Question

The value of $\underset{0}{\overset{16\pi /3}{\int }}|\sin x|dx$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{11}{2}$
• C. $\frac{21}{4}$
• D. $\frac{21}{2}$

Answer 1

The correct option is D $\frac{21}{2}$

Let $I=\underset{0}{\overset{16\pi /3}{\int }}|\sin x|dx$
$\Rightarrow I=\underset{0}{\overset{5\pi }{\int }}|\sin x|dx+\underset{5\pi }{\overset{5\pi +\pi /3}{\int }}|\sin x|dx$
Using the properties:
$\left(i\right).\underset{0}{\overset{nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx\left(ii\right).\underset{nT}{\overset{a+nT}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(x\right)dx\left[T\text{is the period of}f\right(x\left)\right]$
Now,
$I=5\underset{0}{\overset{\pi }{\int }}\sin xdx+\underset{0}{\overset{\pi /3}{\int }}\sin xdx=-5(\cos x{)}_0^{\pi }-(\cos x{)}_0^{\pi /3}\therefore I=10+\frac{1}{2}=\frac{21}{2}$