MCQ
$\int_0^{2n\pi } {\left( {|\sin x| - \left. {\left| {\frac{1}{2}\sin x} \right.} \right|} \right)} \;dx$ equals
- A$n$
- ✓$2n$
- C$-2n$
- DNone of these
✓
Answer
Correct option: B.
$2n$
b
(b) $\int_0^{2n\pi } {\left( {|\sin x| - \frac{1}{2}|\sin x|} \right)} \;dx$
(b) $\int_0^{2n\pi } {\left( {|\sin x| - \frac{1}{2}|\sin x|} \right)} \;dx$
$= \frac{1}{2}\int_0^{2n\pi } {\;\;\;\;|\sin x|dx} $
$ = \frac{{2n}}{2} \times 2\int_0^{\pi /2} {\sin x\;dx }$
$={ 2n} [ - \cos x]_0^{\pi /2} = 2n.$
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