Solution manual calculus 8th edition varberg, purcell, rigdon ch04

4

CHAPTER

4.1 Concepts Review
1. 2 ⋅

The Definite Integral
8

5.

m =1

5(6)
= 30; 2(5) = 10
2

= (−1)1 2−1 + (−1) 2 20 + (−1)3 21
+(−1) 4 22 + (−1)5 23 + (−1)6 24

2. 3(9) – 2(7) = 13; 9 + 4(10) = 49

+(−1)7 25 + (−1)8 26

3. inscribed; circumscribed

1
= − + 1 − 2 + 4 − 8 + 16 − 32 + 64
2
85

=
2

4. 0 + 1 + 2 + 3 = 6

Problem Set 4.1
1.

6

6

6

k =1

k =1

k =1

∑ (k − 1) = ∑ k − ∑ 1

∑ i2 =
i =1

1
1
1
1
3. ∑

=
+
+
k =1 k + 1 1 + 1 2 + 1 3 + 1

1
1
1
1
+
+
+
4 +1 5 +1 6 +1 7 +1
1 1 1 1 1 1 1
= + + + + + +
2 3 4 5 6 7 8
1443
=
840
481
=
280

(−1)5 25 (−1)6 26 (−1)7 27
+
+
6
7
8
1154

=−
105

7.

8

∑ (l + 1)2 = 42 + 52 + 62 + 72 + 82 + 92 = 271

l =3

6

6

n =1

n =1

∑ n cos(nπ) = ∑ ( −1)

n

⋅n

= –1 + 2 – 3 + 4 – 5 + 6
=3

+

4.

(−1)3 23 (−1) 4 24
+
4
5
+

6(7)(13)
= 91
6

7

∑

=

6(7)
− 6(1)
2
= 15
6

(−1)k 2k
k =3 ( k + 1)
7

6.

=

2.

∑ (−1)m 2m−2

⎛ kπ ⎞
k sin ⎜ ⎟
⎝ 2 ⎠
k =−1
6

8.

∑

⎛ π⎞
⎛π⎞
= − sin ⎜ − ⎟ + sin ⎜ ⎟ + 2sin(π)
⎝ 2⎠
⎝2⎠
π
3
⎛ ⎞
⎛ 5π ⎞
+3sin ⎜ ⎟ + 4sin(2π) +5sin ⎜ ⎟ + 6sin(3π)
⎝ 2 ⎠
⎝ 2 ⎠
=1+1+0–3+0+5+0
=4

41

9. 1 + 2 + 3 + " + 41 = ∑ i
i =1

25

10. 2 + 4 + 6 + 8 + " + 50 = ∑ 2i
i =1

11. 1 +

1 1
1 100 1
+ +" +
=∑
2 3
100 i =1 i

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1 1 1
1 100 (−1)i +1
12. 1 − + − + " −
=∑

2 3 4
100 i =1 i

10

20.

i =1
10

= ∑ (4i 2 − i − 3)

50

13. a1 + a3 + a5 + a7 + " + a99 = ∑ a2i −1
i =1

14.

∑ [(i − 1)(4i + 3)]
i =1
10

10

10

i =1

i =1

i =1

= 4∑ i 2 − ∑ i − ∑ 3

f ( w1 )Δx + f ( w2 )Δx + " + f ( wn )Δx

= 4(385) – 55 – 3(10)
= 1455

n

= ∑ f ( wi )Δx
i =1

10

21.

10

15.

∑ (ai + bi )
i =1
10

10

i =1

i =1

k =1

22.

∑ (3an + 2bn )
10

10

n =1

n =1

= 3∑ an + 2 ∑ bn

23.

= 3(40) + 2(50)
= 220

∑ (a p +1 − b p +1 )

p =0

10

p =1

= 40 − 50
= –10

24.

∑ (aq − bq − q)

n

n

n

i =1

i =1

i =1

i =1

=

2n(n + 1)(2n + 1) 3n(n + 1)
−
+n
6
2

=

2n3 + 3n 2 + n 3n 2 + 3n
−
+n
3
2

=

4n3 − 3n 2 − n
6

n

n

10

10

q =1

q =1

q =1

i =1

∑ aq − ∑ bq − ∑ q

10(11)
2

= −65

25.

100

∑ (3i − 2)
i =1

i =1

i =1

∑ (2i − 3)2 = ∑ (4i 2 − 12i + 9)
i =1
n

n

= 4∑ i 2 − 12∑ i + ∑ 9

10

100

k =1

n

q =1

100

k =1

i =1

10

= 40 − 50 −

10

n

10

∑ a p − ∑ bp
p =1

k =1

10

∑ (2i 2 − 3i + 1) = 2∑ i 2 − 3∑ i + ∑1

9

19.

10

∑ 5k 2 (k + 4) = ∑ (5k 3 + 20k 2 )

= 5(3025) + 20(385)
= 22,825

n =1

=

k =1

= 5 ∑ k 3 + 20 ∑ k 2

10

18.

k =1

k =1

= 90

=

10

∑ k3 −∑ k2

10

= 40 + 50

17.

10

= 3025 − 385
= 2640

= ∑ ai + ∑ bi

16.

∑ (k 3 − k 2 ) =

= 3∑ i − ∑ 2
= 3(5050) − 2(100)
= 14,950

i =1

i =1

=

4n(n + 1)(2n + 1) 12n(n + 1)
−
+ 9n
6
2

=

4n3 − 12n 2 + 11n
3

S = 1 + 2 + 3 + " + (n − 2) + (n − 1) + n
+ S = n + (n − 1) + (n − 2) + " + 3 + 2 + 1
2S = (n + 1) + (n + 1) + (n + 1) + " + (n + 1) + (n + 1) + (n + 1)
2S = n(n + 1)
n(n + 1)
S=
2

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26. S − rS = a + ar + ar 2 + " + ar n
− (ar + ar 2 + " + ar n + ar n +1 )

27. a.

= a − ar n +1

(1)

1−
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 12
k =0
2
10

k

10

k

11

10

⎛1⎞
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 1 − ⎜⎝ 2 ⎟⎠
k =1

n +1

a − ar
= S (1 − r ); S =
1− r

10

b.

∑ 2k =

k =0
10

10

⎛1⎞
= 2 − ⎜ ⎟ , so
⎝2⎠

=

1023
.
1024

1 − 211
= 211 − 1, so
−1

∑ 2k = 211 − 2 = 2046 .

k =1

28.

S = a + (a + d ) + (a + 2d ) + "" + [ a + (n − 2)d ] + [ a + (n − 1)d ] + (a + nd )

+ S = (a + nd ) + [ a + (n − 1)d ] + [ a + (n − 2)d ] + " + (a + 2d ) + (a + d ) + a

2S = (2a + nd ) + (2a + nd ) + (2a + nd ) + " + (2a + nd ) + (2a + nd ) + (2a + nd )
2S = (n + 1)(2a + nd)
(n + 1)(2a + nd )
S=
2

( i + 1)3 − i3 = 3i 2 + 3i + 1

29.
n

∑ ⎡⎣( i + 1)
i =1

3

n

(

)

− i 3 ⎤ = ∑ 3i 2 + 3i + 1
⎦ i =1
n

n

n

i =1

i =1

i =1

( n + 1)3 − 13 = 3∑ i 2 + 3∑ i + ∑1
n

n ( n + 1)

i =1

2

n3 + 3n 2 + 3n = 3∑ i 2 + 3

+n

n

2n3 + 6n 2 + 6n = 6∑ i 2 + 3n 2 + 3n + 2n
i =1

2n + 3n + n
= ∑ i2
6
i =1
3

2

n ( n + 1)( 2n + 1)
6

n

n

= ∑ i2
i =1

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( i + 1)4 − i 4 = 4i3 + 6i 2 + 4i + 1

30.

∑ ⎡⎣(i + 1)4 − i 4 ⎤⎦ =∑ ( 4i3 + 6i 2 + 4i + 1)
n

n

i =1

i =1

n

n

n

n

i =1
n

i =1

i =1

i =1

(n + 1)4 − 14 = 4∑ i3 + 6∑ i 2 + 4∑ i + ∑1
n 4 + 4 n3 + 6 n 2 + 4 n = 4∑ i 3 + 6

i =1

n(n + 1)(2n + 1)
n(n + 1)
+4
+n
6
2

n

Solving for

∑ i3 gives
i =1

(

n

) (

)

4∑ i3 = n 4 + 4n3 + 6n 2 + 4n − 2n3 + 3n 2 + n − 2n 2 + 2n − n
i =1
n

4∑ i 3 = n 4 + 2 n3 + n 2
i =1

n 4 + 2n3 + n 2 ⎡ n ( n + 1) ⎤
=⎢
⎥
∑i =
4
⎣ 2 ⎦
i =1
n

2

3

( i + 1)5 − i5 = 5i 4 + 10i3 + 10i 2 + 5i + 1

31.
n

∑ ⎡⎣⎢( i + 1)

5

i =1

n

n

n

n

n

i =1
2

i =1

i =1

− i5 ⎤⎥ =5∑ i 4 + 10∑ i3 + 10∑ i 2 + 5∑ i + ∑ 1
⎦
i =1

n

i =1

( n + 1)5 − 15 = 5∑ i 4 + 10

n

2

( n + 1)
4

i =1

n

n5 + 5n 4 + 10n3 + 10n 2 + 5n = 5∑ i 4 + 52 n 2 ( n + 1)
i =1

2

+ 10

n(n + 1)(2n + 1)
n(n + 1)
+5
+n
6
2

5
+ 10
n ( n + 1) (2n + 1) + n(n + 1) + n
6
2

n

Solving for

∑ i4

yields

i =1

n

∑ i 4 = 15 ⎡⎣n5 + 52 n4 + 53 n3 − 16 n ⎤⎦ =
i =1

n(n + 1)(2n + 1)(3n 2 + 3n − 1)
30

32. Suppose we have a ( n + 1) × n grid. Shade in
n + 1 – k boxes in the kth column. There are n columns, and the shaded area is 1 + 2 + " + n . The shaded area is
n(n + 1)
n(n + 1)
. Thus, 1 + 2 + " + n =
.
2
2
n(n + 1)
. From the diagram the area is
Suppose we have a square grid with sides of length 1 + 2 + " + n =
2

also half the area of the grid or

2

2

⎡ n(n + 1) ⎤

⎡ n(n + 1) ⎤
13 + 23 + " + n3 or ⎢
. Thus, 13 + 23 + " + n3 = ⎢
⎥ .
⎥
⎣ 2 ⎦
⎣ 2 ⎦

33. x =

1
55
(2 + 5 + 7 + 8 + 9 + 10 + 14) =
≈ 7.86
7
7
2

1 ⎡⎛
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞
⎛
s = ⎢⎜ 2 − ⎟ + ⎜ 5 − ⎟ + ⎜ 7 − ⎟ + ⎜ 8 − ⎟ + ⎜ 9 − ⎟ + ⎜ 10 − ⎟ + ⎜ 14 − ⎟
7 ⎢⎝
7 ⎠ ⎝

