Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.
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 :-
Definite Integral
Properties of definite Integral
Even function / Odd function
But try the problem first...
Answer: $ \displaystyle\int_0^a f(x) \mathrm d x $
IIT JAM 2018, Problem 4
Definite and Integral calculus : R Courant
First hint
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 !!!
Second Hint
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 !!!
Final Step
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]
Try this beautiful problem from IIT JAM 2018 which requires knowledge of the properties of Definite integral.
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 :-
Definite Integral
Properties of definite Integral
Even function / Odd function
But try the problem first...
Answer: $ \displaystyle\int_0^a f(x) \mathrm d x $
IIT JAM 2018, Problem 4
Definite and Integral calculus : R Courant
First hint
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 !!!
Second Hint
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 !!!
Final Step
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]
