Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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)

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}

 

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE OSM controversy deepens as students allege fresh errors, seek grace marks
anam-khan

CBSE to issue revised Class-12 physical marksheets through regional offices
anam-khan

NEET aspirant Aditya Mishra becomes CBSE Science topper after re-evaluation

Latest Question