# Example 15 - Chapter 8 Class 12 Application of Integrals (Term 2)

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

Examples

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Example 4

Example 5 Important

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Important Deleted for CBSE Board 2022 Exams

Example 8 Important Deleted for CBSE Board 2022 Exams

Example 9 Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11

Example 12

Example 13 Important

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Important You are here

Chapter 8 Class 12 Application of Integrals (Term 2)

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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.