Get Answers to all your Questions

header-bg qa

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


\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 \\



Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support