Question

If $a$ is a fixed real number such that $f(a-x)+f(a+x)=0$,

then value of ${\int }_0^{2a}f\left(x\right)dx$ is?

• A. $\frac{1}{2}a$
• B. $0$
• C. $-\frac{1}{2}a$
• D. None of the above

Answer 1

The correct option is B $0$

$f(a-x)+f(a+x)=0$

Substitute $x=a-x$

$\Rightarrow f(a+x-a)+f(a-x+a)\Rightarrow f(+x)+f(2a-x)=0\Rightarrow f(2a-x)=-f\left(x\right)I={\int }_0^{2a}f\left(x\right)dx={\int }_0^{2a}f(2a-x)dx=-{\int }_0^{2a}f\left(x\right)dx=-I\Rightarrow 2I=0\therefore I=0$