Question

If $f$ and $g$ are continuous functions in $[0,a]$ satisfying $f\left(x\right)=f(a-x)$ and $g\left(x\right)+g(a-x)=a$, then $\underset{0}{\overset{a}{\int }}f\left(x\right)\cdot g\left(x\right)dx$ is equal to

• A. $\frac{a}{2}$
• B. $\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
• C. $\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
• D. $a\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

Answer 1

The correct option is B $\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

$\begin{array}{cc}\text{Let}& I=\underset{0}{\overset{a}{\int }}f\left(x\right)\cdot g\left(x\right)dx\\ \therefore I& =\underset{0}{\overset{a}{\int }}f(a-x)g(a-x)dx=\underset{0}{\overset{a}{\int }}f\left(x\right)\{a-g(x\left)\right\}dx\\ & =a\underset{0}{\overset{a}{\int }}f\left(x\right)dx-\underset{0}{\overset{a}{\int }}f\left(x\right)\cdot g\left(x\right)=a\underset{0}{\overset{a}{\int }}f\left(x\right)dx-I\\ \therefore & I=\frac{a}{2}\underset{0}{\overset{a}{\int }}f\left(x\right)dx\end{array}$