Question

If $a>0$ and ${\int }_0^a[f\left(x\right)+f(-x)]dx={\int }_{-a}^a\varphi \left(x\right)dx$ then one of the possible values of $\varphi \left(x\right)$ can be

• A. $f(-x)$
• B. $-f\left(x\right)$
• C. $\frac{1}{2}f\left(x\right)$
• D. none of these

Answer 1

The correct option is A $f(-x)$

$Supposef\left(x\right)isanevenfunctionthen{\int }_0^a\left[f\right(x)+f(-x\left)\right]dx={\int }_0^a\left[f\right(x)+f(x\left)\right]dx=2{\int }_0^a\left[f\right(x\left)\right]dx={\int }_{-a}^a\varphi \left(x\right)dx\varphi \left(x\right)=f\left(x\right)andiff\left(x\right)isanoddfunctionthen{\int }_0^a\left[f\right(x)+f(-x\left)\right]dx={\int }_0^a\left[f\right(x)-f(x\left)\right]dx=0={\int }_{-a}^a\varphi \left(x\right)dx\varphi \left(x\right)isalsoanoddfunctionsopossiblevalueof\varphi \left(x\right)=f(-x)$