Question

What is ${\int }_0^a\frac{f(a-x)}{f(x)+f(a-x)}dx$ equal to?

• (A) a
• (B) 2a
• (C) 0
• (D) $\frac{a}{2}$

Answer 1

The Correct Option is D

Explanation:
As we know that ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$ $I={\int }_0^a\frac{f(a-x)}{f(x)+f(a-x)}dx\dots \dots$ (1)So, we have lower limit $a=0$, upper limit $b=a$$\Rightarrow I={\int }_0^a\frac{f[(0+a)-(a-x)]}{f[(a+0)-x]+f[(a+0)-(a-x)]}dx$$\Rightarrow I={\int }_0^a\frac{f(x)}{f(a-x)+f(x)}dx\dots \dots \dots$ (2)Now adding equation (1) and equation (2),$\Rightarrow 2I={\int }_0^a\frac{f(a-x)}{f(x)+f(a-x)}dx+{\int }_0^a\frac{f(x)}{f(a-x)+f(x)}dx$$\Rightarrow 2I={\int }_0^{a}1dx$$\Rightarrow 2I=a$$\therefore I=\frac{a}{2}$Hence, the correct option is (D).