Thomas Calculus 13th [Solutions]

610 Chapter 8 Techniques of Integration
1 ( 1)
II. (a) For n 4, x b a
2 1 x 1 ;
n 4 4 2 3 6
mf ( t ) 12 1
i S (12) 2;
6
(3) (4)
f ( t) 6 f ( t) 0 M 0 E s 0
1 3 1 3
(b) t 1 dt 2 E 1
1
s t dt S
1
2 2 0
(c)
E s
True Valule
100 0%
t i f ( ti
) m mf ( t i )
t0
1 0 1 0
t1
1/2 7/8 4 7/2
t2
0 1 2 2
t3
1/2 9/8 4 9/2
t4
1 2 1 2
2
7. 1 ds
s i f ( s ) m
i
mf ( s
1
2
i )
s
I. (a) For n 4, x b a 2 1 1 x 1 ;
s0
1 1 1 1
n 4 4 2 8
179,573 1 179,573 179,573 s
mf ( si
)
T
1 5/4 16/25 2 32/25
44,100 8 44,100 352,800
0.50899; f ( s) 1 f ( s)
2
s2
3/2 4/9 2 8/9
2 3
s
s
s
f ( s) 6
3 7/4 16/49 2 32/49
M 6 f (1)
4
s
s4
2 1/4 1 1/4
2 1 1
2
E (6) 1
T
0.03125
12 4 32
2 2 2
2
2
(b) 1 1 1 1 1 1 1 0.50899 0.00899
1 ds 2
1 s ds s 2 1 2 E T
1
ds T 2
s
1
s
2
E T 0.00899
E T
(c) 100 0.00899 100 2%
True Value 0.5
II. (a) For n 4, x b a 2 1 1 x 1 ;
n 4 4 3 12
264,821 1 264,821
mf ( si
)
S
44,100 12 44,100
264,821
0.50042;
529,200
(3) (4)
f ( s) 24 f ( s) 120 M 120
5 6
s
s
2 1 1
4
E 1
s
(120) 0.00260
180 4 384
s i f ( si
) m mf ( s i )
s0
1 1 1 1
s1
5/4 16/25 4 64/25
s2
3/2 4/9 2 8/9
s3
7/4 16/49 4 64/49
s4
2 1/4 1 1/4
2
(b) 1
2
ds 1 E 1 1
1
2 2 s ds S 0.50042 0.00042 E 0.00042
1
2 2
s
s
s
E s
(c) 100 0.0004 100 0.08%
True Value 0.5
4
8. 1 ds
2
2
( s 1)
I. (a) For n 4, x b a 4 2 1 x 1 ;
n 4 2 2 4
mf ( s 1269 1 1269 1269
i ) T
0.70500;
450 4 450 1800
2
f ( s) ( s 1) f ( s) 2 f ( s)
6
3 4
( s 1) ( s 1)
ET
4 2 1
2
(6) 1 0.25 M 6
12 2 4
s i f ( si
) m mf ( s i )
s0
2 1 1 1
s1
5/2 4/9 2 8/9
s2
3 1/4 2 1/2
s3
7/2 4/25 2 8/25
s4
4 1/9 1 1/9
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