Question

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|dx=?$

• A. ${\pi }^2$
• B. $\frac{{\pi }^2}{2}$
• C. $\sqrt{2{\pi }^2}$
• D. $2{\pi }^2$

Answer 1

The correct option is A

${\pi }^2$

Explanation For The Correct Option:

Evaluating the integral:

${\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|dx$

consider $f\left(x\right)=\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|$

$$\begin{gathered} f(-x)=\left|\mathrm{\pi }-\left|-\mathrm{x}\right|\right| \\ =\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|[\because \left|-\mathrm{x}\right|=\left|\mathrm{x}\right|=x] \\ =f\left(x\right) \end{gathered}$$

Thus, it is an even function.

We know the property

${\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{l}2{\int }_0^{a}f\left(x\right)dx:\text{if}f\left(x\right)\text{is an even funtion}\\ 0:\text{if}f\left(x\right)\text{is an odd funtion}\end{array}\right.$

$$\begin{gathered} \Rightarrow {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\left|\mathrm{\pi }-\left|\mathrm{x}\right|\right|dx=2{\int }_0^{\mathrm{\pi }}\left|\mathrm{\pi }-x\right|dx \\ =2{\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-x\right)dx \\ =2{\left[\mathrm{\pi x}-\frac{{\mathrm{x}}^2}{2}\right]}_0^{\mathrm{\pi }} \\ =2\left[{\mathrm{\pi }}^2-\frac{{\mathrm{\pi }}^2}{2}-\left(0\right)\right] \\ ={\mathrm{\pi }}^2 \end{gathered}$$

Hence, the correct answer is option (A).