7 ⎠ ⎝
7 ⎠ ⎝
7 ⎠
7
7
7 ⎠
⎝
⎠ ⎝
⎠ ⎝
⎣
2

2

2

2

2

2

2⎤

608
⎥ =
≈ 12.4
49
⎥⎦

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34. a.

x = 1, s 2 = 0

b.

x = 1001, s 2 = 0

c.

x=2
2
s 2 = 13 ⎡(1 − 2)2 + (2 − 2) 2 + (3 − 2)2 ⎤ = 13 ⎡⎢( −1) + 02 + 12 ⎤⎥ = 13 ( 2 ) = 32
⎣
⎦
⎣
⎦

d.

x = 1, 000, 002
s 2 = 13 ⎡(−1)2 + 02 + 12 ⎤ = 32
⎣

⎦

35. a.

b.

n

n

n

i =1

i =1

i =1

∑ ( xi − x ) = ∑ xi − ∑ x = nx − nx = 0
1 n
1 n
( xi − x )2 = ∑ ( xi2 − 2 x xi + x 2 )
∑
n i =1
n i =1
1 n
2x n
1 n
= ∑ xi2 −
xi + ∑ x 2

∑
n i =1
n i =1
n i =1

s2 =

=

1 n 2 2x
1
xi −
(nx ) + (nx 2 )
∑
n i =1
n
n

⎛1 n
⎞
⎛1 n
⎞
= ⎜ ∑ xi2 ⎟ − 2 x 2 + x 2 = ⎜ ∑ xi2 ⎟ − x 2
⎜n
⎟
⎜n
⎟
⎝ i =1 ⎠
⎝ i =1 ⎠

36. The variance of n identical numbers is 0. Let c be the constant. Then
2
2
s 2 = 1n ⎡⎢( c − c ) + ( c − c ) + " + (c − c)2 ⎤⎥ = 0
⎣
⎦
n

37. Let S (c) = ∑ ( xi − c)2 . Then
i =1

d n
( xi − c )2
∑
dc i =1

S '(c) =

n

d
( xi − c )2
dc
i =1

=∑
n

= ∑ 2( xi − c)(−1)
i =1

n

= −2∑ xi + 2nc
i =1

S ''(c) = 2n
Set S '(c) = 0 and solve for c :
n

−2∑ xi + 2nc = 0
c

i =1
n
= 1n xi
i =1

∑

=x

Since S ''( x) = 2n > 0 we know that x minimizes S (c) .

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i

The number of gifts given on the nth day is

38. a.

∑m=

m =1

i (i + 1)
.
2

i (i + 1)
= 364 .
i =1 2
12

The total number of gifts is

∑

i (i + 1)
.
i =1 2
n

b. For n days, the total number of gifts is

∑

i (i + 1) n i 2 n i 1 n 2 1 n
1 ⎡ n(n + 1)(2n + 1) ⎤ 1 ⎡ n(n + 1) ⎤
∑ 2 = ∑ 2 +∑ 2 = 2 ∑ i + 2 ∑ i = 2 ⎢⎣
⎥+ 2⎢ 2 ⎥
6
⎦
⎣
⎦
i =1
i =1
i =1
i =1
i =1
1
1
⎛ 2n + 1 ⎞ 1
= n(n + 1) ⎜
+ 1⎟ = n(n + 1)(2n + 4) = n(n + 1)(n + 2)
4
6
⎝ 3
⎠ 12
n

39. The bottom layer contains 10 · 16 = 160 oranges, the next layer contains 9 · 15 = 135 oranges, the third layer
contains 8 · 14 = 112 oranges, and so on, up to the top layer, which contains 1 · 7 = 7 oranges. The stack contains
1 · 7 + 2 · 8+ ... + 9· 15 + 10 · 16

10

=

∑ i(6 + i) = 715 oranges.
i =1

50

40. If the bottom layer is 50 oranges by 60 oranges, the stack contains

∑ i(10 + i) = 55, 675.
i =1

41. For a general stack whose base is m rows of n oranges with m ≤ n, the stack contains
m

m

m

i =1

i =1

i =1

∑ i ( n − m + i ) = ( n − m)∑ i + ∑ i 2
m(m + 1) m(m + 1)(2m + 1)
= ( n − m)

+
2
6
m(m + 1)(3n − m + 1)
=
6
42.

1
1
1
1
+
+
+" +
1⋅ 2 2 ⋅ 3 3 ⋅ 4
n(n + 1)
1 ⎞
⎛ 1⎞ ⎛ 1 1⎞ ⎛1 1⎞
⎛1
= ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + " + ⎜ −
⎟
2
2
3
3
4
n
n
+1⎠

⎝
⎠ ⎝
⎠ ⎝
⎠
⎝
1
= 1−
n +1

43. A =

1⎡ 3
5⎤ 7
1+ + 2 + ⎥ =
2 ⎢⎣ 2
2⎦ 2

44. A =

1⎡ 5 3 7
9 5 11 ⎤ 15
1+ + + + 2 + + + ⎥ =
⎢
4⎣ 4 2 4
4 2 4⎦ 4

45. A =

1 ⎡3
5 ⎤ 9

+ 2 + + 3⎥ =
⎢
2 ⎣2
2 ⎦ 2

46. A =

1 ⎡5 3 7
9 5 11 ⎤ 17
+ + + 2 + + + + 3⎥ =
4 ⎢⎣ 4 2 4
4 2 4
⎦ 4

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47. A =

2
2
⎞⎤
⎞ ⎛1
1 ⎡⎛ 1 2 ⎞ ⎛ 1 ⎛ 1 ⎞
1 ⎛ 9 3 17 ⎞ 23
⎞ ⎛1 ⎛3⎞

2
⎜
⎜
⎟
+
⋅
+ 1⎟ ⎥ = ⎜ 1 + + + ⎟ =
⋅
+
+
⋅
+
+
⋅
+
0
1
1
1
1
⎜
⎟
⎜
⎟ ⎜ ⎜ ⎟
⎜
⎟
⎢
⎜
⎟
⎟

2⎝ 8 2 8 ⎠ 8
2
2
2 ⎣⎝ 2
2
2
2
⎠ ⎝ ⎝ ⎠
⎠ ⎝ ⎝ ⎠
⎠ ⎦⎥
⎠ ⎝

48. A =

2
⎤
⎛1 3 2 ⎞ 1
⎞ 1
1 ⎡⎛ 1 ⎛ 1 ⎞
1 9 3 17
31
⎢⎜ ⋅ ⎜ ⎟ + 1⎟ + ⎛⎜ ⋅12 + 1⎞⎟ + ⎜ ⋅ ⎜⎛ ⎟⎞ + 1⎟ + ⎜⎛ ⋅ 22 + 1⎟⎞ ⎥ = ⎛⎜ + + + 3 ⎞⎟ =
⎜
⎟
⎜
⎟
2⎝8 2 8
2⎢ 2 ⎝2⎠
2
2

⎠ ⎥⎦
⎠ ⎝2 ⎝2⎠
⎠ 8
⎠ ⎝
⎠ ⎝
⎣⎝

49.

A = 1(1 + 2 + 3) = 6
50.

A=

1 ⎡⎛ 3 ⎞
⎛ 5 ⎞
⎜ 3 ⋅ − 1⎟ + ( 3 ⋅ 2 − 1) + ⎜ 3 ⋅ − 1⎟ + (3 · 3 – 1)] =
2 ⎢⎣⎝ 2 ⎠
⎝ 2 ⎠

1⎛7
13 ⎞ 23
⎜ + 5 + + 8⎟ =
2⎝2
2
⎠ 2

51.

2

⎤
⎞ ⎛ 7 2 ⎞ ⎛ 5 2 ⎞ ⎛ 8 2 ⎞ ⎛ 17 2 ⎞
1 ⎡⎛ ⎛ 13 ⎞
⎢⎜ ⎜ ⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + (32 − 1) ⎥
⎟ ⎜⎝ 3 ⎠
⎟ ⎜⎝ 2 ⎠
⎟ ⎜⎝ 3 ⎠
⎟ ⎜⎝ 6 ⎠
⎟
6 ⎢⎜ ⎝ 6 ⎠
⎥⎦
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
⎣⎝
1 ⎛ 133 40 21 55 253 ⎞ 1243
= ⎜
+
+ + +
+ 8⎟ =
6 ⎝ 36 9
4 9
36
⎠ 216

A=

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52.

2
2
2
⎛
⎞ ⎛
⎞ ⎛
⎞
1⎡
4
4
3
3
2
2
⎢(3(−1)2 + (−1) + 1) + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + (3(0) 2 + 0 + 1)
⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟
5⎢
⎝
⎠ ⎝
⎠ ⎝
⎠
⎣

2
2
2
2
⎤
⎛ ⎛1⎞ 1 ⎞ ⎛ ⎛2⎞
2 ⎞ ⎛ ⎛ 3⎞
3 ⎞⎛ ⎛ 4 ⎞
4 ⎞
+ ⎜ 3 ⎜ ⎟ + + 1⎟ + ⎜ 3 ⎜ ⎟ + + 1⎟ + ⎜ 3 ⎜ ⎟ + + 1⎟ ⎜ 3 ⎜ ⎟ + + 1⎟ + (3(1) 2 + 1 + 1) ⎥
⎜ 5
5 ⎟ ⎜ ⎝5⎠
5 ⎟ ⎜ ⎝5⎠
5 ⎟⎜ ⎝ 5 ⎠
5 ⎟
⎥⎦
⎝ ⎝ ⎠
⎠ ⎝
⎠ ⎝
⎠⎝
⎠
1
= [3 + 2.12 + 1.48 + 1.08 + 1 + 1.32 + 1.88 + 2.68 + 3.72 + 5] = 4.656
5

A=

1
i
, xi =

n
n
i
⎛
⎞⎛ 1 ⎞ i 2
f ( xi )Δx = ⎜ + 2 ⎟ ⎜ ⎟ =
+
⎝n
⎠ ⎝ n ⎠ n2 n
⎡⎛ 1 2 ⎞ ⎛ 2 2 ⎞
1 5
⎛ n 2 ⎞⎤
1
n(n + 1)
+
+
+
+" + ⎜
+
(1 + 2 + 3 + " + n) + 2 =
=
+2 =
+
A( Sn ) = ⎢⎜
2 n⎟ ⎜ 2 n⎟
2 n ⎟⎥
2
2
2n 2
2n

