Title Solucionario - Zill 3.7 MB 340
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Page 1

298 CHAPTER 5. INTEGRALS

Given f(x) =

x4 + 1 = (x4 + 1)1/2, we have f ′(x) = 2x3(x4 + 1)−1/2. Thus,

f ′′(4x) dx =

1

4
f ′(4x) + C =

1

4
{2(4x)3[(4x)4 + 1]−1/2}+ C = 32x

3

256x4 + 1

+ C.

To check this, take the derivative of the above function, yielding
96x2√

256x4 + 1
− 16384x

6√
(256x4 + 1)3

,

which should be the same as f ′′(4x). Since f ′′(x) =
6x2√
x4 + 1

− 4x
6√

(x4 + 1)3
, we have f ′′(4x) =

6(4x)2√
(4x)4 + 1

− 4(4x)
6√

[(4x)4 + 1]3
=

96x2√
256x4 + 1

− 16384x
6√

(256x4 + 1)3
.

76. First evaluating

sec2 3x dx, we get

sec2 3x dx =

1

3

(sec2 3x)(3 dx) u = 3x, du = 3 dx

=
1

3

sec2 u du =

1

3
tanu+ C =

1

3
tan 3x+ C

Next, evaluating

∫ (∫
sec2 3x dx

)
dx =

∫ (
1

3
tan 3x+ C

)
dx, we get

∫ (
1

3
tan 3x+ C

)
dx = (Cx+ C1) +

1

3

tan 3x dx

= (Cx+ C1) +
1

9

(tan 3x)(3 dx) u = 3x, du = 3 dx

= (Cx+ C1) +
1

9

tanu du = (Cx+ C1)−

1

9
ln | cosu|+ C2

= Cx− 1
9
ln | cos 3x|+ C3.

5.3 The Area Problem

1. 3 + 6 + 9 + 12 + 15

2. −1 + 1 + 3 + 5 + 7
3. 2 + 2 + 8/3 + 4

4.
3

10
+

9

100
+

27

1000
+

81

10, 000

5. −1
7
+

1

9
− 1

11
+

1

13
− 1

15
+

1

17
− 1

19
+

1

21
− 1

23
+

1

25

6. 1− 1
4
+

1

9
− 1

16
+

1

25
− 1

36
+

1

49
− 1

64
+

1

81
− 1

100

PROBLEMAS 1.1

CÁLCULO INTEGRAL

MATEMÁTICAS 2

MANUAL DE SOLUCIONES

TEOREMA FUNDAMENTAL DEL CÁLCULO

Page 2

5.3. THE AREA PROBLEM 299

7. 0 + 3 + 8 + 15

8. 1 + 4 + 9 + 16 + 25

9. −1 + 1− 1 + 1− 1

10. 1 + 0− 1
3
+ 0 +

1

5

11.

7∑
k=1

(2k + 1)

12.

6∑
k=1

2k

13.

13∑
k=1

(3k − 2)

14.

10∑
k=1

(4k − 2)

15.

5∑
k=1

(−1)k+1
k

16.

5∑
k=1

(−1)kk
k + 1

17.

8∑
k=1

6

18.

9∑
k=1

k

19.

4∑
k=1

(−1)k+1
k2

cos

p
x

20.

5∑
k=1

(−1)k+1f (k)(1)
2k − 1 (x− 1)

k

21.

20∑
k=1

2k = 2

20∑
k=1

k = 2

(
20 · 21

2

)
= 420

22.

