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  • Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 21 maths textbook solution

Please solve RD Sharma class 12 chapter Areas of Bounded Region exercise 20.1 question 21 maths textbook solution

Answers (1)

Answer:

\frac{\pi }{2}(\pi +2) s q \cdot { unit }

Hint:

use definite integral.

Given:

Draw rough sketch of curve y=\frac{\pi}{2}+2 \sin ^{2} x Find area between x-axis the curve and 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
   2\sin ^{2}x 1.57 2.07 3.57 2.07 1.5

y=\frac{\pi}{2}+2 \sin ^{2} x is an arc cutting y-axis at \left ( 1.57,0 \right )

 x=\pi  at (\pi ,1.57)

Area of shaded region

\begin{aligned} &A=\int_{0}^{\pi}|y| d x \\\\ &A=\int_{0}^{\pi} y d x \\\\ &A=\int_{0}^{\pi}\left[\frac{\pi}{2}+2 \sin ^{2} x\right] d x \end{aligned}

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

\begin{aligned} &A=\frac{\pi}{2}[\pi]+\left[\pi-\frac{\sin 2 \pi}{2}-0\right] \\\\ &A=\pi\left(\frac{\pi}{2}+1\right) \\\\ &A=\pi\left(\frac{\pi+2}{2}\right) \end{aligned}

A=\frac{\pi }{2}(\pi +2) s q \cdot{ unit }

 

Posted by

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