⎠ ⎝n
⎠
⎝n
⎠⎦ n
⎣⎝ n
⎛ 1 5⎞ 5
lim A( Sn ) = lim ⎜ + ⎟ =
n →∞
n→∞ ⎝ 2n 2 ⎠ 2

53. Δx =

1
i
, xi =
n
n
⎡ 1 ⎛ i ⎞2 ⎤ ⎛ 1 ⎞ i 2
1
f ( xi )Δx = ⎢ ⋅ ⎜ ⎟ + 1⎥ ⎜ ⎟ =
+
3
2
n
n
n
⎢⎣ ⎝ ⎠
⎥⎦ ⎝ ⎠ 2n
⎡⎛ 12
⎛ n2 1 ⎞⎤

1 ⎞ ⎛ 22 1 ⎞
1 2
+ ⎟+⎜
+ ⎟ +" + ⎜
+ ⎟⎥ =
(1 + 22 + 32 + " + n 2 ) + 1
A( Sn ) = ⎢⎜
3
3 n⎟ ⎜
3 n⎟
3 n⎟
⎜
⎜
⎢⎣⎝ 2n
⎠ ⎝ 2n
⎠
⎝ 2n
⎠ ⎥⎦ 2n
1 ⎡ n(n + 1)(2n + 1) ⎤
1 ⎡ 2n3 + 3n 2 + n ⎤
1 ⎡
3 1 ⎤
=
=
+
1
⎢
⎥ + 1 = ⎢2 + + 2 ⎥ + 1
⎥
3 ⎢⎣

3
6
12 ⎢⎣
n n ⎦
12 ⎣
⎦
n
2n
⎥⎦
⎡1⎛
3 1 ⎞ ⎤ 7
lim A( Sn ) = lim ⎢ ⎜ 2 + +
⎟ + 1⎥ =
n n2 ⎠ ⎦ 6
n →∞
n →∞ ⎣12 ⎝

54. Δx =

256
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Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


2
2i
, xi = −1 +
n

n
⎡ ⎛
2i ⎞ ⎤ ⎛ 2 ⎞ 8i
f ( xi )Δx = ⎢ 2 ⎜ −1 + ⎟ + 2 ⎥ ⎜ ⎟ =
n ⎠ ⎦ ⎝ n ⎠ n2
⎣ ⎝

1
i
, xi =
n
n
⎡ ⎛ i ⎞3 i ⎤ ⎛ 1 ⎞ i 3
i
f ( xi )Δx = ⎢⎜ ⎟ + ⎥ ⎜ ⎟ =
+
4
n
n
n
⎝
⎠
⎝
⎠
n
n2
⎣⎢
⎦⎥

55. Δx =

58. Δx =

⎡⎛ 8 ⎞ ⎛ 16 ⎞
⎛ 8n ⎞ ⎤
A( Sn ) = ⎢⎜ ⎟ + ⎜ ⎟ + " + ⎜ ⎟ ⎥
2
2
⎝ n2 ⎠⎦
⎣⎝ n ⎠ ⎝ n ⎠
8
8 ⎡ n(n + 1) ⎤
=
(1 + 2 + 3 + " + n) =
⎢
⎥
2
n
n2 ⎣ 2 ⎦
⎡ n2 + n ⎤
4
= 4⎢
⎥ = 4+
2
n
⎣⎢ n ⎦⎥

A( Sn ) =

59.

60.
1
i
, xi =
n
n
3

3
⎛ i ⎞ ⎛1⎞ i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n4
⎡1
1 3
1 3 ⎤
A( Sn ) = ⎢ (13 ) +
(2 ) + " +
(n ) ⎥
4
4
n
n4
⎣n
⎦

n
=

1 ⎡ n(n + 1) ⎤

⎢
⎥
n4 ⎣ 2 ⎦

1 ⎡ n(n + 1) ⎤
1 ⎡ n(n + 1) ⎤
⎥ + 2⎢ 2 ⎥
4 ⎢⎣
2
⎦
⎦
n
n ⎣

2

i 2
⎡i
⎤1
+
f (ti )Δt = ⎢ + 2 ⎥ =
⎣n
⎦ n n2 n

⎛1 1 ⎞
=⎜ + ⎟+2
⎝ 2 2n ⎠
1
5
lim A( Sn ) = + 2 =

2
2
n →∞
1
The object traveled 2 ft.
2

⎛8 4
4 ⎞ 8
lim A( Sn ) = lim ⎜ + +
⎟= .
n →∞
n →∞ ⎝ 3 n 3n 2 ⎠ 3
⎛ 8 ⎞ 16
By symmetry, A = 2 ⎜ ⎟ =
.
⎝3⎠ 3

(13 + 23 + " + n3 ) =

(1 + 2 + " + n)

⎡ n2 + n ⎤
=⎢
⎥+2
2
⎣⎢ 2n ⎦⎥

4 ⎡ 2n3 + 3n 2 + n ⎤ 8 4
4

= ⎢
⎥= + + 2
3
n
3 ⎣⎢
3
n
3n
⎦⎥

4

n2

n
n
2
⎛ i 2⎞ 1 n
A( Sn ) = ∑ ⎜
+ ⎟ = ∑i + ∑
2 n
2
⎠ n i =1 i =1 n
i =1 ⎝ n
1 ⎡ n(n + 1) ⎤
=
⎢
⎥+2
n2 ⎣ 2 ⎦

2

⎛ 2i ⎞ ⎛ 2 ⎞ 8i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n3
⎡⎛ 8 ⎞ ⎛ 8(22 ) ⎞
⎛ 8n 2 ⎞ ⎤
A( Sn ) = ⎢⎜ ⎟ + ⎜
+ ⋅⋅⋅+ ⎜
⎟
⎟⎥
3
⎜ 3 ⎟
⎜ n3 ⎟ ⎥
⎝
⎠⎦
⎣⎢⎝ n ⎠ ⎝ n ⎠
8 2
8 ⎡ n(n + 1)(2n + 1) ⎤
=
(1 + 22 + ⋅⋅⋅ + n 2 ) =
⎢
⎥
3
6
⎦
n
n3 ⎣

1

1

n 2 + 2n + 1 n 2 + n 3n 2 + 4n + 1 3 1
1
+
=
= + +
2
2
2
4
n
4n
2n
4n
4n 2
3
lim A( Sn ) =
4
n →∞

56. First, consider a = 0 and b = 2.
2
2i
Δx = , xi =
n
n

=

(13 + 23 + " + n3 ) +

=

4⎞
⎛
lim A( Sn ) = lim ⎜ 4 + ⎟ = 4
n⎠
n→∞ ⎝

57. Δx =

n

4

2

=

n →∞

2

1

⎡ 1 ⎛ i ⎞2 ⎤ 1
i2
1

+
f (ti )Δt = ⎢ ⎜ ⎟ + 1⎥ =
3
n
⎢⎣ 2 ⎝ n ⎠
⎥⎦ n 2n
n ⎛ 2
1i
1⎞
1 n 2 n 1
A( Sn ) = ∑ ⎜
i +∑
+ ⎟=
⎜ 3 n ⎟ 2 n3 ∑
i =1 ⎝ 2n
i =1
i =1 n
⎠
1 ⎡ n(n + 1)(2n + 1) ⎤
1 ⎡
3 1 ⎤
=
⎥ + 1 = 12 ⎢ 2 + n + 2 ⎥ + 1
3 ⎢⎣
6
⎦
2n
n ⎦
⎣

1
7
(2) + 1 = ≈ 1.17
12
6
n →∞
The object traveled about 1.17 feet.
lim A( Sn ) =

1⎡ 2 1 ⎤
1 ⎡ n 4 + 2n3 + n 2 ⎤
⎢
⎥ = ⎢1 + + 2 ⎥
4
4
4⎣ n n ⎦
n ⎣⎢
⎦⎥
1⎡ 2 1 ⎤ 1
⎢1 + n + 2 ⎥ = 4
n →∞ 4 ⎣
n ⎦

lim A( Sn ) = lim

n →∞

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


65. a.

A02 ( x3 ) =

23+1
=4
3 +1

b.

A12 ( x3 ) =

23+1 13+1
1 15
−
= 4− =
3 +1 3 +1
4 4

c.

A12 ( x5 ) =

25+1 15+1 32 1 63
−
=

− =
5 +1 5 +1 3 6 6

2

61. a.

3 2
⎛ ib ⎞ ⎛ b ⎞ b i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n3

A0b =
b3
=
6

b3 n

∑ i2
n3

=

i =1

b3 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥
6

⎦
n3 ⎣

3 1 ⎤
⎡
⎢2 + n + 2 ⎥
n ⎦
⎣

lim A0b =

n →∞

2b3 b3
=
6
3

=

21
= 10.5
2

b. Since a ≥ 0, A0b = A0a + Aab , or
Aab

=

A0b

−

A0a

b3 a 3
=
− .
3
3

d.

A05 =

53 125
=
3
3

b.

A14 =

43 13 63
− =
= 21
3 3
3

c.

A25 =

53 23 117
−
=
= 39
3
3
3

64. a.

Δx =

b
bi
, xi =
n
n

m
m +1 m
⎛ bi ⎞ ⎛ b ⎞ b i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝n⎠
n m +1
b m +1 n m
A( Sn ) =

∑i
n m +1 i =1

=

⎤
b m +1 ⎡ n m +1
+ Cn ⎥
⎢
m +1 m + 1
n
⎢⎣
⎥⎦

b m +1 b m +1Cn
=
+
m +1
n m +1
A0b ( x m ) = lim A( Sn ) =
n →∞

lim

Cn

n →∞ n m +1

29+1 1024
=

= 102.4
9 +1
10

66. Inscribed:
Consider an isosceles triangle formed by one side
of the polygon and the center of the circle. The
2π
. The length of the base
angle at the center is
n
π
π
is 2r sin . The height is r cos . Thus the area
n
n
π
π 1 2
2π
2
of the triangle is r sin cos = r sin
.
n
n 2
n
2π ⎞ 1
2π
⎛1
An = n ⎜ r 2 sin ⎟ = nr 2 sin
n ⎠ 2

n
⎝2

53 33 98
62.
=
−
=
≈ 32.7
3 3
3
The object traveled about 32.7 m.
A35

63. a.

A02 ( x9 ) =

Circumscribed:
Consider an isosceles triangle formed by one side
of the polygon and the center of the circle. The
2π
. The length of the base
angle at the center is
n
π
is 2r tan . The height is r. Thus the area of the
n
π
triangle is r 2 tan .

n
π
π
⎛
⎞
Bn = n ⎜ r 2 tan ⎟ = nr 2 tan
n⎠
n
⎝
⎛ sin 2π ⎞
1 2
2π
n ⎟
nr sin
= lim πr 2 ⎜
⎜ 2π ⎟
n n→∞
n→∞ 2
⎝ n ⎠

lim An = lim

n →∞

b m +1
m +1

= πr 2

= 0 since Cn is a polynomial in n

of degree m.

lim Bn = lim nr 2 tan

n →∞

n→∞

π
π
πr 2 ⎛ sin n ⎞
⎜
⎟
= lim
n n→∞ cos π ⎜ π ⎟
n⎝ n ⎠

= πr 2

b. Notice that Aab ( x m ) = A0b ( x m ) − A0a ( x m ) .

