Integral Calculus Question 129
Question: Let $ f:R\to R $ is differentiable function and $ f(1)=4, $ then the value of $ \underset{x\to 1}{\mathop{\lim }},\int\limits_0^{f(x)}{\frac{2tdt}{x-1}} $ is
Options:
A) $ 8f’(1) $
B) $ 4f’(1) $
C) $ 2f’(1) $
D) $ f’(1) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \underset{x\to 1}{\mathop{\lim }},\int\limits_0^{f(x)}{\frac{2t}{x-1}dt} $ $ =\underset{x\to 1}{\mathop{\lim }},\frac{1}{x-1}[ t^{2} ]_0^{f(x)}=\underset{x\to 1}{\mathop{\lim }},\frac{f^{2}(x)-f^{2}(0)}{x-1} $ $ =\underset{x\to 1}{\mathop{\lim }},\frac{2f(x)f’(x)}{1} $ (L? Hospital Rule) $ =2f(1)f’(1)=8f’(1) $
BETA
