Question

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

• A. $0$
• B. $4$
• C. $8$
• D. $1$

Answer 1

The correct option is B

$4$

Explanation for correct answer:

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

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

Using the property of mod we can write the above integral in a simplified way shown below

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

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

Hence, option (B) is the correct answer.