Question

Solve ${\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx$

Answer 1

$I={\int }_a^b\frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx$ $..........\left(1\right)$

$Replacef\left(x\right)\to f\left(a+b-x\right)$

Therefore,

$I={\int }_a^b\frac{f\left(a+b-x\right)}{f\left(a+b-x\right)+f\left(a+b-\left(a+b-x\right)\right)}dx$

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

On adding equation (1) and (2), we get

$2I={\int }_a^b\frac{f\left(a+b-x\right)+f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx\Rightarrow 2I={\int }_a^{b}dx={\left[x\right]}_a^b\Rightarrow 2I=b-a\Rightarrow I=\frac{b-a}{2}$

Hence, this is the answer.