The integral _-1^(3)/(2)(| ^2x ( x)|)dx is equal to:

Question

The integral $\int _{-1}^{\frac{3}{2}}(|{\pi }^{2}x\sin (\pi x)|)dx$ is equal to:

Accepted Answer

$$\begin{aligned}
&\text { Let, } \mathrm{I}=\pi^2 \int_{-1}^{3 / 2}|\mathrm{x} \sin \pi \mathrm{x}| \mathrm{dx}\
&\begin{aligned}
& =\pi^2\left\{\int_{-1}^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_1^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\} \
& =\pi^2\left\{2 \int_0^1 \mathrm{x} \sin \pi \mathrm{xdx}-\int_{-1}^{3 / 2} \mathrm{x} \sin \pi \mathrm{xdx}\right\}
\end{aligned}
\end{aligned}$$

$$\begin{aligned}
&\text { Consider }\
&\begin{aligned}
& \int \mathrm{x} \sin \pi \mathrm{xdx} \
& -\mathrm{x} \cdot \frac{1}{\pi} \cos \pi \mathrm{x}+\int 1 \cdot \frac{1}{\pi} \cos \pi \mathrm{xdx} \
& =--\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2} \
& \mathrm{I}=\pi^2\left\{2\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_0^1-\left(-\frac{\mathrm{x}}{\pi} \cos \pi \mathrm{x}+\frac{\sin \pi \mathrm{x}}{\pi^2}\right)_1^{3 / 2}\right\} \
& =\pi^2\left\{\frac{2}{\pi}-\left(-\frac{1}{\pi^2}-\frac{1}{\pi}\right)\right\} \
& =\pi^2\left\{\frac{3}{\pi}+\frac{1}{\pi^2}\right\} \
& =3 \pi+1
\end{aligned}
\end{aligned}$$

Solutions · 1