Question

The value of ${\int }_0^{16\pi /3}|sinx|dx$, is?

• A. $21$
• B. $\frac{21}{2}$
• C. $10$
• D. $11$

Answer 1

Answer: (B)

$\frac{21}{2}$

$I={\int }_0^{16\pi /3}\left|sinx\right|dx$

$={\int }_0^{\pi }\left|sinx\right|dx+{\int }_{\pi }^{2\pi }\left|sinx\right|dx+{\int }_{2\pi }^{3\pi }\left|sinx\right|dx$

$+{\int }_{3\pi }^{4\pi }\left|sinx\right|dx+{\int }_{4\pi }^{5\pi }\left|sinx\right|dx+{\int }_{5\pi }^{16\pi /3}\left|sinx\right|dx$
As is first and second quadrant sine function is positive and in third and fourth quadrant sin function is negative
$I={\int }_0^{\pi }\left(sinx\right)dx-{\int }_{\pi }^{2\pi }\left(sinx\right)dx+{\int }_{2\pi }^{3\pi }\left(sinx\right)dx$

$-{\int }_{3\pi }^{4\pi }\left(sinx\right)dx+{\int }_{4\pi }^{5\pi }\left(sinx\right)dx-{\int }_{5\pi }^{16\pi /3}\left(sinx\right)dx$

$=3{\int }_0^{\pi }\left(sinx\right)dx-2{\int }_{\pi }^{2\pi }\left(sinx\right)dx-{\int }_{5\pi }^{16\pi /3}\left(sinx\right)dx$

$=3{[-cosx]}_0^{\pi }-2{[-cosx]}_{\pi }^{2\pi }-{[-cosx]}_{5\pi }^{16\pi /3}$

$=3(1+1)-2(-1-1)-(\frac{1}{2}-1)=6+4+\frac{1}{2}=\frac{21}{2}$

Hence, option 'B' is correct.