Thus, using part a, Aab ( x m ) =

b m +1 a m +1
−
.
m +1 m +1

258

Section 4.1
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


4.2 Concepts Review
1. Riemann sum
2. definite integral;

b

∫a

f ( x )dx

3. Aup − Adown
4. 8 −

1 15
=
2 2

Problem Set 4.2
1. RP = f (2)(2.5 − 1) + f (3)(3.5 − 2.5) + f (4.5)(5 − 3.5) = 4(1.5) + 3(1) + (–2.25)(1.5) = 5.625
2. RP = f(0.5)(0.7 – 0) + f(1.5)(1.7 – 0.7) + f(2)(2.7 – 1.7) + f(3.5)(4 – 2.7)
= 1.25(0.7) + (–0.75)(1) + (–1)(1) + 1.25(1.3) = 0.75
5

3. RP = ∑ f ( xi )Δxi = f (3)(3.75 − 3) + f (4)(4.25 − 3.75) + f (4.75)(5.5 − 4.25) + f (6)(6 − 5.5) + f (6.5)(7 − 6)

i =1

= 2(0.75) + 3(0.5) + 3.75(1.25) + 5(0.5) + 5.5(1) = 15.6875
4

4. RP = ∑ f ( xi )Δxi = f (−2)(−1.3 + 3) + f (−0.5)(0 + 1.3) + f (0)(0.9 − 0) + f (2)(2 − 0.9)
i =1

= 4(1.7) + 3.25(1.3) + 3(0.9) + 2(1.1) = 15.925
8

5. RP = ∑ f ( xi )Δxi = [ f (−1.75) + f (−1.25) + f (−0.75) + f (−0.25) + f (0.25) + f (0.75) + f (1.25) + f (1.75) ] (0.5)
i =1

= [–0.21875 – 0.46875 – 0.46875 – 0.21875 + 0.28125 + 1.03125 + 2.03125 + 3.28125](0.5) = 2.625
6

6. RP = ∑ f ( xi )Δxi = [ f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) + f (3) ] (0.5)
i =1

= [1.5 + 5 + 14.5 + 33 + 63.5 + 109](0.5) = 113.25

7.
8.

3

x3 dx

2

( x + 1)3 dx

∫1

∫0

11. Δx =

2
2i
, xi =
n
n

f ( xi ) = xi + 1 =
n

9.

10.

1

x

2

∫−1 1 + x dx
π

∫0 (sin x)

dx

n

⎡

⎛ 2 ⎞⎤ 2

∑ f ( xi )Δx = ∑ ⎢⎣1 + i ⎜⎝ n ⎟⎠⎥⎦ n
i =1

i =1
n

2
4
2
4 ⎡ n(n + 1) ⎤
∑1 + ∑ i = n (n) + n2 ⎢⎣ 2 ⎥⎦
n i =1 n 2 i =1
⎛ 1⎞
= 2 + 2 ⎜1 + ⎟
⎝ n⎠
2
⎡
⎛ 1 ⎞⎤
⎢ 2 + 2 ⎜1 + n ⎟ ⎥ = 4

∫0 ( x + 1)dx = nlim
→∞ ⎣
⎝
⎠⎦
=

2

2i
+1
n

n

Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


12. Δx =

2
2i
, xi =
n
n

14. Δx =

2

2

2

3i ⎞
36i 27i 2
⎛
f ( xi ) = 3 ⎜ −2 + ⎟ + 2 = 14 −
+
n⎠
n
⎝
n2
n
n ⎡
⎛ 36 ⎞ ⎛ 27 ⎞ ⎤ 3
∑ f ( xi )Δx = ∑ ⎢14 − ⎜⎝ n ⎟⎠ i + ⎜⎝ n2 ⎟⎠ i 2 ⎥ n
⎦
i =1
i =1 ⎣

4i
⎛ 2i ⎞
f ( xi ) = ⎜ ⎟ + 1 =
+1
⎝n⎠
n2

n
n ⎡
⎛ 4 ⎞⎤ 2
∑ f ( xi )Δx = ∑ ⎢1 + i 2 ⎜⎝ n2 ⎟⎠⎥ n
⎦
i =1
i =1 ⎣
=

2 n
8 n 2 2
8
+
i = ( n) +
1
∑
∑
3
n i =1 n i =1
n
n3

⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥
6
⎣
⎦

4⎛

3 1 ⎞
= 2+ ⎜2+ +
⎟
3⎝
n n2 ⎠
2

∫0

3
3i
, xi = −2 +
n
n

=

3 n
108 n
81 n 2
−
i
+
14
∑ n 2 ∑ n3 ∑ i
n i =1
i =1
i =1

= 42 −

⎡
4⎛
3 1 ⎞ ⎤ 14
( x 2 + 1) dx = lim ⎢ 2 + ⎜ 2 + +
⎟⎥ =
n n2 ⎠ ⎦ 3
3⎝
n →∞ ⎣

3 1 ⎞
⎛ 1 ⎞ 27 ⎛
= 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟
n n2 ⎠
⎝ n⎠ 2 ⎝
1

∫−2 (3x

3
3i
13. Δx = , xi = −2 +
n
n
3i ⎞
6i
⎛
f ( xi ) = 2 ⎜ −2 + ⎟ + π = π − 4 +
n⎠

n
⎝
n
6i ⎤ 3
⎡
Δ
=
f
(
x
)
x
∑ i
∑ ⎢⎣ π − 4 + n ⎥⎦ n
i =1
i =1
n
3
18 n
18 ⎡ n(n + 1) ⎤
= ∑ (π − 4) + ∑ i = 3(π − 4) +
⎢
⎥
2
n i =1
n i =1
n2 ⎣ 2 ⎦

⎛ 1⎞
= 3π − 12 + 9 ⎜ 1 + ⎟

⎝ n⎠
⎡

⎛

1 ⎞⎤

⎢3π − 12 + 9 ⎜ 1 + n ⎟ ⎥
∫−2 (2 x + π) dx = nlim
→∞ ⎣
⎝
⎠⎦

= 3π − 3

2

+ 2) dx

⎡
3 1 ⎞⎤
⎛ 1 ⎞ 27 ⎛
= lim ⎢ 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟ ⎥ = 15
n n2 ⎠ ⎦
n→∞ ⎣
⎝ n⎠ 2 ⎝

n

1

108 ⎡ n(n + 1) ⎤ 81 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥+ ⎢
⎥
6
⎦
n 2 ⎣ 2 ⎦ n3 ⎣

5
5i
, xi =
n
n
5i
f ( xi ) = 1 +
n

15. Δx =

n
⎡
⎛ 5 ⎞⎤ 5
f
x
x
(
)
Δ

=
∑ i
∑ ⎢⎣1 + i ⎜⎝ n ⎟⎠⎥⎦ n
i =1
i =1
n
5
25 n
25 ⎡ n(n + 1) ⎤
= ∑1 +
∑ i = 5 + n2 ⎢⎣ 2 ⎥⎦
n i =1 n 2 i =1
n

= 5+

25 ⎛ 1 ⎞
⎜1 + ⎟
2 ⎝ n⎠

5

⎡

⎢5 +
∫0 ( x + 1) dx = nlim
→∞ ⎣

16. Δx =

25 ⎛ 1 ⎞ ⎤ 35
⎜1 + ⎟ =
2 ⎝ n ⎠ ⎥⎦ 2

20
20i
, xi = −10 +
n
n
2

20i ⎞ ⎛
20i ⎞
380i 400i 2
⎛
f ( xi ) = ⎜ −10 +
+
⎟ + ⎜ −10 +
⎟ = 90 −
n ⎠ ⎝
n ⎠
n
⎝
n2
n
n ⎡
20 n
7600 n
8000 n 2
⎛ 380 ⎞ 2 ⎛ 400 ⎞ ⎤ 20

f
x
x
i
i
=
−
i
+
(
)
90
90
Δ
=
−
+
∑ i
∑⎢
∑
∑ n3 ∑ i
⎜ 2 ⎟⎥
⎜
⎟
n i =1
⎝ n ⎠
n 2 i =1
⎝ n ⎠⎦ n
i =1
i =1

i =1 ⎣
= 1800 −
10

∫−10 ( x

2

7600 ⎡ n(n + 1) ⎤ 8000 ⎡ n(n + 1)(2n + 1) ⎤
3 1 ⎞
⎛ 1 ⎞ 4000 ⎛
⎥+ 3 ⎢
⎥ = 1800 − 3800 ⎜ 1 + n ⎟ + 3 ⎜ 2 + n + 2 ⎟
2 ⎢⎣
2
6
⎝
⎠
⎦
⎣
⎦
n
n
n ⎠
⎝
⎡
3 1 ⎞ ⎤ 2000
⎛ 1 ⎞ 4000 ⎛
+ x) dx = lim ⎢1800 − 3800 ⎜1 + ⎟ +
⎜ 2 + + 2 ⎟⎥ =

n n ⎠⎦
3 ⎝
3
n→∞ ⎣
⎝ n⎠

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21. The area under the curve is equal to the area of a

17.

semi-circle:

A

∫− A

A2 − x 2 dx = 12 π A2 .

5

∫0 f ( x) dx
1
1

27
(1)(2) + 1(2) + 3(2) + (3)(3) =
2
2
2

=

18.

22. The area under the curve is equal to the area of a
triangle:

y
4

2

⎛1⎞

∫−4 f ( x ) dx = 2 ⎜⎝ 2 ⎟⎠ 4 ⋅ 4 = 16
4

2

2

∫0

f ( x) dx =

4 x

1
1
9
(1)( 3) + (1)( 2 ) + (1)( 2 ) =
2
2
2

23. s ( 4 ) = ∫ v ( t ) dt =
4

0

1 ⎛ 4 ⎞ 2
4⎜ ⎟ =
2 ⎝ 60 ⎠ 15

4
1
24. s ( 4 ) = ∫ v ( t ) dt = 4 + 4 ( 9 − 1) = 20
0
2

19.

25. s ( 4 ) = ∫ v ( t ) dt =
4

0

1
2 (1) + 2 (1) = 3
2

4
1
2
26. s ( 4 ) = ∫ v ( t ) dt = π ( 2 ) + 0 = π
0
4

27.
2

∫0

f ( x) dx =

1
1
1 π
(π ⋅12 ) + (1)(1) = +
4
2
2 4

20.

t

s(t)

20

40

40

80

60

120

80

160

100

200

120

240

1

1
f ( x) dx = − (π ⋅ 22 ) − (2)(2) − (2)(4)
4
2
= −π − 8
2

∫−2

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28.

t

s(t)

20

10

40

40

60

90

80

160

100

250

120

360

e.

f.

g.

