Thomas Calculus 13th [Solutions]

Section 8.8 Improper Integrals 625
73. (a)
1
1
0 t (1 t) dt
1
With u t and du dt the limits of integration are unchanged.
2 t
0
1 1
1 2
dt
t (1 t) 1 u
0
0
2
1 1
2 lim tan 1 tan
a
du
a
2 4 2
(b)
1
0 t (1 t) dt
1
With u t and du dt the limits of integration are unchanged. We split the integral into two
2 t
integrals, the first of which was evaluated in (a).
0
1
1 2 2
dt du du
t (1 t) 1 u 1 u
2
2 2
0 1
1 1
2 lim tan b tan 1
b
2
2 2 2
74. Let c be any number in (3, ).
c
1 1 1
dx dx dx
x x 9 x x 9 x x 9
2 2 2
3 3
c
provided both integrals on the right converge.
1 1 1 x
Formula 20 in Table 8.1 gives
dx sec .
2
x x 9 3 3
Section 3.9.) Both integrals do converge:
(The definition of the inverse secant is given in
3
c
x
1 1 1 c 1 1 a 1 1 c
dx lim sec sec sec
2
x 9 a 3 3 3 3 3 3 3
c
x
1 1 1 b 1 1 c 1 1 c
dx lim sec sec sec
2
x 9
b 3 3 3 3 6 3 3
Thus
3
x
1
2
x
dx
9
.
6
Copyright
2014 Pearson Education, Inc.