Question

The value of${\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left[2\sin \left(x\right)\right]dx$ is

• A. $\frac{\mathrm{\pi }}{3}$
• B. $-\frac{5\mathrm{\pi }}{3}$
• C. $\frac{4\mathrm{\pi }}{3}$
• D. $-\frac{\mathrm{\pi }}{3}$

Answer 1

The correct option is B

$-\frac{5\mathrm{\pi }}{3}$

Explanation for all correct options

Given: ${\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left[2\sin \left(x\right)\right]dx$

According to integral

$\mathrm{\pi }\le \mathrm{x}\le 2\mathrm{\pi }$

taking sin to both sides

$-1\le \sin \left(\mathrm{x}\right)\le 0$

When $-\frac{1}{2}=\sin \left(\mathrm{x}\right)$

$x=\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}$

$\begin{array}{rcl}{\int }_{\mathrm{\pi }}^{2\mathrm{\pi }}\left[2\sin \left(x\right)\right]dx& =& {\int }_{\mathrm{\pi }}^{\frac{7\mathrm{\pi }}{6}}\left(-1\right)dx+{\int }_{\frac{7\mathrm{\pi }}{6}}^{\frac{11\mathrm{\pi }}{6}}\left(-2\right)dx+{\int }_{\frac{11\mathrm{\pi }}{6}}^{2\mathrm{\pi }}\left(-1\right)dx\\ & =& -{\left|x\right|}_{\mathrm{\pi }}^{\frac{7\mathrm{\pi }}{6}}-2{\left|x\right|}_{\frac{7\mathrm{\pi }}{6}}^{\frac{11\mathrm{\pi }}{6}}-1{\left|x\right|}_{\frac{11\mathrm{\pi }}{6}}^{2\mathrm{\pi }}\\ & =& -\left(\frac{7\mathrm{\pi }}{6}-\mathrm{\pi }\right)-2\left(\frac{11\mathrm{\pi }}{6}-\frac{7\mathrm{\pi }}{6}\right)-\left(2\mathrm{\pi }-\frac{11\mathrm{\pi }}{6}\right)\\ & =& -\frac{7\mathrm{\pi }-6\mathrm{\pi }}{6}-2\left(\frac{11\mathrm{\pi }-7\mathrm{\pi }}{6}\right)-\frac{12\mathrm{\pi }-11\mathrm{\pi }}{6}\\ & =& -\frac{\mathrm{\pi }}{6}-2\times \frac{4\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{6}\\ & =& \frac{-\mathrm{\pi }-8\mathrm{\pi }-\mathrm{\pi }}{6}\\ & =& -\frac{5\mathrm{\pi }}{3}\end{array}$

Hence, option B is correct.