Page 204 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 8] APPLICATIONS OF PARTIAL DERIVATIVES 195

Then
ð u 2 ð Þ u 2 u 1
¼ f ðx; Þ dx þ f ð 2 ; þ Þ f ð 1 ; þ Þ
u 1 ð Þ
Taking the limit as ! 0, making use of the fact that the functions are assumed to have continuous
derivatives, we obtain
d ð u 2 ð Þ du 2 du 1
d ¼ f ðx; Þ dx þ f ½u 2 ð Þ; Š d f ½u 1 ð Þ; Š d
u 1 ð Þ
ð 2

sin x
dx, find ð Þ where 6¼ 0.
0
8.16. If ð Þ¼
x
By Leibnitz’s rule,
ð 2
@ sin x sinð Þ d sinð Þ d
2
0 2
@ x d d
ð Þ¼ dx þ 2 ð Þ ð Þ
ð 2 3 2

2 sin sin
¼ cos xdx þ

2 3 2 3 2

sin x 2 sin sin 3 sin 2 sin

¼ þ ¼

ð dx ð dx
8.17. If ¼ p ffiffiffiffiffiffiffiffiffiffiffiffiffi ; > 1 find . (See Problem 5.58, Chapter 5.)
2
0 cos x 1 0 ð2 cos xÞ 2
ð dx
2
0 cos x
By Leibnitz’s rule, if ð Þ¼ ¼ ð 1Þ 1=2 ; then
ð dx 1
0 2 3=2
2 2 2 3=2
ð Þ¼ ¼ ð 1Þ 2 ¼
0 ð cos xÞ ð 1Þ

ð dx ð dx 2
Thus from which ¼ p :
2 ¼ 2 3=2 2 ffiffiffi
0 ð cos xÞ ð 1Þ 0 ð2 cos xÞ 3 3
INTEGRATION UNDER THE INTEGRAL SIGN
8.18. Prove the result (18), Page 187, for integration under the integral sign.
ð ð
u 2
f ðx; Þ d dx
Consider ð1Þ ð Þ¼
u 1 a
By Leibnitz’s rule,
@
ð ð ð
u 2 u 2
0
@
ð Þ¼ f ðx; Þ d dx ¼ f ðx; Þ dx ¼ ð Þ
u 1 a u 1
ð
Then by integration, ð2Þ ð Þ¼ ð Þ d þ c
a
Since ðaÞ¼ 0 from (1), we have c ¼ 0in (2). Thus from (1) and (2)with c ¼ 0, we find

ð ð ð ð
u 2 u 2
f ðx; Þ dx d
f ðx; Þ dx dx ¼
u 1 a a u 1
Putting ¼ b,the required result follows.