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Q. $ \int_{0}^{1}{|\sin 2\pi x|dx} $ is equal to

Jharkhand CECEJharkhand CECE 2011

Solution:

$ I=\int_{0}^{1}{|\sin 2\pi x|dx} $
$ =\int_{0}^{1/2}{(\sin 2\pi x)dx+\int_{1/2}^{1}{-(\sin 2\pi x)dx}} $
$ =-\frac{1}{2\pi }[\cos 2\pi x]_{0}^{1/2}+\frac{1}{2\pi }[\cos 2\pi x]_{1/2}^{1}=\frac{2}{\pi } $