Integral Calculus Question 342
Question: $ \int_{{}}^{{}}{
[f(x)g’’(x)-f’’(x)g(x)]}\ dx $ is equal to [MP PET 2001]
Options:
A) $ \frac{f(x)}{g’(x)} $
B) $ f’(x)g(x)-f(x)g’(x) $
C) $ f(x)g’(x)-f’(x)g(x) $
D) $ f(x)g’(x)+f’(x)g(x) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{[f(x),{g}’’(x)-{f}’’(x),g(x)],dx} $ $ =\int_{{}}^{{}}{f(x),{g}’’(x),dx}-\int_{{}}^{{}}{{f}’’(x),g(x),dx} $ $ =( f(x),{g}’(x)-\int_{{}}^{{}}{{f}’(x){g}’(x),dx} )-( g(x),{f}’(x)-\int_{{}}^{{}}{{g}’(x),{f}’(x),dx} ) $ $ =f(x),{g}’(x)-{f}’(x),g(x). $
BETA