50∑
k=0

(−3k) = −3
50∑
k=1

k = −3
(
50 · 51

2

)
= −3825

Page 170

6.2. AREA REVISITED 345

2 4

3

6

9

32. A =

∫ 9
1

(√
y − 1√

y

)
dy =

∫ 9
1

(y1/2 − y−1/2) dy =
(
2

3
x3/2 − 2y1/2

)]9
1

= 12−
(
−4
3

)
=

40

3

-3 3

-4

433. A =

∫ 1
−3

[(−x2 + 6)− (x2 + 4x)] dx =
∫ 1
−3

(6− 4x− 2x2) dx

=

(
6x− 2x2 − 2

3
x3
)]1
−3

=
10

3
− (−18) = 64

3

-1 1 2

-1

1

2

34. A =

∫ 3/2
0

[(−x2 + 3x)− x2] dx =
∫ 3/2
0

(3x− 2x2) dx

=

(
3

2
x2 − 2

3
x3
)]3/2

0

=
9

8

-8 8

435. A =

∫ 8
−8

(4− x2/3) dx =
(
4x− 3

5
x5/3
)]8
−8

=
64

5

(
−64

5

)
=

128

5

-2 2

-2

2

36. A =

∫ 1
−1

[(1− x2/3)− (x2/3 − 1)] dx =
∫ 1
−1

(2− 2x2/3) dx

=

(
2x− 6

5
x5/3
)]1
−1

=
4

5

(
−4
5

)
=

8

5

-2 2 4 6

5

10

15

20

37. A =

∫ 5
−1

[(2x+ 2)− (x2 − 2x− 3)] dx+
∫ 6
5

[(x2 − 2x− 3)− (2x+ 2)] dx

=

∫ 5
−1

(5 + 4x− x2) dx+
∫ 6
5

(x2 − 4x− 5) dx

=

(
5x+ 2x2 +

1

3
x3
)]5
−1

+

(
1

3
x3 − 2x2 − 5x

)]6
5

=
100

3

(
−8
3

)
+ (−30)−

(
−100

3

)
=

118

3

-2 2 4

-2

2

4
38. A =

∫ 5/2
0

[
(−x2 + 4x)− 3

2
x

]
dx =

∫ 5/2
0

(
5

2
x− x2

)
dx

=

(
5

4
x2 − 1

3
x3
)]5/2

0

=
125

48

-4 -2 2 4

2

4

6

8

39. A =

∫ 0
−4

(
x+ 6 +

1

2
x

)
dx+

∫ 2
0

(x+ 6− x3) dx

=

(
3

4
x2 + 6x

)]0
−4

+

(
1

2
x2 + 6x− 1

4
x4
)]2

0

= (0 + 12) + (10− 0) = 22

Page 171

346 CHAPTER 6. APPLICATIONS OF THE INTEGRAL

-1 1 2

-1

1

2

40. A =

∫ 1
0

y2 dy =
1

3
y3
]1
0

=
1

3

-2 -1 1 2

-2

-1

1

2

41. A =

∫ 2
−1

[(2− y2)− (−y)] dy =
∫ 2
−1

(2 + y − y2) dy

=

(
2y +

1

2
y2 − 1

3
y3
)]2
−1

=
10

3

(
−7
6

)
=

9

2

3 6

-3

3

42. A =

∫ √3

3

[(6− y2)− y2] dy =
∫ √3

3

(6− 2y2) dy =
(
6y − 2

3
y3
)]√3

3

= 4

3− (−4

3) = 8

3

2 4

-2

43. A =

∫ 0
−2

[(−y2 − 2y + 2)− (y2 + 2y + 2)] dy =
∫ 0
−2

(−2y2 − 4y) dy

=

(
−2
3
y3 − 2y2

)]0
−2

= 0−
(
−8
3

)
=

8

3

-8 -4

-4

444. A =

∫ 4
0

[(−y2 + 2y + 1)− (y2 − 6y + 1)] dy =
∫ 4
0

(8y − 2y2) dy

=

(
4y2 − 2

3
y3
)]4

0

=
64

3

-2 2

2

445. A =

∫ 1
−1

[(x+ 4)− (x3 − x)] dx =
∫ 1
−1

(4 + 2x− x3) dx

=

(
4x− x2 − 1

4
x4
)]1
−1

=
19

4

(
−13

4

)
= 8

-2 2

-2

2

46. A =

∫ 0
−1

(y3 − y) dy +
∫ 1
0

−(y3 − y) dy

=

(
1

4
y4 − 1

2
y2
)]0
−1

+

(
−1
4
y4 +

1

2
y2
)]1

0

= 0−
(
−1
4

)
+

1

4
− 0 = 1

2

-2

2

47. A =

∫ π/4
0

(cosx− sinx) dx+
∫ π/2
π/4

(sinx− cosx) dx

= (sinx+ cosx)]
π/4
0 + (− cosx− sinx)]

π/2
π/4

=

2− 1 + (−1)− (−

2) = 2

2− 2

Page 339

7.7. IMPROPER INTEGRALS 495

For all x in [1,∞], 1
x

=
x2

x3
=

x4

x3
<

x4 + 1

x3
. By Problem 75,

∫ ∞
1

1

x
dx diverges.

