Question

If $f\left(x\right)$ is integrable on $[0,a]$ then ${\int }_0^a\frac{f\left(x\right)}{f\left(x\right)+f(a-x)}dx=?$

• A. $0$
• B. $1$
• C. $a$
• D. $\frac{a}{2}$

Answer 1

The correct option is D $\frac{a}{2}$

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

${\int }_0^{a}g\left(x\right)dx={\int }_0^{a}g(a-x)dx$

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

$2I={\int }_0^{a}1\cdot dx$

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