Question

The value of integral ${\int }_0^{16\pi /3}\left|\sin x\right|dx$

• A. $21/2$
• B. $11/2$
• C. $9/2$
• D. $19/2$

Answer 1

Answer: (D)

$19/2$

Solution:- (D) None of these

${\int }_0^{\frac{16\pi }{3}}\left|\sin x\right|dx$

$\Rightarrow {\int }_0^{\frac{16\pi }{3}}\left|\sin x\right|dx={\int }_0^{\pi }\sin xdx-{\int }_{\pi }^{2\pi }\sin xdx+{\int }_{2\pi }^{3\pi }\sin xdx-{\int }_{3\pi }^{4\pi }\sin xdx+{\int }_{4\pi }^{\frac{16\pi }{3}}\sin xdx$

$\Rightarrow ={[-\cos x]}_0^{\pi }-{[-\cos x]}_{\pi }^{2\pi }+{[-cosx]}_{2\pi }^{3\pi }-{[-\cos x]}_{3\pi }^{4\pi }+{[-\cos x]}_{4\pi }^{\frac{16\pi }{3}}$

$\Rightarrow =-(\cos \pi -\cos 0)+(\cos 2\pi -\cos \pi )-(\cos 3\pi -\cos 2\pi )+(\cos 4\pi -\cos 3\pi )-(\cos \frac{16\pi }{3}-\cos 4\pi )$

$\Rightarrow =\cos 0-2(\cos \pi -\cos 2\pi +\cos 3\pi -\cos 4\pi )-\cos \frac{16\pi }{3}$

$\Rightarrow =1-2(-1-1-1-1)-(-\frac{1}{2})$

$\Rightarrow =1+8+\frac{1}{2}$

$\Rightarrow =\frac{19}{2}$

Hence ${\int }_0^{\frac{16\pi }{3}}\left|\sin x\right|dx=\frac{19}{2}$.