Mathematics - Definite Integration Question with Solution | TestHub

$f(-a)+f(a)=1$ $I_1={\int }_{f(-a)}^{f(a)}xg\{x(1-x)\}dx={\int }_{f(-a)}^{f(a)}(1-x)g\{x(1-x)\}dx\left(\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx\right)$ $2I_1={\int }_{f(-a)}^{f(a)}g\{x(1-x)\}dx=I_2\Rightarrow I_2/I_1=2$.