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  • Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 22

Provide solution for RD Sharma maths class 12 chapter Areas of Bounded Region exercise 20.1 question 22

Answers (1)

best_answer

Answer:

\frac{3\pi }{2} s q \text { unit }

Hint:

You know about integral of \sin ^{2}x

Given:

Draw rough curve y=\frac{x}{\pi}+2 \sin ^{2} x . Find area between x-axis with ordinate x=0, x=\pi

Solution:

 

  x 0 \frac{\pi }{6} \frac{\pi }{2} \frac{5\pi }{6} \pi
  \sin x 0 \frac{1}{2} 1 \frac{1}{2} 0
  \frac{x}{\pi }+2\sin ^{2}x 0 \frac{2}{3} \frac{5}{2} \frac{4}{3} 1

A=\int_{0}^{\pi}|y| d x \quad\quad\quad\quad[y>0 \Rightarrow|y|=y]

\begin{aligned} &A=\int_{0}^{\pi}\left(\frac{x}{\pi}+2 \sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi} \int_{0}^{\pi}(x) d x+2 \int_{0}^{\pi}\left(\sin ^{2} x\right) d x \\\\ &A=\frac{1}{\pi}\left[\frac{x^{2}}{2}\right]_{0}^{\pi}+2\left[\frac{x}{2}-\frac{1}{2} \sin x \cos x\right] \end{aligned}

\begin{aligned} &A=\frac{\pi^{2}}{2 \pi}+\frac{2}{2}\left[\pi-\frac{1}{2} \sin x \cos x-0\right] \\\\ &A=\frac{\pi}{2}+\pi \\\\ &A=\frac{3 \pi}{2} s q \cdot u n i t \end{aligned}

 

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