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# Definite Integral | IIT JAM 2018 | Problem 4

Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of definite integral. We have sequential hints.

Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.

## Properties of Definite Integral -IIT JAM2018 (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 :-

• $\displaystyle\int_0^a f(x) \mathrm d x$
• $2\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$
• $2\displaystyle\int_0^a f(x) \mathrm d x$
• $2a\displaystyle\int_0^a \frac{f(x)}{1+e^x}\mathrm d x$

### Key Concepts

Definite Integral

Properties of definite Integral

Even function / Odd function

Answer: $\displaystyle\int_0^a f(x) \mathrm d x$

IIT JAM 2018, Problem 4

Definite and Integral calculus : R Courant

## Try with Hints

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]

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