Definite Integration Question 348
Question: If $ f(x)=\frac{e^{x}}{1+e^{x}},\ I_1=\int _{f(-a)}^{f(a)}{xg{x(1-x)}dx} $ , and $ I_2=\int _{f(-a)}^{f(a)}{g{x(1-x))}dx} $ , then the value of $ \frac{I_2}{I_1} $ is
[AIEEE 2004]
Options:
A) 1
B) -3
C) -1
D) 2
Show Answer
Answer:
Correct Answer: D
Solution:
Given, $ f(x)=\frac{e^{x}}{x+1},I_1=\int _{f(-a)}^{f(a)}{xg{x(1-x)}dx} $
and $ I_2=\int _{f(-a)}^{f(a)}{g{x(1-x)}dx} $
$ f(a)=\frac{e^{a}}{1+e^{a}},f(-a)=\frac{{e^{-a}}}{1+{e^{-a}}} $
$ f(a)+f(-a)=1 $
Now, $ 2I_1=\int _{f(-a)}^{f(a)}{xg{x(1-x)}dx} $ + $ \int _{f(-a)}^{f(a)}{{f(a)+f(-a)-x}g{(1-x)(x)}}dx $
therefore $ 2I_1=\int _{f(-a)}^{f(a)}{g{x(1-x)}dx=I_2} $ , $ (\because f(a)+f(-a)=1) $
$ 2I_1=I_2 $
therefore $ \frac{I_2}{I_1}=2 $ .
BETA
