Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 29.

Need solution for RD Sharma Maths Class 12 Chapter 20 Areas of Bounded Regions exercise 20.3 question 29.

Answers (1)

Answer:

\frac{9}{2}\text{sq. units}

Hint:

 Use concept.

Given:

The given equations are y=2-x^2 and  y+x=0 .

Solution:

 To find area bounded by

y=2-x^2            ......(1)

y+x=0        .......(2)

Equation (1) represents  a parabola with vertex (0,2)  and downward, meets axes at (\pm 0, \sqrt{2}) .

Equation (2) represents a line passing through (0,0)   and (2,-2) .The points of intersection of line  and parabola are  (2,-2)  and (1,-1) .

A rough sketch of curve is as follows:

Shaded region is sliced into rectangles with area =(y_2-y_1)_{\Delta }x. It slides from x =-1  to x = 2 so,

Required area =Region ABPCOA

\begin{aligned} &A=\int_{-1}^{2}\left(y_{1}-y_{2}\right) d x \\ &=\int_{-1}^{2}\left(2-x^{2}+x\right) d x \\ &=\left[2 x-\frac{x^{3}}{3}+\frac{x^{2}}{2}\right]_{-1}^{2} \\ &=\left[\left(4-\frac{8}{3}+2\right)-\left(-2+\frac{1}{3}+\frac{1}{2}\right)\right] \\ &=\left[\frac{10}{3}+\frac{7}{6}\right] \\ &=\frac{27}{6} \\ &=\frac{9}{2} \text { sq.units } \end{aligned}

Posted by

infoexpert24

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question