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Question 74 of 90
The value of ${\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\sin \left|x\right| dx$ is equal to :
1. 2
2. - 2
3. 1
4. 0

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{I}={\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \sin \left|\mathrm{x}\right|\mathrm{dx} \\ \sin \left|\mathrm{x}\right| \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{fucntion}, \mathrm{hence} \mathrm{using} \mathrm{property} \\ {\int }_{-\mathrm{a}}^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right).\mathrm{dx}=2{\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right).\mathrm{dx} \\ {\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}} \sin \left|\mathrm{x}\right|\mathrm{dx}=2{\int }_0^{\frac{\mathrm{\pi }}{2}} \sin \left|\mathrm{x}\right|\mathrm{dx} \\ \mathrm{As} \mathrm{x}>0, \mathrm{hence} \left|\mathrm{x}\right|=\mathrm{x} \\ \mathrm{I}=2{\int }_0^{\frac{\mathrm{\pi }}{2}} \mathrm{sinxdx} \\ =-2{\left[\mathrm{cosx}\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =-2\left[\cos \frac{\mathrm{\pi }}{2}-\cos 0\right] \\ =-2\times \left(-1\right) \\ =2 \left[\mathrm{Answer}\right] \end{gathered}$$

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