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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 32

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Answer:  \frac{2}{\pi}

Hint: To solve this question we will do integration by separation

Given:   \int_{0}^{1}|\sin 2 \pi x| d x


\int_{0}^{1}|\sin 2 \pi x| d x

\begin{aligned} &=\int_{0}^{\frac{1}{2}}(\sin 2 \pi x) d x+\int_{\frac{1}{2}}^{1}-\sin 2 \pi x d x \\ & \end{aligned}

=\left[\frac{-\cos 2 \pi x}{2 \pi}\right]_{0}^{\frac{1}{2}}+\left[\frac{\cos 2 \pi x}{2 \pi}\right]_{\frac{1}{2}}^{1}

\begin{aligned} &=\frac{-1}{2 \pi}[\cos \pi-\cos 0]+\frac{1}{2 \pi}[\cos 2 \pi-\cos \pi] \\ & \end{aligned}

=\frac{-1}{2 \pi}[-1-1]+\frac{1}{2 \pi}[1-(-1)] \\

=\frac{-1}{2 \pi} \times-2+\frac{1}{2 \pi} \times 2 \\

=\frac{1+1}{\pi} \\



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