Question

If $f\left(x\right)=f(a-x)$ and $g\left(x\right)+g(a-x)=2$, then the value of ${\int }_0^{a}f\left(x\right)g\left(x\right)dx$ is

• A. ${\int }_0^{a}f\left(x\right)dx$
• B. ${\int }_0^{a}g\left(x\right)dx$
• C. ${\int }_0^a[g(x)–f(x\left)\right]dx$
• D. ${\int }_0^a[g(x)+f(x\left)\right]dx$

Answer 1

The correct option is A

${\int }_0^{a}f\left(x\right)dx$

Explanation for the correct option:

Find the value of ${\int }_0^{a}f\left(x\right)g\left(x\right)dx$:

We have,

$I={\int }_0^{a}f\left(x\right)g\left(x\right)dx...\left(i\right)$

for $x\to 0+a-x$, we will get,

$$\begin{gathered} I={\int }_0^{a}f(a-x)g(a-x)dx \\ I={\int }_0^{a}f\left(x\right)(2-g(x\left)\right)dx...\left(ii\right) \end{gathered}$$

Now we will add equation $\left(i\right)$ and $\left(ii\right)$ we get:-

$$\begin{gathered} 2I={\int }_0^{a}2f\left(x\right)dx \\ I={\int }_0^{a}f\left(x\right)dx \end{gathered}$$

Hence, the correct option is A.