Integral Calculus Question 103
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) $ -2 $
C) 2
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ f(x)=\frac{e^{x}}{1+e^{x}} $
$ \therefore f(a)=\frac{e^{q}}{1+e^{a}} $ And $ f(-a)=\frac{{e^{-a}}}{1+{e^{-a}}}=\frac{{e^{-a}}}{1+\frac{1}{e^{a}}}=\frac{1}{1+e^{a}} $
$ \therefore f(a)+f(-a)=\frac{e^{a}+1}{1+e^{q}}=1 $ Let $ f(-a)=\alpha $ or $ f(a)=1-\alpha $ Now, $ I_1=\int\limits_{\alpha }^{1-\alpha }{xg(x(1-x))dx} $ $ =\int\limits_{\alpha }^{1-\alpha }{(1-x)g((1-x)(1-(1-x))dx} $ $ =\int\limits_{\alpha }^{1-\alpha }{(1-x)g(x(1-x)dx} $
$ \therefore 2I_1=\int\limits_{\alpha }^{1-\alpha }{g(x(1-x))dx}=I_2 $ Or $ \frac{I_2}{I_1}=2 $
BETA
