Question

$\int_{-1}^{\frac{3}{2}}|x \sin (\pi x)| d x=$

• 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 =\int\limits_{-1}^{3 / 2}|x \,\sin (\pi x)| d x $
$ I= \int\limits_{-1}^{1} x \,\sin \,\pi \,x d x-\int\limits_{1}^{3 / 2} x \,\sin \,\pi x d \,x $
$ I=2 \int\limits_{0}^{1} x \,\sin \,\pi \,x d x-\int\limits_{1}^{3 / 2} \,x\,\sin\,pi x \,d x $
$ I=2\left[\frac{-x \,\cos \,\pi \,x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{0}^{1} $
$-\left[\frac{-x \cos \pi x}{\pi}+\frac{\sin \pi x}{\pi^{2}}\right]_{1}^{3 / 2}$
$I=2\left[\left(\frac{1}{\pi}+0\right)-(0)\right]-\left[0-\frac{1}{\pi^{2}}-\frac{1}{\pi}\right]$
$I=\frac{3}{\pi}+\frac{1}{\pi^{2}}$