Question

The integral $\underset{0}{\stackrel{a}{\int }}\frac{g\left(x\right)}{f\left(x\right)+f(a-x)}dx$ vanishes, if

• A. $g\left(x\right)$ is odd
• B. $f\left(x\right)=f(a-x)$
• C. $g\left(x\right)=-g(a-x)$
• D. $f(a-x)=-g\left(x\right)$

Answer 1

The correct option is C $g\left(x\right)=-g(a-x)$

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

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

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

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

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

Value of $I$ will be zero if,

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

$g\left(x\right)=-g(a-x)$