1
1
(3)(3) + (3)(3) = 9
2
2

∫−3

3

x dx =

3

x x dx =

∫−3

(−3)3 (3)3
+
=0
3
3

2

0

1

2

∫−1 x a x b dx = − ∫−1 x dx + 0∫0 x dx + ∫1

x dx

1
1

= − (1)(1) + 1(1) + (1)(1) = 1
2
2

h.

2

0

1

2
2
2
∫−1 x a xb dx = − ∫−1 x dx + 0∫0 x dx
2

29.

30.

t

s(t)

20

20

40

80

60

160

80

240

100

320

120

400

+ ∫ x 2 dx
1

=−

t

s(t)

20

20

32. a.

1

∫−1

13 ⎛ 23 13 ⎞
+⎜ − ⎟ = 2
3 ⎜⎝ 3 3 ⎟⎠
f ( x) dx = 0 because this is an odd

function.

40

60

60

80

80

60

100

0

120

-100

b.

∫−1

1

g ( x ) dx = 3 + 3 = 6

c.

∫−1

1

f ( x) dx = 3 + 3 = 6

d.

∫−1 [ − g ( x)] dx = −3 + (−3) = −6

e.

∫−1

1

1

xg ( x) dx = 0 because xg(x) is an odd

function.
f.

1

∫−1

f 3 ( x) g ( x) dx = 0 because f 3 ( x) g ( x)

is an odd function.

31. a.

b.

33. RP =

3

∫−3a x b dx = (−3 − 2 − 1 + 0 + 1 + 2)(1) = −3
3

∫−3a xb

2

+(−1) 2 + 0 + 1 + 4](1) = 19
⎡1

⎤

c.

∫−3 ( x − a x b) dx = 6 ⎢⎣ 2 (1)(1) ⎥⎦ = 3

d.

∫−3 ( x − a x b)
3

2

1

dx = 6 ∫ x 2 dx = 6 ⋅
0

(

1 n 2
∑ xi − xi2−1
2 i =1

)

1⎡ 2
( x1 − x02 ) + ( x22 − x12 ) + ( x32 − x22 )
⎣
2
+ " + ( xn2 − xn2−1 ) ⎤
⎦
1 2
= ( xn − x02 )
2
1 2
= (b − a 2 )
2
1
1
lim (b 2 − a 2 ) = (b 2 − a 2 )
2
n →∞ 2
=

dx = [(−3)2 + (−2) 2

3

=

1 n
∑ ( xi + xi −1 )( xi − xi −1 )
2 i =1

13
=2
3

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(

)

12

⎡1
⎤
34. Note that xi = ⎢ xi2−1 + xi −1 xi + xi2 ⎥
⎣3
⎦
1/ 2

⎡1
⎤
≥ ⎢ ( xi2−1 + xi2−1 + xi2−1 ⎥
⎣3
⎦

= xi −1 and
1/ 2

⎡1
⎤
xi = ⎢ ( xi2−1 + xi −1 xi + xi2 ⎥
⎣3
⎦
1/ 2

⎡1
⎤
≤ ⎢ ( xi2 + xi2 + xi2 ) ⎥
⎣3
⎦

= xi .

n

R p = ∑ xi2 Δxi
i =1

n

1
= ∑ ( xi2 + xi −1 xi + xi2−1 )( xi − xi −1 )
i =1 3
=

n

1
∑ ( xi3 − xi3−1 )
3 i =1

1
= ⎡ ( x13 − x03 ) + ( x23 − x13 ) + ( x33 − x23 )
3⎣
+ " + ( xn3 − xn3−1 ) ⎤
⎦
1 3
1
= ( xn − x03 ) = (b3 − a3 )
3
3

35. Left:

2

∫0

Right:

( x3 + 1) dx = 5.24
2

∫0

Midpoint:

36. Left:

( x + 1) dx = 6.84
3

2

∫0

37. Left:

1

∫0 cos x dx ≈ 0.8638

Right:

1

∫0 cos x dx ≈ 0.8178

Midpoint:

1

∫0 cos x dx ≈ 0.8418

⎛1⎞

⎜ ⎟ dx ≈ 1.1682
⎝ x⎠
3 ⎛1⎞
Right: ∫ ⎜ ⎟ dx ≈ 1.0349
1 ⎝ x⎠
3 ⎛1⎞
Midpoint: ∫ ⎜ ⎟ dx ≈ 1.0971
1 ⎝ x⎠

38. Left:

3

∫1

39. Partition [0, 1] into n regular intervals, so
1
P = .
n
i 1
If xi = + , f ( xi ) = 1 .
n 2n
n

∑
P →0
lim

i =1

n

1

∑ n =1
n →∞

f ( xi )Δxi = lim

i =1

i 1
If xi = + , f ( xi ) = 0 .
n πn
n

n

i =1

i =1

∑ f ( xi )Δxi = nlim
∑0 = 0
→∞
P →0
lim

Thus f is not integrable on [0, 1].

( x3 + 1) dx = 5.98

1

∫0 tan x dx ≈ 0.5398

Right:

1

∫0 tan x dx ≈ 0.6955

Midpoint:

1

∫0 tan x dx ≈ 0.6146

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4.3 Concepts Review

5. A( x ) =

1

ax 2
x ( ax ) =
2
2

6. A( x ) =

1
1
2
( x − 2)(−1 + x / 2) = ( x − 2 ) , x ≥ 2
2
4

1. 4(4 – 2) = 8; 16(4 – 2) = 32
2. sin 3 x
3.

4

∫1

f ( x) dx ;

5

∫2

x dx

4. 5

Problem Set 4.3
1. A( x) = 2 x

7.

2. A( x) = ax

3. A( x) =

1
2

( x − 1)2 ,

⎧2 x
⎪
⎪⎪2 + ( x − 1)
A( x ) = ⎨3 + 2( x − 2)
⎪5 + ( x − 3)
⎪
⎪⎩etc.

0 ≤ x ≤1
1< x ≤ 2
2< x≤3
3< x≤ 4

x ≥1

8.

4. If 1 ≤ x ≤ 2 , then A( x) =

If 2 ≤ x , then A( x) = x −

1
2
3
2

( x − 1)2 .

⎧ 1 x2
⎪2
⎪ 1 + 1 (3 − x)( x − 1)
⎪2 2
⎪1 + 1 ( x − 2) 2
⎪
A( x ) = ⎨ 2
⎪ 3 + 1 (5 − x)( x − 3)
⎪2 2
⎪2+ 1 ( x − 4)2
⎪ 2
⎪⎩etc.

0 ≤ x ≤1
1< x ≤ 2
2< x≤3

3< x≤ 4
4< x≤5

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9.
10.

2

∫1
2

∫0

2 f ( x) dx = 2∫

2

f ( x) dx = 2(3) = 6

1

2 f ( x) dx = 2∫

2

0

f ( x) dx

1
2
= 2 ⎡⎢ ∫ f ( x) dx + ∫ f ( x)dx ⎤⎥ = 2(2 + 3) = 10
1
⎣ 0
⎦

11.

∫0 [ 2 f ( x) + g ( x)] dx = 2∫0
2

2

2

f ( x) dx + ∫ g ( x) dx
0

1
2
2
= 2 ⎡⎢ ∫ f ( x) dx + ∫ f ( x) dx ⎤⎥ + ∫ g ( x) dx
0

1
0
⎣
⎦
= 2(2 + 3) + 4 = 14

12.

1

1

x
x
22. G ′( x) = Dx ⎡⎢ ∫ xt dt ⎤⎥ = Dx ⎡⎢ x ∫ t dt ⎤⎥
⎣1
⎦
⎣ 1
⎦
x
⎡ ⎡ 2⎤ ⎤
⎡ ⎛ x2 − 1 ⎞⎤
t
= Dx ⎢ x ⎢ ⎥ ⎥ = Dx ⎢ x ⎜
⎟⎥
⎜ 2 ⎟⎥
⎢ ⎢2⎥ ⎥
⎢
⎣
⎦

⎝
⎠
⎣
⎦
1
⎣⎢
⎦⎥

⎛ x3 x ⎞ 3
1
= Dx ⎜ − ⎟ = x 2 −
⎜ 2 2⎟ 2
2
⎝
⎠

1

∫0 [2 f ( s) + g ( s)] ds = 2∫0 f (s) ds + ∫0 g (s) ds
= 2(2) + (–1) = 3

13.

π/4
21. G ′( x) = Dx ⎡⎢ ∫
( s − 2) cot(2 s )ds ⎤⎥
x
⎣
⎦
x

⎡
⎤
= Dx ⎢ − ∫
( s − 2) cot(2 s )ds ⎥
⎣ π/4
⎦
= −( x − 2) cot(2 x)

1

2

∫2 [2 f ( s) + 5 g ( s)] ds = −2∫1

2

f ( s ) ds − 5∫ g ( s ) ds

⎡ x2
⎤
23. G ′( x ) = Dx ⎢ ∫ sin t dt ⎥ = 2 x sin( x 2 )
1
⎢⎣
⎥⎦

1

2
1
= −2(3) − 5 ⎡⎢ ∫ g ( s ) ds − ∫ g ( s ) ds ⎤⎥

0
⎣ 0
⎦
= –6 – 5[4 + 1] = –31

⎡ x2 + x
24. G ′( x ) = Dx ⎢ ∫
1
⎣⎢

⎤
2 z + sin z dz ⎥
⎦⎥

= (2 x + 1) 2( x 2 + x) + sin( x 2 + x)

14.
15.

∫1 [3 f ( x) + 2 g ( x)] dx = 0
1

25.

∫0 [3 f (t ) + 2 g (t )] dt
2

t2

x

G ( x) = ∫ 2
−x

1+ t2
t2

0

=∫ 2
−x

1
2
2
= 3 ⎡⎢ ∫ f (t ) dt + ∫ f (t ) dt ⎤⎥ + 2 ∫ g (t ) dt
1
0
⎣ 0
⎦
= 3(2 + 3) + 2(4) = 23

= −∫

1+ t2

2

∫0

2

+π∫ dt

=

0

= 3 (2 + 3) + 2(4) + 2π = 5 3 + 4 2 + 2π

(

)

x
19. G ′( x) = Dx ⎡⎢ ∫ 2t 2 + t dt ⎤⎥ = 2 x 2 + x
0
⎣
⎦
x
20. G ′( x) = Dx ⎡⎢ ∫ cos3 (2t ) tan(t ) dt ⎤⎥
⎣1
⎦

= cos3 (2 x) tan( x)

0

1+ t2

1+ t2

dt

x

t2

0

1+ t2

dt + ∫

dt

2

1
2
2
= 3 ⎡⎢ ∫ f (t ) dt + ∫ f (t ) dt ⎤⎥ + 2 ∫ g (t ) dt
0
1
0
⎣
⎦

1
x

18. G ′( x) = Dx ⎡⎢ ∫ 2t dt ⎤⎥ = Dx ⎡⎢ − ∫ 2t dt ⎤⎥ = −2 x
x
1
⎣
⎦
⎣
⎦

t2

− x2 )
(
x2
G '( x) = −
−2 x ) +
(
2
1 + x2
1 + ( − x2 )

⎡ 3 f (t ) + 2 g (t ) + π ⎤ dt
⎣
⎦

x
17. G ′( x) = Dx ⎡⎢ ∫ 2t dt ⎤⎥ = 2 x
⎣1
⎦

x

dt + ∫

t2

− x2

0

16.

dt

2 x5
1 + x4

+

x2
1 + x2

sin x 5 ⎤
26. G ( x) = Dx ⎡⎢ ∫
t dt ⎥
cos
x
⎣
⎦
sin x 5
0

⎡
= Dx ⎢ ∫
t dt + ∫
t 5 dt ⎤⎥
0
cos
x
⎣
⎦
sin x 5
cos x 5 ⎤
⎡
= Dx ⎢ ∫
t dt − ∫
t dt ⎥
0
⎣ 0
⎦

= sin 5 x cos x + cos5 x sin x
27.

f ′( x) =

x
1 + x2

; f ′′ ( x ) =

1

( x + 1)
2

3/ 2

So, f(x) is increasing on [0, ∞) and concave up
on (0, ∞ ).

