Mathematics - Definite Integration Question with Solution | TestHub

Given $I_1={\int }_0^{2\pi }xf(x)dx$. Using the property ${\int }_0^{a}g(x)dx={\int }_0^{a}g(a-x)dx$, we have $I_1={\int }_0^{2\pi }(2\pi -x)f(2\pi -x)dx$. Since $f(x)=f(2\pi -x)$, we get $I_1={\int }_0^{2\pi }(2\pi -x)f(x)dx=2\pi {\int }_0^{2\pi }f(x)dx-{\int }_0^{2\pi }xf(x)dx$. This simplifies to $I_1=2\pi I_2-I_1$, which means $2I_1=2\pi I_2$. Therefore, $\frac{I_1}{I_2}=\pi$ (approximately 3.14159).