# Example 15 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 12, 2019 by Teachoo

Last updated at Dec. 12, 2019 by Teachoo

Transcript

Example 15 Find the area of the region {(๐ฅ, ๐ฆ) : 0 โค ๐ฆ โค ๐ฅ2 + 1, 0 โค ๐ฆ โค ๐ฅ + 1, 0 โค ๐ฅ โค 2} Here, ๐โค๐โค๐^๐+๐ ๐ฆโฅ0 So it is above ๐ฅโ๐๐ฅ๐๐ ๐ฆ=๐ฅ^2+1 i.e. ๐ฅ^2=๐ฆโ1 So, it is a parabola ๐โค๐โค๐+๐ ๐ฆโฅ0 So it is above ๐ฅโ๐๐ฅ๐๐ ๐ฆ=๐ฅ+1 It is a straight line Also ๐โค๐โค๐ Since ๐ฆโฅ0 & 0โค๐ฅโค2 We work in First quadrant with 0โค๐ฅโค2 So, our figure is Finding point of intersection P & Q Here, P and Q are intersection of parabola and line Solving ๐ฆ=๐ฅ^2+1 & ๐ฆ=๐ฅ+1 ๐ฅ^2+1=๐ฅ+1 ๐ฅ^2โ๐ฅ+1โ1=0 ๐ฅ^2โ๐ฅ+0=0 ๐ฅ(๐ฅโ1)=0 So, ๐ฅ=0 , ๐ฅ=1 For ๐ = 0 ๐ฆ=๐ฅ+1=0+1=1 So, P(0 , 1) For ๐ = 1 ๐ฆ=๐ฅ+1=1+1=2 So, Q(1 , 2) Finding area Area required = Area OPQRST Area OPQRST = Area OPQT + Area QRST Area OPQT Area OPQT =โซ_0^1โใ๐ฆ ๐๐ฅใ ๐ฆโ equation of Parabola PQ ๐ฆ=๐ฅ^2+1 โด Area OPQT =โซ_0^1โ(๐ฅ^2+1) =[๐ฅ^3/3+๐ฅ]_0^1 =[1^3/3+1]โ[0^3/3+0] =1/3+1 =4/3 Area QRST Area QRST=โซ_1^2โใ๐ฆ ๐๐ฅใ Here, ๐ฆโ equation of line QP ๐ฆ=๐ฅ + 1 โด Area QRST=โซ_1^2โ(๐ฅ+1) ๐๐ฅ =[๐ฅ^2/2+๐ฅ]_1^2 =(2^2/2+2)โ(1^2/2+1) =2+2โ(1/2+1) =4โ3/2 =5/2 Thus, Area Required = Area OPQT + Area QPST = 4/3+5/2 = (8 + 15)/6 = ๐๐/๐ square units

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.