Definite Integral | IIT JAM 2018 | Problem 4

Let $a$ be positive real number. If $f$ is a continuous and even function defined on the interval $[-a,a]$, then $\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x$ is equal to :-

In this first I will give you the properties we need to solve this problem :

Property 1 : $\displaystyle\int_a^b f(x) \mathrm d x = \displaystyle\int_a^b f(a+b-x) \mathrm d x $

[Where $f$ is continuous on $[a,b]$]

Property 2 : If $f$ is an even function i.e., $f(x)=f(-x)$ then

$ \displaystyle\int_{-a}^{a} f(x) \mathrm d x = 2 \displaystyle\int_{0}^{a} f(x) \mathrm d x $

Can you drive it from here !!!! Give it a try !!!

Let $I=\displaystyle\int_{-a}^a \frac{f(x)}{1+e^x} \mathrm d x \quad \ldots (i)$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(a-a-x)}{1+e^{(a-a-x)}} \mathrm d x $

[Since, $f$ is continuous then $\displaystyle\int_{a}^b f(x) \mathrm{d}x = \displaystyle\int_{a}^b f(a+b-x) \mathrm{d} x $]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(-x)}{1+e^{-x}} \mathrm d x$

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{f(x)}{1+\frac{1}{e^x}} \mathrm d x$ [Since $f(x)$ is even]

$\Rightarrow I= \displaystyle\int_{-a}^a \frac{e^x.f(x)}{1+e^{x}} \mathrm d x \quad \ldots (ii) $

Adding $(i)$ and $(ii)$ we can get some interesting result !!!

Adding $(i)$ and $(ii)$ we get ,

$2I= \displaystyle\int_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm d x + \displaystyle\int_{-a}^a \frac{e^x .f(x)}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{[f(x)+e^x.f(x)]}{1+e^{x}} \mathrm d x$

$\Rightarrow 2I = \displaystyle\int_{-a}^a \frac{f(x)[1+e^x]}{[1+e^{x}]}$

$ \Rightarrow 2I= \displaystyle\int_{-a}^a f(x) \mathrm d x$

$\Rightarrow 2I = 2\displaystyle\int_0^a f(x) \mathrm d x $ [Since $f(x) $ is even ]

$\Rightarrow I = \displaystyle\int_0^a f(x) \mathrm d x $ [ANS]