Integral Calculus Question 184
Question: $ \int_0^{1}{
[f(x)g’’(x)-f’’(x)g(x)]dx} $ is equal to: [Given f(0) = g (0) = 0]
Options:
A) $ f(1)g(1)-f(1)g’(1) $
B) $ f(1)g’(1)+f’(1)g(1) $
C) $ f(1)g’(1)-f’(1)g(1) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Integrating by parts. $ \int{f(x)g’’(x)dx-\int{f’’(x)g(x)dx}} $ $ =f(x)g’(x)-\int{f’(x)g’(x)dx} $ $ -f’(x)g(x)+\int{f’(x)g’(x)dx} $ $ =f(x)g’(x)-f’(x)g(x) $ Hence, $ \int_0^{1}{f(x)g’’(x)dx-\int_0^{1}{f’’(x)g(x)dx}} $ $ =f(1)g’(1)-f’(1)g(1)-f(0)g’(0)+f’(0)g(0) $ $ =f(1)g’(1)-f’(1)g(1) $
BETA
