3+ C
) 13
and y = eln y =
e(x3
3+C)
13 . For the particular solution, we need 1 = e( 8
3+C)
13 and hence that
(83
+ C) 1
3 = 0. So
C = −83. So the particular solution is
y = e
(x3
3− 8
3
) 13
.
13P. Chemical Reactions
100. i. Turning to the case a = b first, we get∫dy
(a− y)2=
∫kdt .
For the integral on the left, we turn to the substitution u = a− y. So dy = −du and∫dy
(a− y)2= −
∫u−2du = −u−1
−1+ C1
= (a− y)−1 + C1.
Therefore,
(a− y)−1 + C1 = kt + C2
and hence (a−y)−1 = kt+C. So a−y = 1kt+C
and hence y = a− 1kt+C
. Refer back to the
narrative of the problem, and notice that y(0) = 0. Therefore, 1c
= a and C = 1a. Hence
y = a− 1
kt + 1a
= a− 1akt+1
a
= a− a
akt + 1=
a(akt + 1) − a
akt + 1.
So y(t) = a2ktakt+1
.
Observe that limt→∞
y(t) = a. By the quotient rule,
dy
dt=
a2k(akt + 1) − (a2kt)(ak)
(akt + 1)2
35
y = a
t
y
=a2k
(akt + 1)2= a2k(akt + 1)−2, and
d2y
dt2= −2a2k(akt + 1)−3(ak)
=−2a3k2
(akt + 1)3.
We now have the following information about the graph of y(t): it is increasing, concave
down, and has y = a as horizontal asymptote. So the graph has the shape
Note that the number of molecules produced at time t is always less that a. Suppose that
this were a commercial production run and that the manufacturer of the compound is
satisfied with a production of 0.99a molecules. How long should the reaction be allowed
to run? This is simple: put a2ktakt+1
= 0.99a = 99100
a and solve for t:
100a2kt = 99a(akt + 1), so 100a2kt− 99a2kt = 99a, hence a2kt = 99a and t =99
ak.
So if a and k are both large, then the run will end relatively soon.
ii. Now the case a = b. As before,∫dy
(a− y)(b− y)=
∫kdt.
By “reversing” common denominators, we get
1
(a− y)(b− y)=
A
a− y+
B
b− y
=A(b− y) + B(a− y)
(a− y)(b− y)
=−(A + B)y + Ab + Ba
(a− y)(b− y).
So A+B = 0 and Ab+Ba = 1. Since B = −A, we get Ab−Aa = 1 and hence A = 1b−a
and B = −1b−a
. (Notice that a = b is needed here.) Therefore,
1
(a− y)(b− y)=
1b−a
a− y+
− 1b−a
b− y=
(1
b− a
)1
a− y−
(1
b− a
)1
b− y.
36
So ∫dy
(a− y)(b− y)=
1
b− a
∫dy
a− y− 1
b− a
∫dy
b− y
=1
b− a[− ln(a− y) + ln(b− y) + C1]
=1
b− a
[ln
(b− y
a− y
)+ C1
].
[Refer to the narrative of the problem and notice that a−y > 0 and b−y > 0 throughout
the reaction. Why is this relevant in the computations above?] Because∫kdt = kt + C2,
we now get
1
b− a
[ln
(b− y
a− y
)+ C1
]= kt + C2
and, after combining constants, that
ln
(b− y
a− y
)= (b− a)kt + C.
Therefore, b−ya−y
= e(b−a)kt+C = De(b−a)kt, where D = eC . Because y = 0 when t = 0 (see
the narrative of the problem), we see that D = ba. So
b− y
a− y=
b
ae(b−a)kt.
Finally, solve for y. After multiplying through and simplifying,
b− y = (a− y)b
ae(b−a)kt
y − yb
ae(b−a)kt = b− be(b−a)kt
y
(1 − b
ae(b−a)kt
)= b− be(b−a)kt
y =b(1 − e(b−a)kt)
1 − bae(b−a)kt
This is the expression we needed to find.
We will now assume that a > b. This is no restriction, for as a = b, we could have
arranged this at the beginning of the narrative of the problem. What about limt→∞
y(t) = ?
Since one molecule of the compound is formed by a combination of one molecule of each
37
of the reacting chemicals, it is clear that no more than b molecules of the compound can
be formed. Because a > b, limt→∞
e(b−a)kt = 0, and it follows that
limt→∞
y(t) = limt→∞
b(1 − e(b−a)kt)
1 − bae(b−a)kt)
= b.
Because dydt
= k(a− y)(b− y) > 0, y is an increasing function of t. What about the
matter of concavity? The second derivative d2ydt2
is best computed with the chain rule
as follows: Let r (for rate) be the function dydt
of t. By the chain rule,
d2y
dt2=
dr
dt=
dr
dy· dydt
.
Because r = k(ab− (a + b)y + y2), we get drdy
= k(2y − (a + b)). So
dr
dt= −k((a− y) + (b− y))[k(a− y)(b− y)]
= −k2((a− y) + (b− y))(a− y)(b− y).
Because a − y > 0 and b − y > 0, it follows that drdt
= d2ydt2
is negative throughout, and
hence that the graph of y(t) is concave down. So the graph of y(t) has the same shape
as that sketched earlier for the case a = b.
