Question

Let $f$ and $g$ be continuous function on $[0,a]$ such that $f\left(x\right)=f(a-x)$ and $g\left(x\right)+g(a-x)=4,$ then $\underset{0}{\overset{a}{\int }}f\left(x\right)g\left(x\right)dx$ is equal to :

• A. $\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
• B. $2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
• C. $4\underset{0}{\overset{a}{\int }}f\left(x\right)dx$
• D. $-3\underset{0}{\overset{a}{\int }}f\left(x\right)dx$

Answer 1

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

Given $f\left(x\right)=f(a-x)$
$g\left(x\right)+g(a-x)=4\Rightarrow g(a-x)=4-g\left(x\right)$
$I=\underset{0}{\overset{a}{\int }}f\left(x\right).g\left(x\right)dx$
$=\underset{0}{\overset{a}{\int }}f(a-x).g(a-x)dx$
$=\underset{0}{\overset{a}{\int }}f\left(x\right).(4-g(x\left)\right)dx$
$\Rightarrow I=\underset{0}{\overset{a}{\int }}4f\left(x\right)dx-\underset{0}{\overset{a}{\int }}f\left(x\right).g\left(x\right)dx$
$\Rightarrow 2I=\underset{0}{\overset{a}{\int }}4f\left(x\right)dx$
$\Rightarrow I=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx$