Document Text Contents
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
UNIDAD 1
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
kπ
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
[π
2
−
(
−π
2
)]
=
aπ√
c− b2
.
We then set
aπ√
c− b2
= 8900 or
aπ
√
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
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
UNIDAD 1
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
kπ
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
[π
2
−
(
−π
2
)]
=
aπ√
c− b2
.
We then set
aπ√
c− b2
= 8900 or
aπ
√
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
