Question

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

$\frac{a}{2}$

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

$I={\int }_0^a\frac{dx}{1+e^{f(a+0-x)}}$

adding both the equations,

$2I={\int }_0^a\frac{1+e^{f(a-x)}+1+e^{f\left(x\right)}}{(1+e^{f\left(x\right))}(1+e^{f(a-x)})}dx$

$={\int }_0^a\frac{2+e^{f\left(x\right)}+e^{f(a-x)}}{1+e^{f(a-x)}+e^{f\left(x\right)}+e^{f\left(x\right)+f(a-x)}}$

$f\left(x\right)+f(a-x)=0$

$2I={\int }_0^a\frac{2+e^{f\left(x\right)}+e^{f(a-x)}}{2+e^{f\left(x\right)}+e^{f(a-x)}}dx$

$2I=a$

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