Question

$\int\limits^{2}_{-2} |x \, \cos \, \pi x | dx$ is equal to

• A. $\frac{8}{\pi}$
• B. $\frac{4}{\pi}$
• C. $\frac{2}{\pi}$
• D. $\frac{1}{\pi}$

Answer 1

Answer: (A)

$I = \int\limits^{2}_{-2} \left|x \,\cos\, \pi x\right|dx $
$ = 2\int\limits^{2}_{0} \left|x\, \cos\,\pi x\right|dx $
$ = 2\left[\int\limits^{1/2}_{0} \left(x \cos\pi x\right)dx - \int\limits^{3/2}_{1/2} \left(x \cos\pi x\right)dx + \int\limits^{2}_{3/2}\left(x\cos\pi x\right)dx \right] $
Now $ \int x \cos\pi x dx = \frac{x \sin\pi x}{\pi} + \frac{\cos\pi x}{\pi^{2}} $
$ \therefore I = 2 \left[\left(\frac{1}{2\pi} - \frac{1}{\pi^{2}}\right) - \left(- \frac{3}{2\pi} - \frac{1}{2\pi}\right) + \left(\frac{1}{\pi^{2}} + \frac{3}{2\pi}\right)\right] $
$= 2 \left[\frac{1}{2\pi} - \frac{1}{\pi^{2}} + \frac{3}{2\pi} + \frac{1}{2\pi} + \frac{1}{\pi^{2}} + \frac{3}{2\pi}\right] $
$= 2 \left(\frac{8}{2\pi}\right) = \frac{8}{\pi}$