Question

${\int }_{-1}^{\frac{3}{2}}|x\sin (\pi x)|dx=$

• A. $\frac{1}{\pi }-\frac{1}{{\pi }^2}$
• B. $\frac{2}{\pi }+\frac{1}{{\pi }^2}$
• C. $\frac{3}{\pi }-\frac{1}{{\pi }^2}$
• D. $\frac{3}{\pi }+\frac{1}{{\pi }^2}$

Answer 1

Answer: (D)

Let
$I=\underset{-1}{\overset{3/2}{\int }}|x\sin (\pi x)|dx$
$I=\underset{-1}{\overset{1}{\int }}x\sin \pi xdx-\underset{1}{\overset{3/2}{\int }}x\sin \pi xdx$
$I=2\underset{0}{\overset{1}{\int }}x\sin \pi xdx-\underset{1}{\overset{3/2}{\int }}x\sin pixdx$
$I=2{\left[\frac{-x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\right]}_0^1$
$-{\left[\frac{-x\cos \pi x}{\pi }+\frac{\sin \pi x}{{\pi }^2}\right]}_1^{3/2}$
$I=2\left[\left(\frac{1}{\pi }+0\right)-(0)\right]-\left[0-\frac{1}{{\pi }^2}-\frac{1}{\pi }\right]$
$I=\frac{3}{\pi }+\frac{1}{{\pi }^2}$