Thomas Calculus 13th [Solutions]

Section 10.3 The Integral Test 725
46. No, if
n 1
n
a is a convergent series of positive numbers, then 2 a 2a also converges, and
n
n
n 1 n 1
2 a a . There is no largest convergent series of positive numbers.
n
47. (a) Both integrals can represent the area under the curve f ( x )
1
, and the sum s 50 can be considered an
(b)
x 1
approximation of either integral using rectangles with x 1. The sum
51
1
1 x 1
n
50
50
n 1
1
n 1
s is an overestimate
of the integral dx . The sum s 50 represents a left-hand sum (that is, the we are choosing the
left-hand endpoint of each subinterval for c i ) and because f is a decreasing function, the value of f is
a maximum at the left-hand endpoint of each subinterval. The area of each rectangle overestimates the true
area, thus
51
50
1 1
1 x 1 n 1
n 1
dx . In a similar manner, s 50 underestimates the integral
50
1 dx . In
0 x 1
this case, the sum s 50 represents a right-hand sum and because f is a decreasing function, the value of f is
a minimum at the right-hand endpoint of each subinterval. The area of each rectangle underestimates the
true area, thus
50
n 1
2 52 2 2 11.6 and
1
50
1
n 1 0 x 1
dx . Evaluating the integrals we find
50
1
50
0 x 1 0
51
1
51
dx 2 x 1
1 x 1 1
dx 2 x 1 2 51 2 1 12.3. Thus, 11.6 12.3.
n 1
1
n 1
sn
dx x n n
1 x 1 1
n
1000 2 1 2 1 2 2 1000 500 2 1 251414.2
251415.
2
50
n 1
1
n 1
30
1
n 1 n
1
n
n 1
1
n
n 31
48. (a) Since we are using s 30 to estimate , the error is given by . We can consider this
4
4
4
sum as an estimate of the area under the curve f ( x ) when x 30 using rectangles with x 1 and
4
x
c is the right-hand endpoint of each subinterval. Since f is a decreasing function, the value of f is a
i
minimum at the right-hand endpoint of each subinterval, thus dx lim dx
4 4 4
lim lim 1.23 10 . Thus the error
b
1 1 b 1
n 30 x 30 x
n 31
b
1 b
1 1
5
5
1.23 10 .
3 3 3
3x
30 b 3b
3(30)
1 1 b
1 1 b
n x n x b n x b 3x
n
(b) We want S sn
0.000001 dx 0.000001 dx lim dx lim
4 4 4 3
1 1 1 n 3 1000000
3 3 3
3b 3n 3n
3
lim 0.000001 69.336 n 70.
b
1 1 b
1 1 b
1 1
n x n x b n x b 2x n b 2b 2n
8
1 0.01 n 50 7.071 n 8 S s 1
2 3
8 1.195
2n
n
n 1
49. We want S sn
0.01 dx 0.01 dx lim dx lim lim
3 3 3 2 2 2
1
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