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  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 13

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise 19.3 question 13

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Answer:  2-\sqrt{2}

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

Given:  \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}|\sin x| d x

Solution:

\mathrm{I}=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}|\sin x| d x

                                        

                                                                           

\begin{gathered} I=\int_{\frac{-\pi}{4}}^{0}(-\sin x) d x-\int_{0}^{\frac{\pi}{4}} \sin x d x \\ \end{gathered}

\begin{gathered} \left\{\begin{array}{l} \sin x>0,\left[0, \frac{\pi}{4}\right] \\\\ \sin x<0\left[-\frac{\pi}{4}, 0\right] \end{array}\right\} \end{gathered}

\begin{gathered} I=(\cos x) \frac{0}{\frac{0 \pi}{4}}+(\cos x)_{0}^{\frac{\pi}{4}} \\ \end{gathered}
I=1-\cos \left(\frac{-\pi}{4}\right)+\left(\cos \frac{\pi}{4}-\cos 0\right) \\
I=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}+1 \\

I=2-\sqrt{2}

 

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