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

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

Answers (1)

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

Solution:

\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} \\

=\frac{2}{\pi}

 

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