Mathematics - Definite Integration Question with Solution | TestHub

Let, $\mathrm{I}={\pi }^2{\int }_{-1}^{3/2}|\mathrm{x}\sin \pi \mathrm{x}|\mathrm{dx}$
$\begin{array}{rl}& ={\pi }^2\left\{{\int }_{-1}^1\mathrm{x}\sin \pi \mathrm{xdx}-{\int }_1^{3/2}\mathrm{x}\sin \pi \mathrm{xdx}\right\}\\ & ={\pi }^2\left\{2{\int }_0^1\mathrm{x}\sin \pi \mathrm{xdx}-{\int }_{-1}^{3/2}\mathrm{x}\sin \pi \mathrm{xdx}\right\}\end{array}$
Consider
$\begin{array}{rl}& \int \mathrm{x}\sin \pi \mathrm{xdx}\\ & -\mathrm{x}\cdot \frac{1}{\pi }\cos \pi \mathrm{x}+\int 1\cdot \frac{1}{\pi }\cos \pi \mathrm{xdx}\\ & =-\frac{\mathrm{x}}{\pi }\cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{{\pi }^2}\end{array}$
$\begin{array}{rl}\mathrm{I}& ={\pi }^2\left\{2{\left(-\frac{\mathrm{x}}{\pi }\cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{{\pi }^2}\right)}_0^1-{\left(-\frac{\mathrm{x}}{\pi }\cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{{\pi }^2}\right)}_1^{3/2}\right\}\\ & ={\pi }^2\left\{\frac{2}{\pi }-\left(-\frac{1}{{\pi }^2}-\frac{1}{\pi }\right)\right\}\\ & ={\pi }^2\left\{\frac{3}{\pi }+\frac{1}{{\pi }^2}\right\}\\ & =3\pi +1\end{array}$