MCQ
$\int_{\,\pi }^{\,10\pi } {\,|\sin x|dx} $ is
- A$20$
- B$8$
- C$10$
- ✓$18$
✓
Answer
Correct option: D.
$18$
d
(d) $\int_\pi ^{10\pi } {|\sin x|dx = \int_0^\pi {|\sin x|dx + \int_\pi ^{10\pi } {\,\,|\sin x|dx} } } - \int_0^\pi {\,|\sin x|dx} $
(d) $\int_\pi ^{10\pi } {|\sin x|dx = \int_0^\pi {|\sin x|dx + \int_\pi ^{10\pi } {\,\,|\sin x|dx} } } - \int_0^\pi {\,|\sin x|dx} $
$ = \int_0^{10\pi } {|\sin x|dx - \int_0^\pi {\,|\sin x|dx} } $
$ = 10\int_{\,0}^{\,\pi } {|\sin x|dx - \int_{\,0}^{\,\pi } {\,|\sin x|dx} } $
$ = 9\int_{\,0}^{\,\pi } {\sin x\,dx} $
$[\because \,|\sin x|$ is periodic with period $\pi $ and in $[0,\pi ],\sin x \ge 0]$
$ = 9\,[ - \cos x]_0^\pi = 9\,( - \cos \pi + \cos 0)$
$ = 9\,(1 + 1) = 18$.
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