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28.

f ′( x) =

1+ x
1+ x

35.

2

(1 + x ) − (1 + x ) 2 x = − x + 2 x − 1
f ′′ ( x ) =
( x + 1)

( x + 1)
2

2

2

2

2

2

So, f(x) is increasing on [0, ∞ ) and concave up on

( 0, −1 + 2 ) .

29.

f ′ ( x ) = cos x; f ′′ ( x ) = − sin x

4

⎡ π ⎤ ⎡ 3π 5π ⎤
So, f(x) is increasing on ⎢0, ⎥ , ⎢ , ⎥ ,... and
⎣ 2⎦ ⎣ 2 2 ⎦
concave up on (π , 2π ) , ( 3π , 4π ) ,... .
30.

∫0

2

4

0

2

f ( x) dx = ∫ (2 − x)dx + ∫ ( x − 2) dx
= 2+2 = 4

36.

f ′ ( x ) = x + sin x; f ′′ ( x ) = 1 + cos x

So, f(x) is increasing on ( 0, ∞ ) and concave up
on ( 0, ∞ ) .
31.

1
1
; f ′′ ( x ) = − 2
x
x
So, f(x) is increasing on (0, ∞) and never
concave up.

f ′( x) =

32. f(x) is increasing on x ≥ 0 and concave up on
( 0,1) , ( 2,3) ,...

∫0 ( 3 + x − 3 ) dx
3
4
= ∫ ( 3 + x − 3 ) dx + ∫ ( 3 + x − 3 ) dx
0
3
4

3

4

0

3

= ∫ ( 6 − x ) dx + ∫ x dx =

37. a.

33.

27 7
+ = 17
2 2

Local minima at 0, ≈ 3.8, ≈ 5.8, ≈ 7.9,

≈ 9.9;
local maxima at ≈ 3.1, ≈ 5, ≈ 7.1, ≈ 9, 10

b. Absolute minimum at 0, absolute maximum
at ≈ 9
c.

≈ (0.7, 1.5), (2.5, 3.5), (4.5, 5.5), (6.5, 7.5),
(8.5, 9.5)

d.
4

∫0

2

4

0

2

f ( x) dx = ∫ 2 dx + ∫ x dx = 4 + 6 = 10

34.

38. a.

Local minima at 0, ≈ 1.8, ≈ 3.8, ≈ 5.8;

local maxima at ≈ 1, ≈ 2.9, ≈ 5.2, ≈ 10

b. Absolute minimum at 0, absolute maximum
at 10
c.
4

∫0

1

2

4

0

1

2

f ( x) dx = ∫ dx + ∫ x dx + ∫ (4 − x) dx

(0.5, 1.5), (2.2, 3.2), (4.2,5.2), (6.2,7.2),
(8.2, 9.2)

= 1 + 1.5 + 2.0 = 4.5

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d.

39.

a.

f.

0

0

b.

(

)

F (0) = ∫ t 4 + 1 dt = 0

41.

1 ≤ 1 + x4 ≤ 1 + x4 .

y = F ( x)

1

)

y = 15 x5 + x + C

c.

0=

42.

1

+0+C

40.

a.

1

4

∫0 ( 4 + x
1

)

1

G (0) = ∫ sin t dt = 0
G (2π ) = ∫

0

1

(

)

4 + x 2 dx ≤ ∫ 4 + x 2 dx
0

4 + x 2 dx ≤

= 3 + 65 =

43.

21
5

5 ≤ f ( x) ≤ 69 so

( 5 + x3 ) dx ≤ 4 ⋅ 69
20 ≤ ∫ ( 5 + x3 ) dx ≤ 276
0

4⋅5 ≤ ∫

0

2π

6
5

) dx = ∫01( 3 + 1 + x4 ) dx
1
1
= ∫ 3 dx + ∫ (1 + x 4 ) dx
0
0

4

x

G ( x) = ∫ sin t dt

1 + x 4 dx ≤

21
0
5
Here, we have used the result from problem 39:

1

6
+ 1 dx = F (1) = 15 + 1 = .
5
5

0
0

b.

1

2≤∫

Thus y = F ( x) = 15 x5 + x

∫0 ( x

1

On the interval [0,1], 2 ≤ 4 + x 4 ≤ 4 + x 4 .
Thus

∫0 2 dx ≤ ∫0

C=0

d.

0

0

Now apply the initial condition y (0) = 0 :
1 05
5

1

1 + x 4 dx ≤ ∫ (1 + x 4 ) dx

By problem 39d, 1 ≤ ∫

dy = x + 1 dx
4

1

∫0 dx ≤ ∫0

dy
= F '( x) = x 4 + 1
dx

(

t ≤ t . Since 1 + x 4 ≥ 1 for all x,

For t ≥ 1 ,

4

0
4

sin t dt = 0

Let y = G ( x) . Then
dy
= G '( x) = sin x .
dx
dy = sin x dx
y = − cos x + C

c.

d.

e.

Apply the initial condition
0 = y (0) = − cos 0 + C . Thus, C = 1 ,
and hence y = G ( x) = 1 − cos x .
π

∫0 sin x dx = G (π ) = 1 − cos π = 2

44.

On [2,4], 85 ≤ ( x + 6 ) ≤ 105 . Thus,

5

2 ⋅ 85 ≤ ∫

4

2

( x + 6 )5 dx ≤ 2 ⋅105

65,536 ≤ ∫

4

2

( x + 6 )5 dx ≤ 200, 000

G attains the maximum of 2 when
x = π ,3π .
G attains the minimum of 0 when
x = 0, 2π , 4π
Inflection points of G occur at
π 3π 5π 7π
x= , , ,
2 2 2 2

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


45.

On [1,5],
2
2
2
3+ ≤ 3+ ≤ 3+
5
1
x
5⎛
2⎞
⎛ 17 ⎞
4 ⎜ ⎟ ≤ ∫ ⎜ 3 + ⎟ dx ≤ 4 ⋅ 5
1
5
x⎠
⎝ ⎠
⎝
5⎛
68
2⎞
≤ ∫ ⎜ 3 + ⎟ dx ≤ 20
1
5
x⎠

⎝

48.

On [0.2,0.4],
0.002 + 0.0001cos 2 0.4 ≤ 0.002 + 0.0001cos 2 x
≤ 0.002 + 0.0001cos 2 0.2

(

0.2 0.002 + 0.0001cos 2 0.4

)

( 0.002 + 0.0001cos2 x ) dx
≤ 0.2 ( 0.002 + 0.0001cos 2 0.2 )
≤∫

0.4

0.2

Thus,
0.000417 ≤ ∫

0.4

0.2

( 0.002 + 0.0001cos2 x ) dx

≤ 0.000419

46.

On [10, 20],
5

5

1 ⎞
1⎞
⎛
⎛ 1⎞
⎛
⎜1 + ⎟ ≤ ⎜ 1 + ⎟ ≤ ⎜1 + ⎟
20
x
10
⎝
⎠
⎝
⎠
⎝
⎠
5

5

5

20 ⎛
1⎞
⎛ 21 ⎞
⎛ 11 ⎞
10 ⎜ ⎟ ≤ ∫ ⎜ 1 + ⎟ dx ≤ 10 ⎜ ⎟
10
⎝ 20 ⎠
⎝ x⎠
⎝ 10 ⎠

5

5

20 ⎛
4, 084,101
1⎞
161, 051
≤ ∫ ⎜ 1 + ⎟ dx ≤
10
320, 000
10, 000
⎝ x⎠
20 ⎛

49.

47.

1

∫

x 1+ t

51.

∫1

8π

π
(5 + 201 sin 2 x ) dx ≤ 101
5

dt . Then

dt
2+t
1 1+ t
1 ⎡ x 1+ t
⎤
= lim
dt − ∫
dt
⎢∫
0 2 + t ⎥⎦
x →1 x − 1 ⎣ 0 2 + t
F ( x) − F (1)

= lim
x −1
1
→
x
1+1 2
= F '(1) =
=
2 +1 3

( 4π ) (5) ≤ ∫4π ( 5 + 201 sin 2 x ) dx ≤ ( 4π ) ( 5 + 201 )
4π

2+t

50.

1 sin 2 x ≤ 5 + 1
5 ≤ 5 + 20
20

8π

0

lim

On [ 4π ,8π ]

20π ≤ ∫

x 1+ t

1 x 1+ t
F ( x) − F (0)
dt = lim
∫
x−0
x →0 x 0 2 + t
x →0
1+ 0 1
= F '(0) =
=
2+0 2

5

1⎞
1 + ⎟ dx ≤ 16.1051
⎜
10 ⎝
x⎠

12.7628 ≤ ∫

Let F ( x) = ∫

lim

x →1 x − 1 1

x

f (t ) dt = 2 x − 2

Differentiate both sides with respect to x:
d x
d
f (t ) dt = ( 2 x − 2 )
dx ∫1
dx
f ( x) = 2
If such a function exists, it must satisfy
f ( x) = 2 , but both sides of the first equality
may differ by a constant yet still have equal
derivatives. When x = 1 the left side is
1

∫1 f (t ) dt = 0 and the right side is 2 ⋅1 − 2 = 0 .
Thus the function f ( x) = 2 satisfies
x

∫1

f (t ) dt = 2 x − 2 .

268
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


52.

x

∫0

f (t ) dt = x 2

59.

Differentiate both sides with respect to x:
d x
d 2
f (t ) dt =
x
∫
0
dx
dx
f ( x) = 2 x

53.

x2

∫0

f (t ) dt =

60.