How long will it take for 0.99b molecules of the compound to be formed? Set
y(t) =b(1 − e(b−a)kt)
1 − bae(b−a)kt
=99
100b
and solve for t. Doing this, we get
100(1 − e(b−a)kt) = 99(1 − b
ae(b−a)kt)
−100e(b−a)kt = −1 − 99b
ae(b−a)kt .
Therefore,(100 − 99
b
a
)e(b−a)kt = 1, and hence
(100a− 99b
a
)e(b−a)kt = 1.
So e(b−a)kt =a
100a− 99band (b− a)kt = ln
(a
100a− 99b
). Now solving for t we get
t =1
(b− a)kln
(a
100a− 99b
)=
1
−(a− b)kln
(a
100a− 99b
)
=1
(a− b)kln
(a
100a− 99b
)−1
=1
(a− b)kln
(100a− 99b
a
)
=1
(a− b)kln
(100 − 99b
a
).
This will happen quickly if a ≈ b and if k is large.
38
13Q. First-Order Linear Differential Equations
101. The equation y′ +y = t fits into the scheme of equation (2) with p(t) = 1 and q(t) = t. Taking
P (t) = t, we need to solve ∫tetdt.
This integral succumbs to integration by parts after taking u = t and dv = etdt. Since du = dt
and v = et, we get that ∫tetdt =
∫udv = uv −
∫vdu
= tet −∫
etdt = tet − et + C.
Therefore, the general solution that we are looking for is
y(t) = e−t[tet − et + C
]= t− 1 + Ce−t.
Because y′(t) = 1 − Ce−t,
y′(t) + y(t) = t
as required. To find the particular solution f(t) = t − 1 + Ce−t with f(0) = 2 we need
2 = 0 − 1 + C and hence C = 3. So f(t) = t− 1 + 3e−t is the required particular solution.
102. The scheme applies to this equation with p(t) = 2t and q(t) = 2t. Taking P (t) = t2, we get
y(t) = e−t2[∫
2tet2dt + C
]
= e−t2[et2 + C
]= 1 + Ce−t2 .
To get the particular solution, we solve
0 = y(0) = 1 + Ce◦ = 1 + C
for C, to get C = −1. So y(t) = 1 − e−t2 .
103. Take p(t) = 1t
and q(t) = cos t. With P (t) = ln t, we get
y(t) = e− ln t
[∫cos t eln tdt + C
]
= t−1
[∫t cos t dt + C
].
An integration by parts handles∫t cos t dt. Let u = t and dv = cos t dt. So du = dt, v = sin t,
and hence ∫t cos tdt =
∫udv = uv −
∫vdu
= t sin t−∫
sin t dt
= t sin t + cos t + C.
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So
y(t) = t−1[t sin t + cos t + C] = sin t + t−1 cos t + Ct−1.
To get y(π) = 0, we need 0 = 0 + π−1(−1) + Cπ−1; hence Cπ−1 = π−1 and C = 1. So the
particular solution is y(t) = sin t + t−1 cos t + t−1.
104. i. We rewrite the equation as
dy
dx− x−1 + x−2 = 0.
So dydx
= x−1−x−2, and y = lnx+x−1+C. To get y(1) = 3, we need 3 = ln 1+1+C = 1+C.
So C = 2 and the answer is
y(x) = lnx + x−1 + 2.
ii. This time,
dy
dx− x−1y = −x−2.
This fits into the scheme discussed earlier with p(x) = −x−1 and q(x) = −x−2. Taking
P (x) = − lnx, we get
y(x) = eln x
[∫−x−2e− ln xdx + C
]= x
[∫−x−2eln x−1
dx + C
]
= x
[∫−x−3dx + C
]= x
[1
2x−2 + C
]
=1
2x−1 + Cx.
To get y(1) = 3, we solve 3 = y(1) = 12
+ C for C. So C = 52
and y(x) = 12x−1 + 5
2x is
the particular solution.
iii. This time,
dy
dx+ x−2y = −x−2
and we are “in the scheme” by taking p(x) = x−2 and q(x) = −x−2. Take P (x) = −x−1
to get
y(x) = ex−1
[∫−x−2e−x−1
dx + C
]= e−x−1
[−e−x−1
+ C]
= −e−2x−1
+ Ce−x−1
.
To get the particular solution, we need 3 = −e−2 + Ce−1; hence Ce−1 = 3 + e−2 and
C = 3e + e−1. Therefore,
y(x) = −e−2x−1
+ (3e + e−1)e−x−1
.
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13R. Application to Radioactive Decay
105. Notice that there are x0 = 0.9928 grams of 23892U in the sample at time t = 0. Since the
half-life of 23892U is λ1 = 4.5 × 109 years, we convert that of 234
90Th into years as well. This is
λ2 = 24.1365.25
= 0.066 = 6.6 × 10−2. Since we are dealing with a sample of pure uranium at
time t = 0, the amount of 23490Th at t = 0 is y0 = 0. Therefore by the formula, the amount (in
grams) of 23490Th at any time t ≥ 0, (in years) is
y(t) ≈ (0.9928)(4.5 × 109)
(−4.5 × 109)(0 − e−0.066t)
≈ 0.9928e−0.066t.
With t = 109 this is approximately
0.9928e−66×106 ≈ 0.
So the amount is negligible. Think of it this way: Given the long half-life of 23892U, the daughter
23490Th is produced very slowly; given the relatively short half-life of 234
90Th, the small amounts
of thorium produced vanish rapidly.
106. This problem is virtually identical to the previous problem and will be omitted.
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