Question

${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin \left|x\right|dx=$

• A. $0$
• B. $1$
• C. $2$
• D. $\mathrm{\pi }$

Answer 1

The correct option is C

$2$

Explanation for the correct option:

The value for the given integral is:

Let $I={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\sin \left|x\right|dx$

We know that, $$\begin{gathered} f\left(x\right)=\left\{x,ifx\ge 0 \\ -x,ifx\le 0\right\} \end{gathered}$$

$$\begin{gathered} =-{\int }_{-\frac{\mathrm{\pi }}{2}}^0\sin xdx+{\int }_0^{\frac{\mathrm{\pi }}{2}}\sin xdx \\ ={\left[\cos x\right]}_{-\frac{\mathrm{\pi }}{2}}^0-{\left[\cos x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\left[\cos 0-\cos \left(-\frac{\mathrm{\pi }}{2}\right)\right]-\left[\cos \frac{\mathrm{\pi }}{2}-\cos 0\right] \\ =\left[1-0\right]-\left[0-1\right] \\ =2 \end{gathered}$$

Thus, option C is the correct answer.