Since 0 ≤ 1
x

x4 + 1

x3
for all x in [1,∞),

∫ ∞
1

x4 + 1

x3
dx diverges.

90. (a) The peak death rate, occurring when t = 17, is R = 890.

(b)

∫ ∞
−∞

890 sech2(0.2t− 3.4) dt

= 890 lim
k→−∞

1

0.2
tanh(0.2t− 3.4)

]0
k

+ 890 lim
p→∞

1

0.2
tanh(0.2t− 3.4)

]p
0

= 4450[0− (−1)] + 4450(1− 0) = 8900

(c)

∫ 34
0

890 sech2(0.2t− 3.4) dt = 890
[

1

0.2
tanh(0.2t− 3.4)

]∣∣∣∣
34

0

= 4450[tanh 3.4− tanh(−3.4)]
= 8900 tanh 3.4

The percentage of total deaths is 100 tanh 3.4 ≈ 99.8%.

(d) To find the peak death rate we solve
dR0
dt

= − a(2t− 2b)
(t2 − 2bt+ c)2 = 0, obtaining t = b, at

which time R0 =
a

c− b2 . Thus, we set b = 17 and
a

c− b2 = 890. The total number of
deaths is

∫ ∞
−∞

a

t2 − 2bt+ c dt =
∫ ∞
−∞

a

(t− b)2 + (c− b2) dt c > b
2

=
a√

c− b2

(
lim

k→−∞
tan−1

t− b√
c− b2

]0
k

+ lim
p→∞

tan
t− b√
c− b2

]p
0

)

=
a√

c− b2

2

(
−π
2

)]
=

aπ√
c− b2

.

We then set
aπ√
c− b2

= 8900 or

c− b2

c− b2 = 8900. Using
a

c− b2 = 890 we obtain

π

c− b2 = 10 or c = 100

π2
+ b2 =

100

π2
+ 172 ≈ 299.13. Then a = 890(c− b2) ≈ 9017.59.

(e) We note that 34 = 2b. Then the number of deaths in the first 34 weeks is

∫ 2b
0

a

t2 − 2bt+ c dt =
∫ 2b
0

a

(t− b)2 + (c− b2) dt =
a√

c− b2
tan−1

t− b√
c− b2

]2b
0

=
a√

c− b2
(
tan−1

b√
c− b2

− tan−1 −b√
c− b2

)

=
2a√
c− b2

tan−1
b√

c− b2

Page 340

496 CHAPTER 7. TECHNIQUES OF INTEGRATION

Using part (d) we see that the fraction of total deaths occurring in the first 34 weeks is

∫ 2b
0

a

t2 − 2bt+ c dt∫ ∞
−∞

a

t2 − 2bt+ c dt
=

2a√
c− b2

tan−1
b√

c− b2
aπ√
c− b2

=
2

π
tan−1

b√
c− b2

.

With b = 17 and c = 299.13 we find the percentage of total deaths within the first 34
weeks is

2

π
tan−1

17√
299.13− 172

× 100 = 88.22%.

7.8 Approximate Integration

1. Midpoint Rule
k 1 2 3
xk 3/2 5/2 7/2

f(xk) 39/4 95/4 175/4∫ 4
1

(3x2 + 2x) dx ≈ 4− 1
3

(
39

4
+

95

4
+

175

4

)
=

309

4
= 77.25

∫ 4
1

(3x2 + 2x) dx = (x3 + x2)
]4
1
= 80− 2 = 78

2. Midpoint Rule
k 1 2 3 4
xk π/48 π/16 5π/48 7π/48

f(xk) 0.997859 0.980785 0.94693 0.896873∫ �/6
0

cosx dx ≈ π/6− 0
4

(0.997859 + 0.980785 + 0.94693 + 0.896873) ≈ 0.500357
∫ �/6
0

cosx dx = sinx
]�/6
0

=
1

2
− 0 = 1

2

3. Trapezoidal Rule
k 0 1 2 3 4
xk 1 3/2 2 5/2 3

f(xk) 2 35/8 9 133/8 28∫ 3
1

(x3 + 1) dx ≈ 3− 1
8

[
2 + 2

(
35

8

)
+ 2(9) + 2

(
133

8

)
+ 28

]
=

45

2
≈ 22.5

∫ 3
1

(x3 + 1) dx =

(
x4

4
+ x

)]3
1

=
93

4
− 5

4
= 22