Question

${\int }_0^{2\pi }\left|\sin \left(x\right)\right|dx=$

• A. $2$
• B. $\sqrt{3}$
• C. $4$
• D. $0$

Answer 1

The correct option is C

$4$

Explanation for correct option:

Integrating ${\int }_0^{2\pi }\left|\sin \left(x\right)\right|dx=$:

Let, $I={\int }_0^{2\pi }\left|\sin \left(x\right)\right|dx$

Now we know the property of a mod, that is,

$\left|x\right|=\left\{\begin{array}{l}xx\ge 0\\ -xx<0\end{array}\right.$

Therefore, applying the same property for $\sin \left(x\right)$ function.

We know, $\sin \left(x\right)$ is greater than zero for $\left(0,\pi \right)$ and less than zero for $\left(\pi ,2\pi \right)$.

Therefore,

$\begin{array}{rcl}I& =& {\int }_0^{2\pi }\left|\sin \left(x\right)\right|dx\\ & =& {\int }_0^{\pi }\sin \left(x\right)dx+{\int }_{\pi }^{2\pi }-\sin \left(x\right)dx\\ & =& {\left[-\cos \left(x\right)\right]}_0^{\pi }-{\left[-\cos \left(x\right)\right]}_{\pi }^{2\pi }\\ & =& \left[-\cos \left(\pi \right)-\left(-\cos \left(0\right)\right)\right]-\left[-\cos \left(2\pi \right)-\left(-\cos \left(\pi \right)\right)\right]\\ & =& \left[-\left(-1\right)-\left(-1\right)\right]-\left[-\left(1\right)-\left(1\right)\right]\\ & =& \left[1+1+1+1\right]\\ I& =& 4\end{array}$

Therefore the value of the given integral ${\int }_0^{2\pi }\left|\sin \left(x\right)\right|dx=4$

Hence, option (C) is the correct answer.