Question

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

• A. ${\int }_0^{a}f(a-x)dx$
• B. $a$
• C. ${\int }_0^{a}f\left(x\right)dx$
• D. $a^2$

Answer 1

Answer: (A), (C)

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

Let , $I={\int }_0^{a}f\left(x\right)g\left(x\right)dx$
$={\int }_0^{a}f(a-x)g(a-x)dx$
since , $g\left(x\right)+g\left(ax\right)=2$
$I={\int }_0^{a}f(a-x)\{2-g\left(x\right)\}dx$
$=2{\int }_0^{a}f(a-x)dx-{\int }_0^{a}f(a-x)g\left(x\right)dx$
since , $f(a-x)=f\left(x\right)$
$2I=2{\int }_0^{a}f(a-x)dx$
$I={\int }_0^{a}f(a-x)dx$
$I={\int }_0^{a}f\left(x\right)dx$