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Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 12

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Answer:  4

Hint: You must know the rules of solving definite integral.

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

Solution:

                \begin{aligned} &I=\int_{0}^{2 \pi}|\sin x| d x=\int_{0}^{\pi} \sin x d x-\int_{\pi}^{2 \pi} \sin x d x \\ & \end{aligned}

                I=\int_{0}^{\pi} \sin x d x-\int_{\pi}^{2 \pi} \sin x d x \\

               I=(-\cos x)_{0}^{\pi}-(-\cos x)_{\pi}^{2 \pi} \\

               I=(1+1)-(-1-1) \\

               I=2-(-2) \\

               I=4

 

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