Question

Let $f\left(x\right)$ be a continuous function such that $f(a-x)+f\left(x\right)=0$ for all $xϵ[0,a].$ Then ${\int }_0^a\frac{dx}{1+e^{f\left(x\right)}}$ is equal to

• A. $a$
• B. $\frac{a}{2}$
• C. $f\left(a\right)$
• D. $\frac{1}{2}f\left(a\right)$

Answer 1

Answer: (B)

$\frac{a}{2}$

Let $f\left(x\right)$ be a continuous function such that $f(a-x)+f\left(x\right)=0$ $\left(i\right)$ for all $x[0,a]$.

Then,

${\int }_0^a\frac{dx}{1+e^{f\left(x\right)}}$

Let $y={\int }_0^a\frac{dx}{1+e^{f\left(x\right)}}$

We can also write $y={\int }_0^a\frac{dx}{1+e^{f(a-x)}}$

Using $\left(i\right)$ we can write $f(a-x)=-f\left(x\right)$

$1+e^{f(a-x)}=1+e^{-f\left(x\right)}$

$1+e^{f(a-x)}=\frac{e^{f\left(x\right)}}{1+e^{f\left(x\right)}}$

$y={\int }_0^a\frac{e^{f\left(x\right)}}{1+e^{f\left(x\right)}}$

$y={\int }_0^a\frac{dx}{1+e^{f\left(x\right)}}$

Adding both we get $2y={\int }_0^{a}dx$

$y=\frac{a}{2}$