Thomas Calculus 13th [Solutions]

Chapter 5 Additional and Advanced Exercises 429
46. (a) If f is continuously differentiable on a, b , then so is the function g( x) ( x c) f ( x).
So the two
integrals on the right side exist and
c b b
( x c) f ( x) dx ( x c) f ( x) dx ( x c) f ( x)
dx
a c a
b b b b
x f ( x) dx c f ( x) dx x f ( x) dx c(0) x f ( x)
dx
a a a a
(b) Split the right side in part (a) into two integrals and write it as t f ( c t) dt t f ( c t) dt . For the
0 0
first integral above, use the substitution t c x, x c t, dt dx;
when t 0, x c and when
t ( b a) / 2, x ( a b) / 2 ( b a) / 2 a . (Note that when x is in a, c , c x is positive and thus
c
x agrees in sign with t.)
0
c
c
( ) ( ) ( ) ( )( ) ( ) ( ) .
0
t f c t dt t f c t dt a
x c f x dx a
x c f x dx
c
Thus the first integral above is equal to ( x c) f ( x) dx . For the second integral above, use the
a
substitution t x c,
c t x and dt dx ; when t 0, x c and when t ( b a) / 2,
b
x ( b a) / 2 ( a b) / 2 b . Thus the second integral above is equal to ( x c) f ( x)
dx and
c
b c b
xf ( x) dx ( x c) f ( x) dx ( x c) f ( x) dx t f ( c t) dt t f ( c t)
dt
a a c
0 0
0
t( f ( c t) f ( c t)) dt.
(c) According to the mean value theorem of Section 4.2, for every t in 0,
c t,
c t at which
f ( c t) f ( c t) f ( c t) f ( c t)
( c t) ( c t) 2t
each t, we have f ( q) M , f ( c t) f ( c t) 2tM for all t in 0, . Thus
, there is a point in q in
f ( q).
Since for all these q belonging to
( b a)/2 3
2 3 ( b a)
x f ( x) dx t( f ( c t) f ( c t)) dt ( t)(2 tM ) M t M.
0 0 3 12
b
a
0
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