Question

If $f,g,h$ be continuous functions on $[0,a]$ such that $f(a-x)=f\left(x\right)$, $g(a-x)=-g\left(x\right)$ and $3h\left(x\right)-4h(a-x)=5$, then evaluate ${\int }_0^{a}f\left(x\right).g\left(x\right).h\left(x\right)dx$.

Answer 1

Given $f(a-x)=f\left(x\right)$, $g(a-x)=-g\left(x\right)$ and $3h\left(x\right)-4h(a-x)=5$ &

$I={\int }_0^{a}f\left(x\right).g\left(x\right).h\left(x\right)dx$
$={\int }_0^{a}f(a-x).g(a-x).h(a-x)dx(\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx)$.....(1)
$I=-{\int }_0^{a}f\left(x\right).g\left(x\right)h(a-x)dx$ (by (1))

Since $7I=3I+4I$

$7I={\int }_0^{a}f\left(x\right).g\left(x\right)\left\{\right(3h\left(x\right)-4h(a-x)\}dx$

$7I=5{\int }_0^{a}f\left(x\right)g\left(x\right)dx$ ....(2)

$7I=5{\int }_0^{a}f(a-x)g(a-x)dx$

Since $f(a-x)g(a-x)=-f\left(x\right)g\left(x\right)$

$\Rightarrow 7I=-5{\int }_0^{a}f\left(x\right)g\left(x\right)dx$ ....(3)

Adding (2) and (3),we get

$I=0$