Thomas Calculus 13th [Solutions]

344 Chapter 5 Integration
3. f ( x ) 1
Since f is decreasing on [1, 5], we use left endpoints to
x
obtain upper sums and right endpoints to obtain lower
sums.
2
(a) x
5 1 2 and x 1 1 2
2
i i x i a lower sum is 1
2 2
1 1 16
x i 3 5 15
i 1
4
(b) x
5 1
1 and x 1 1
4
i i x i a lower sum is 1 1 1 1 1 1 1
x i 2 3 4 5
i 1
1
5 1
(c) x 2 and x 1 1 2
2
i i x i an upper sum is
1
2 2 1
1 8
x i
3 3
i 0
3
5 1
(d) x 1 x 1 1
4
i i x i an upper sum is 1
1 1 1 1 1 1
x i
2 3 4
i 0
77
60
25
12
4.
f ( x) 4
2
x Since f is increasing on [ 2, 0] and decreasing on
[0, 2], we use left endpoints on [ 2, 0] and right
endpoints on [0, 2] to obtain lower sums and use right
endpoints on [ 2, 0] and left endpoints on [0, 2] to
obtain upper sums.
(a)
(b)
(c)
(d)
2 ( 2)
2
2 i
x
2 ( 2)
4
1 an i 2
1((4
2
( 2) ) (4
2
( 1) ) (4
2
1 )
x
2 ( 2)
2
2 and i 2
x
2 ( 2)
4
1 and i 2
2 2
x and x 2 i x 2 2i a lower sum is 2 (4 ( 2) ) 2 (4 2 ) 0
2 2 2
1((4 ( 1) ) (4 0 ) (4 0 )
x i x 2 i a lower sum is
2
(4 2 )) 6
i
1
0
2
(4 ( x ) ) 1
x i x 2 2i an upper sum is 2 (4 (0) )
x i x 2 i an upper sum is
2
(4 1 )) 14
2
i 1
i
2
2
(4 ( x ) ) 1
i
i
4
3
2
(4 ( x ) ) 1
2
2 (4 0 ) 16
3
2
xi
i 2
i
(4 ( ) ) 1
5.
2
f ( x)
x 1 0
Using 2 rectangles x
1
2 2
2 2
1
f
1
f
3 1 1 3 10 5
2 4 4 2 4 4 32 16
1 0
x
1
4 4
1 f 1 f 3 f 5 f 7
4 8 8 8 8
Using 4 rectangles
1 1
2
3
2
5
2
7
2
21
4 8 8 8 8 64
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