Differentiate both sides with respect to x:

( )

d x2
d 1 3
f (t ) dt =
x
∫
0
dx
dx 3

( ) ( 2x) = x
x
f ( x2 ) =
2

f x

2

2

f ( x) =

No such function exists. When x = 0 the left
side is 0, whereas the right side is 1

55.

True; by Theorem B (Comparison Property)

56.

False. a = –1, b = 2, f(x) = x is a
counterexample.

57.

False. a = –1, b = 1, f(x) = x is a
counterexample.

62.

b

False. a = 0, b = 1, f(x) = 0, g(x) = –1 is a
counterexample.
⎧⎪2 + ( t − 2 ) , t ≤ 2
v (t ) = ⎨
⎪⎩ 2 − ( t − 2 ) , t > 2
t≤2
⎧ t,
=⎨
−

>2
4
,
t
t
⎩
s ( t ) = ∫ v ( u ) du
t

⎧ t
0≤t ≤2
⎪ ∫0 u du ,
=⎨
t
2
⎪ u du + ( 4 − u ) du, t > 2
∫2
⎩ ∫0
⎧t2
0≤t≤2
⎪ ,
⎪2
=⎨
2
⎪2 + ⎡⎢ 4t − t ⎤⎥ , t > 2
⎪ ⎢
2 ⎥⎦
⎩ ⎣

x

2

False; A counterexample is f ( x ) = 0 for all x,
except f (1) = 1 . Thus,

b

0

54.

58.

∫ a f ( x)dx − ∫ a g ( x)dx

= ∫ ba [ f ( x) − g ( x )]dx

61.

1 x3
3

True.

⎧
t2
,
⎪
⎪
2

=⎨
t2
⎪
⎪⎩−4 + 4t − 2

0≤t≤2
t>2

t2
− 4t + 4 = 0; t = 4 + 2 2 ≈ 6.83
2

∫0 f ( x ) dx = 0 , but f is
2

not identically zero.
⎧ t
⎪ ∫ 5 du,
0 ≤ t ≤ 100
⎪ 0
⎪⎪ 100
t ⎛
u ⎞
du
100 < t ≤ 700
a. s ( t ) = ⎨ ∫ 5 du + ∫ ⎜ 6 −
0
100 ⎝
100 ⎟⎠
⎪

⎪ 100
700 ⎛
t
u ⎞
du + ∫ ( −1) du, t > 700
⎪ ∫ 5 du + ∫ ⎜ 6 −
⎟
0
100
700
100 ⎠
⎝
⎩⎪
⎧
⎪
⎪5t ,
0 ≤ t ≤ 100
⎪
t
⎪
⎡
u2 ⎤
⎪
= ⎨500 + ⎢6u −
100 < t ≤ 700
⎥
200 ⎦⎥
⎪
⎣⎢
100

⎪
700
2
⎪
⎡
u ⎤
⎪500 + ⎢6u −
⎥ − ( t − 700 ) t > 700
200 ⎦⎥
⎪⎩
⎣⎢
100
0 ≤ t ≤ 100
⎧5t ,
⎪
2
t
⎪
= ⎨−50 + 6t −
, 100 < t ≤ 700
200
⎪
⎪2400 − t ,
t > 700
⎩

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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


b. v ( t ) > 0 for 0 ≤ t < 600 and v ( t ) < 0 for
t > 600 . So, t = 600 is the point at which
the object is farthest to the right of the origin.
At t = 600 , s ( t ) = 1750 .

c. s ( t ) = 0 = 2400 − t ; t = 2400
− f ( x ) ≤ f ( x) ≤ f ( x) , so

63.

b

∫a −
b

∫a

b

f ( x) dx ≤ ∫ f ( x) dx ⇒

∫−1 (3x

4.

∫1

5.

3
⎡ 1⎤
⎛ 1⎞
∫1 w2 dw = ⎢⎣− w ⎥⎦1 = ⎜⎝ − 4 ⎟⎠ − (−1) = 4

6.

⎡ 1⎤
8
⎛ 1⎞
∫1 t 3 dt = ⎢⎣ − t 2 ⎥⎦ = ⎜⎝ − 9 ⎟⎠ − (−1) = 9
1

b

f ( x ) dx ≥ ∫ f ( x) dx,
a

we can conclude that
b

∫a

f ( x) dx ≤ ∫

b

f ( x) dx

a

If x > a ,

64.

3.

2

2

2

(4 x3 + 7) dx = ⎡ x 4 + 7 x ⎤
⎣
⎦1
= (16 + 14) – (1 + 7) = 22

x

∫a

f ′( x ) dx ≤ M ( x − a) by the

7.

∫x

f ( x) dx = − ∫

x

a

f ′( x) dx ≥ − M ( x − a ) by

8.

4

1

4

3

2

3

4

4

16
⎡2
⎤
⎛2 ⎞

t dt = ⎢ t 3 / 2 ⎥ = ⎜ ⋅ 8 ⎟ − 0 =
3
⎣3
⎦0 ⎝ 3 ⎠

8 3
1

⎡3
⎤
⎛3
⎞ ⎛ 3 ⎞ 45
w dw = ⎢ w4 / 3 ⎥ = ⎜ ⋅16 ⎟ − ⎜ ⋅1⎟ =
⎝4
⎠ ⎝4 ⎠ 4
⎣4
⎦1

∫0

Boundedness Property. If x < a ,
a

2

− 2 x + 3) dx = ⎡ x3 − x 2 + 3 x ⎤
⎣
⎦ −1
= (8 – 4 + 6) – (–1 –1 – 3) = 15
2

b

f ( x ) dx ≥ − ∫ f ( x) dx

and combining this with
b

⎡ x5 ⎤
32 1 33
4
∫−1 x dx = ⎢⎢ 5 ⎥⎥ = 5 + 5 = 5
⎣ ⎦ −1
2

a

a

∫a

2

2.

∫

8

the Boundedness Property. Thus

x

∫a

f ′( x) dx ≤ M x − a .

From Problem 63,
x

∫a

x

∫a

f ′( x) dx ≥

9.
x

∫a

f ′( x) dx .

f ′( x ) dx = f ( x) − f (a) ≥ f ( x) − f (a)

Therefore, f ( x) − f (a) ≤ M x − a or
f ( x) ≤ f (a ) + M x − a .

10.

−2

⎡ y3
⎛ 2 1 ⎞
1 ⎤
∫−4 ⎜⎜ y + y3 ⎟⎟ dy = ⎢⎢ 3 − 2 y 2 ⎥⎥
⎝
⎠
⎣
⎦ −4
⎛ 8 1 ⎞ ⎛ 64 1 ⎞ 1783
= ⎜− − ⎟−⎜− − ⎟ =
96
⎝ 3 8 ⎠ ⎝ 3 32 ⎠
−2

4

∫1

s4 − 8
s2

4

4

ds = ∫ ( s − 8s
2

−2

1

⎡ s3 8 ⎤
) ds = ⎢ + ⎥
⎢⎣ 3 s ⎥⎦1

⎛ 64
⎞ ⎛1
⎞
= ⎜ + 2 ⎟ − ⎜ + 8 ⎟ = 15
⎝ 3
⎠ ⎝3 ⎠

4.4 Concepts Review

π/2

cos x dx = [sin x ]0

π/2

2sin t dt = [ −2 cos t ]π / 6 = 0 + 3 = 3

11.

∫0

12.

∫π / 6

π/2

=1–0=1

1. antiderivative; F(b) – F(a)
2. F(b) – F(a)
3. F (d ) − F (c )
13.
4.

∫

2

1

1 4
u du
3

Problem Set 4.4
14.
2

⎡ x4 ⎤
1. ∫ x dx = ⎢ ⎥ = 4 − 0 = 4

0
⎣⎢ 4 ⎦⎥ 0
2

3

π/2

1

⎡2 5
⎤
4
2
3
∫0 (2 x − 3x + 5) dx = ⎢⎣ 5 x − x + 5 x ⎥⎦0
22
⎛2
⎞
= ⎜ −1+ 5⎟ − 0 =
5
5
⎝
⎠
1

1

⎡3 7/3 3 4/3⎤
4/3

1/ 3
∫0 ( x − 2 x ) dx = ⎢⎣ 7 x − 2 x ⎥⎦ 0
15
⎛3 3⎞
= ⎜ − ⎟−0 = −
7
2
14
⎝
⎠
1

270
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Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


15. u = 3x + 2, du = 3 dx
1
2 3/ 2
2
3/ 2
∫ u ⋅ 3 du = 9 u + C = 9 (3x + 2) + C
16. u = 2x – 4, du = 2 dx
3 4/3
3
1/ 3 1
4/3

∫ u ⋅ 2 du = 8 u + C = 8 (2 x − 4) + C
17. u = 3x + 2, du = 3 dx
1
1
1
∫ cos(u) ⋅ 3 du = 3 sin u + C = 3 sin(3x + 2) + C
18. u = 2x – 4, du = 2 dx
1
1
∫ sin u ⋅ 2 du = − 2 cos u + C
1
= − cos(2 x − 4) + C
2
19. u = 6x – 7, du = 6dx
1
1
∫ sin u ⋅ 6 du = − 6 cos u + C
1
= − cos(6 x − 7) + C
6
20. u = πv − 7, du = π dv
1
1
1
∫ cos u ⋅ π du = π sin u + C = π sin(πv − 7) + C

25. u = x 2 + 4, du = 2 x dx
1

1

= − cos( x 2 + 4) + C
2

26. u = x3 + 5, du = 3 x 2 dx

1

27. u = x 2 + 4, du =

22. u = x3 + 5, du = 3 x 2 dx
1
1 10
1 3
10
∫ u ⋅ 3 du = 30 u + C = 30 ( x + 5) + C
9

1

x
x +4
2

3

+ 5) + C

dx

∫ sin u du = − cos u + C = − cos

x2 + 4 + C

2z

3

28. u = z 2 + 3, du =

dz
2
3 2
⎛
⎞
3⎜ z + 3 ⎟
⎝
⎠
3
3
3
3 2
∫ cos u ⋅ 2 du = 2 sin u + C = 2 sin z + 3 + C

29. u = ( x3 + 5)9 ,
du = 9( x3 + 5)8 (3x 2 )dx = 27 x 2 ( x3 + 5)8 dx
1

1

∫ cos u ⋅ 27 du = 27 sin u + C

21. u = x 2 + 4, du = 2 x dx

1
1
1
u ⋅ du = u 3 / 2 + C = ( x 2 + 4)3 / 2 + C
2
3
3

1

∫ cos u ⋅ 3 du = 3 sin u + C = 3 sin( x

=

∫

1

∫ sin(u ) ⋅ 2 du = − 2 cos u + C

1
sin ⎡( x3 + 5)9 ⎤ + C
⎣
⎦
27

30. u = (7 x 7 + π)9 , du = 441x 6 (7 x7 + π)8 dx

1

1

∫ sin u ⋅ 441 du = − 441 cos u + C
=−

1
cos(7 x 7 + π)9 + C
441

31. u = sin( x 2 + 4), du = 2 x cos( x 2 + 4) dx

23. u = x + 3, du = 2 x dx
2

7
−12 / 7 1
⋅ du = − u −5 / 7 + C
∫u
2
10
7 2
= − ( x + 3)−5 / 7 + C
10

1
1
u ⋅ du = u 3 / 2 + C
2

3
3/ 2
1⎡
= sin( x 2 + 4) ⎤
+C
⎣
⎦
3

∫

32. u = cos(3 x7 + 9)
24. u = 3 v + π, du = 2 3v dv
2

∫u
=

7/8

4
15

⋅

1
2 3

(
3

du =

4
15 3

3 v2 + π

)

