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  4. ∫ limits02 π| sin x| d x=

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Q. $\int\limits_0^{2 \pi}|\sin x| d x=$

Integrals

Solution:

$I=\int\limits_0^{2 \pi}|\sin x| d x=2 \int\limits_0^\pi|\sin x| d x$
$ \because f(2 \pi-x)=f(x)$
$=4 \int\limits_0^{\pi / 2}|\sin x| d x=4 \int\limits_0^{\pi / 2} \sin x d x=4[-\cos x]_0^{\pi / 2}$
$=4(-0+1)=4$