Thomas Calculus 13th [Solutions]

Chapter 5 Additional and Advanced Exercises 425
15.
16.
2 1 1 2 2
f ( x) dx dx (1 x ) dx 2 dx
2 2 1 1
3 1
[ 1 2
x ] 2 x x
3
[2 x ] 1
1
3 ( 1)
( 1 ( 2)) 1 1 1 3
2(2) 2(1)
3 3
1 2 2 4 2 13
3 3 3
2 0 1 2 2
h( r) dr r dr (1 r ) dr dr
1 1 0 1
2 0 3 1
r
2
r r [ r]
2 3 1
1 0
2
( 1) 3
0 1 1 0 (2 1) 1 2 1 7
2 3
2 3 6
2
17. Ave. value 1
b
( ) 1 ( ) 1
b a a f x dx 2 0 0
f x dx 1 2
2 0 x dx 1 ( x 1) dx 2 1 2
1 x 1 x x
2
2 2
0
2 2
1
2 2 2
1 1 0 2 2 1 1 1
2 2 2 2 2
18. Ave. value
1
b
1
3
( ) ( )
b a a f x dx 3 0 0
f x dx 1
1 2 3
dx 0dx dx 1[1 0 0 3 2] 2
3 0 1 2 3 3
19.
lim b
1
1 1 1 1 1
0 lim [sin ] b
dx x
2
0 lim (sin b sin 0) lim (sin b 0) lim sin b
1
2
b 1 x b 1 b 1 b 1 b 1
20.
x 1
1
x 1
tan t dt
0
lim tan t dt lim form
x
x
0
x
lim tan x
x
1 2
1
x
1 1 1 1 1 1 1 1 1
21. lim
lim
1 2 2 1 1 1
n
n n n n
n 1 n 1 2 n 1
n n n n
which can be interpreted as a
Riemann sum with partitioning
x 1 1 1 1
1
1
1
n lim n 1 n 2 2n 0 1 x
dx [ln(1 x)]
0 ln 2
n
22.
1 1/ n 2/ n 1 (1/ n) 1 2(1/ n) 1 n(1/ n)
lim [ e e e] lim e e e which can be interpreted as a
n
n
n
n n n
1/ 2/ 1
1
Riemann sum with partitioning x 1 lim 1 [ n n ] x
0
[ x
n n
e e e e dx e ] 0 e 1
n
5
23. Let f ( x)
x on [0, 1]. Partition [0, 1] into n subintervals with x 1 0 1 .
n n
Then 1 , 2 , , n are the right-hand
n n n
5
endpoints of the subintervals. Since f is increasing on [0, 1], U j 1
n n
is the upper sum for 5
f ( x)
x on
j 1
j
5 5 5 5
5 5 5
[0, 1] lim 1 lim 1 1 2
n lim 1 2 n
1 6
x 1
5 dx x 1
n n n n n n 6
n n
j 1
n n
0 6
0
6
Copyright
2014 Pearson Education, Inc.