Question

The solution of ${\int }_{-\frac{1}{2}}^{\frac{1}{2}}\left|x\cos \frac{1}{2}\pi x\right|dx$ is equal to

• A. $\frac{1}{{\pi }^2}(\pi \sqrt{2}-4\sqrt{2}-8)$
• B. $\frac{1}{{\pi }^2}(\pi \sqrt{2}+4\sqrt{2}-8)$
• C. $\frac{1}{{\pi }^2}(\pi \sqrt{2}+4\sqrt{2}+8)$
• D. $\frac{1}{{\pi }^2}(\pi \sqrt{2}+2\sqrt{2}-4)$

Answer 1

The correct option is B $\frac{1}{{\pi }^2}(\pi \sqrt{2}+4\sqrt{2}-8)$

$f\left(x\right)=\left|x\cos \frac{1}{2}\pi x\right|$

As $f(-x)=|-x\cos \frac{1}{2}\pi (-x)|=|-x\cos \frac{1}{2}\pi x|=\left|x\cos \frac{1}{2}\pi x\right|=f\left(x\right)$

Then, $f\left(x\right)=2{\int }_0^{\frac{1}{2}}\left|x\cos \frac{1}{2}\pi x\right|dx=2{\int }_0^{\frac{1}{2}}x\cos \frac{1}{2}\pi xdx$
$=2{[\frac{x\sin \frac{1}{2}\pi x}{\frac{1}{2}\pi }+\frac{\cos \frac{1}{2}\pi x}{\frac{1}{4}{\pi }^2}]}^{\frac{1}{2}}$
$=2[\frac{2}{\pi }.\frac{1}{2\sqrt{\left(2\right)}}+\frac{4}{{\pi }^2}.\frac{1}{\sqrt{\left(2\right)}}-\frac{4}{{\pi }^2}]$

$=\frac{1}{{\pi }^2}(\pi \sqrt{2}+4\sqrt{2}-8)$