15 / 8

u15 / 8 + C

+C

du = −21x 6 sin(3 x7 + 9) dx
1
⎛ 1 ⎞
u ⋅ ⎜ − ⎟ du = − u 4 / 3 + C
21
28
⎝
⎠
4/3
1 ⎡
=−
+C
cos(3x 7 + 9) ⎤
⎦

28 ⎣

∫

3

Instructor’s Resource Manual
Section 4.4
271
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


33.

u = cos( x3 + 5), du = −3 x 2 sin( x3 + 5) dx

= 2∫

u = tan( x −3 + 1) , du = −3 x −4 sec2 ( x −3 + 1) dx

⎡4
⎤ ⎡4
⎤
= ⎢ (125) ⎥ − ⎢ (125) ⎥ = 0
⎣3
⎦ ⎣3
⎦

3

∫−3

9

7 + 2t 2 (8t ) dt = 2 ∫
25

25

5
⎛ 1⎞
u ⋅ ⎜ − ⎟ du = − u 6 / 5 + C
3
18
⎝
⎠
6/5
5 ⎡
=−
+C
tan( x −3 + 1) ⎤
⎣
⎦
18

∫

35.

u = 7 + 2t 2 , du = 4t dt

1
⎛ 1⎞
⋅ ⎜ − ⎟ du = − u10 + C
3
30
⎝
⎠
1
= − cos10 ( x3 + 5) + C
30

∫u

34.

41.

42.

x2 + 1

∫1
=

x3 + 3 x

43.

1 16 −1/ 2
⎡2
⎤
u
du = ⎢ u1/ 2 ⎥
∫
4
3
⎣3
⎦4

u = cos x, du = − sin x dx
π/2

∫0

u = x3 + 1, du = 3 x 2 dx

∫−1

x3 + 1 (3x 2 ) dx = ∫

1

0

u = t + 2, du = dt
1

5 −2

u du
1

∫−1 (t + 2)2 dt = ∫

⎡ 1⎤
= ⎢− ⎥
⎣ u ⎦1

π/2

=

39.

9

1 π /2 2
1 −1
sin 3 x ( 3cos 3 x ) dx = ∫ u 2 du
∫
3 0
3 0

⎡ u3 ⎤
1
⎛ 1⎞
= ⎢ ⎥ = ⎜− ⎟−0 = −
9
⎝ 9⎠

⎢⎣ 9 ⎥⎦ 0

9

45.

1

=∫

u = 3x + 1, du = 3 dx
8
1 8
1 25
∫5 3x + 1 dx = 3 ∫5 3x + 1 ⋅ 3dx = 3 ∫16 u du

11

02

2

+ 2 x)2 dx

( x 2 + 2 x)2 2( x + 1) dx
3

=

25

u = 2x + 2, du = 2 dx
7
1
1 7
2
∫1 2 x + 2 dx = 2 ∫1 2 x + 2 dx
16
1 16
= ∫ u −1/ 2 du = ⎡⎣ u ⎤⎦ = 4 − 2 = 2
4
2 4

u = x 2 + 2 x, du = (2 x + 2) dx = 2( x + 1) dx

∫0 ( x + 1)( x

⎡2
⎤
⎡2
⎤ ⎡2
⎤ 122
= ⎢ u 3 / 2 ⎥ = ⎢ (125) ⎥ − ⎢ (64) ⎥ =
9
9
9
9
⎣
⎦16 ⎣
⎦ ⎣

⎦

40.

sin 2 3 x cos 3x dx

−1

⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
1
3
⎣
⎦1
2
2
52
⎡
⎤ ⎡
⎤
= ⎢ (27) ⎥ − ⎢ (1) ⎥ =
⎣3
⎦ ⎣3 ⎦ 3
y − 1 dy = ∫

u = sin 3 x, du = 3cos 3 x dx

∫0

u = y – 1, du = dy
10

cos 2 x ( − sin x ) dx

0

4
⎡ 1⎤
= ⎢ − ⎥ − [ −1] =
5
⎣ 5⎦

∫2

π /2

0

0 2
u du
1

44.

5

cos 2 x sin x dx = − ∫

⎡ u3 ⎤

= −∫
= ⎢− ⎥
⎢⎣ 3 ⎥⎦1
⎛ 1⎞ 1
= 0−⎜− ⎟ =
⎝ 3⎠ 3

1

⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
3
⎣
⎦0

⎛2
⎞ ⎛2 ⎞ 2
= ⎜ ⋅13 / 2 ⎟ − ⎜ ⋅ 0 ⎟ =
⎝3
⎠ ⎝3 ⎠ 3

3

1 3 3x2 + 3
dx
3 ∫1 x3 + 3 x

⎛2 ⎞ ⎛2 ⎞ 8
= ⎜ ⋅6⎟ − ⎜ ⋅ 2⎟ =

⎝3 ⎠ ⎝3 ⎠ 3

2047
⎡1
⎤ ⎡1
⎤
= ⎢ (2)11 ⎥ − ⎢ (1)11 ⎥ =
11
11
11
⎣
⎦ ⎣
⎦

0

dx =

36

u = x 2 + 1, du = 2 x dx
2

38.

⎡4
⎤
u du = ⎢ u 3 / 2 ⎥
3
⎣

⎦ 25

u = x3 + 3x, du = (3 x 2 + 3) dx
3

⎡ u11 ⎤
2 10
2
10
∫0 ( x + 1) (2 x)dx = ∫1 u du = ⎢⎢ 11 ⎥⎥
⎣
⎦1

37.

7 + 2t 2 ⋅ ( 4t ) dt

−3
25

5

1

36.

3

⎡ u3 ⎤
1 3 2

9
u du = ⎢ ⎥ =
∫
2 0
6
2
⎣⎢ ⎦⎥
0

46.

u = x − 1, du =
4

∫1

( x − 1)3
x

1
2 x

dx = 2 ∫

4

1

dx
( x − 1)3

2 x

dx

1

1 3
u du
0

= 2∫

⎡u4 ⎤
1
= 2⎢ ⎥ =
⎢⎣ 4 ⎥⎦ 0 2

272
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Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


47.

u = sin θ , du = cos θ dθ
1/ 2 3
u du
0

∫
48.

3/2

1

50.

⎡u
=⎢ ⎥
⎣⎢ 4 ⎦⎥ 0

=

56.

u = π sin θ , du = π cos θ dθ
1 π
1
π
cos u du = [ sin u ]−π = 0
π ∫−π
π

1
1
−0 =
64

64

u = cos θ , du = − sin θ dθ

−∫
49.

4 ⎤1/ 2

57.

1 cos1 3
1 ⎡u4 ⎤
− ∫
u du = − ⎢ ⎥
2 1
2 ⎣⎢ 4 ⎦⎥

3/2
1
1⎛4 ⎞ 1
u −3 du = ⎡u −2 ⎤
= ⎜ − 1⎟ =
⎣
⎦
1
2
2⎝3 ⎠ 6

u = 3 x − 3, du = 3dx

1 0
1
1
0
cos u du = [sin u ]−3 = (0 − sin(−3))
3 ∫−3
3
3
sin 3
=
3

u = 2πx, du = 2πdx
1 π
1
1
π
sin u du = − [ cos u ]0 = − (−1 − 1)
2π ∫0
2π
2π
1
=
π

u = cos( x 2 ), du = −2 x sin( x 2 )dx
cos1

=−

1

=

58.

cos 4 1 1
+
8
8

1 − cos 1
8
4

u = sin( x3 ), du = 3x 2 cos( x3 )dx
3

1 sin( π3 / 8) 2
1 ⎡ 3 ⎤ sin( π / 8)
3 / 8) u du = ⎣u ⎦
∫
sin(
−
π
− sin( π3 / 8)
3
9
3
2sin 3 ⎛⎜ π8 ⎞⎟

⎝ ⎠
=
9

59.

a. Between 0 and 3, f ( x) > 0 . Thus,
3

51.

∫0 f ( x) dx > 0 .

u = πx 2 , du = 2πx dx
1 π
1
1
π
sin u du = − [ cos u ]0 = − (−1 − 1)
2π ∫0
2π
2π
1
=
π

52.

b. Since f is an antiderivative of f ' ,
3

∫0 f '( x) dx = f (3) − f (0)
= 0 − 2 = −2 < 0

u = 2 x5 , du = 10 x 4 dx

c.

1 2π5
1
2 π5
cos u du = [sin u ]0
∫
10 0
10
1
1
= (sin(2π5 ) − 0) = sin(2π5 )
10
10

53.

54.

u = 2 x, du = 2dx
1 π/ 2
1 π/2
cos u du + ∫
sin u du

2 ∫0
2 0
1
π/2 1
π/2
= [sin u ]0 − [ cos u ]0
2
2
1
1
= (1 − 0) − (0 − 1) = 1
2
2

= −1 − 0 = −1 < 0
d. Since f is concave down at 0, f ''(0) < 0 .
3

∫0 f '''( x) dx = f ''(3) − f ''(0)
= 0 − (negative number) > 0

60.

a. On [ 0, 4] , f ( x) > 0 . Thus,

u = cos x, du = − sin x dx

4

∫0

f ( x) dx > 0 .

b. Since f is an antiderivative of f ' ,
4

∫0

u = 3 x, du = 3dx; v = 5 x, dv = 5dx
1 3π / 2
1 5π / 2
cos u du + ∫
sin v dv
∫
3
/
2
−
π
3
5 −5π / 2
1
1
3π / 2
5π / 2
= [sin u ]−3π / 2 − [ cos v ]−5π / 2
3
5
1
1

2
= [(−1) − 1] − [0 − 0] = −
3
5
3

55.

3

∫0 f ''( x) dx = f '(3) − f '(0)

f '( x) dx = f (4) − f (0)
= 1 − 2 = −1 < 0

c.

4

∫0

f ''( x) dx = f '(4) − f '(0)
=

d.

4

∫0

1
9
− (−2) = > 0
4
4

f '''( x) dx = f ''(4) − f ''(0)
= ( negative ) − ( positive ) < 0

− ∫ sin u du = [ cos u ] = 1 − cos1
0

1

0
1

Instructor’s Resource Manual
Section 4.4
273
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