Integral Calculus Question 136
Question: If $ f(x)=\frac{e^{x}}{1+e^{x}},I_1=\int\limits_{f(-a)}^{f(a)}{xg{x(1-x)}dx} $ and $ I_2=\int\limits_{f(-a)}^{f(a)}{g{x(1-x)}dx,} $ then the value of $ \frac{I_2}{I_1} $ is
Options:
A) 1
B) $ -3 $
C) $ -1 $
D) $ 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ f(x)=\frac{e^{x}}{1+e^{x}}\Rightarrow f(-x)=\frac{{e^{-x}}}{1+{e^{-x}}} $ $ =\frac{1}{e^{x}+1} $
$ \therefore f(x)+f(-x)=1\forall x $ Now $ I_1=\int\limits_{f(-a)}^{f(a)}{xg{x(1-x)}dx} $ $ =\int\limits_{f(-a)}^{f(a)}{(1-x)g{x(1-x)}dx} $ $ =I_2-I_1\Rightarrow 2I_1=I_2 $
BETA
