The value of the integral int_{0}^{pi}|sin 2x | vedclass

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(B) Let $I = \int_{0}^{\pi} |\sin 2x| dx$.Substitute $2x = t$,so $2 dx = dt$ or $dx = \frac{1}{2} dt$.When $x = 0$,$t = 0$. When $x = \pi$,$t = 2\pi$.

Vedclass · Accepted Answer

$2$

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(B) Let $I = \int_{0}^{\pi} |\sin 2x| dx$.Substitute $2x = t$,so $2 dx = dt$ or $dx = \frac{1}{2} dt$.When $x = 0$,$t = 0$. When $x = \pi$,$t = 2\pi$.Thus,$I = \frac{1}{2} \int_{0}^{2\pi} |\sin t| dt$.Using the property $\int_{0}^{2a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(2a-x) = f(x)$,we note that $|\sin(2\pi - t)| = |-\sin t| = |\sin t|$.So,$I = \frac{1}{2} 	imes 2 \int_{0}^{\pi} |\sin t| dt = \int_{0}^{\pi} \sin t dt$.Evaluating the integral: $[-\cos t]_{0}^{\pi} = -(\cos \pi - \cos 0) = -(-1 - 1) = 2$.

The value of the integral $\int_{0}^{\pi}|\sin 2x| dx$ is

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