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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

Answers (1)

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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