Slide46.JPG Slide47.JPG Slide48.JPG Slide49.JPG Slide50.JPG Slide51.JPG

  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise


Example 15 Find the area of the region {(๐‘ฅ, ๐‘ฆ) : 0 โ‰ค ๐‘ฆ โ‰ค ๐‘ฅ2 + 1, 0 โ‰ค ๐‘ฆ โ‰ค ๐‘ฅ + 1, 0 โ‰ค ๐‘ฅ โ‰ค 2} Here, ๐ŸŽโ‰ค๐’šโ‰ค๐’™^๐Ÿ+๐Ÿ ๐‘ฆโ‰ฅ0 So it is above ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  ๐‘ฆ=๐‘ฅ^2+1 i.e. ๐‘ฅ^2=๐‘ฆโˆ’1 So, it is a parabola ๐ŸŽโ‰ค๐’šโ‰ค๐’™+๐Ÿ ๐‘ฆโ‰ฅ0 So it is above ๐‘ฅโˆ’๐‘Ž๐‘ฅ๐‘–๐‘  ๐‘ฆ=๐‘ฅ+1 It is a straight line Also ๐ŸŽโ‰ค๐’™โ‰ค๐Ÿ Since ๐‘ฆโ‰ฅ0 & 0โ‰ค๐‘ฅโ‰ค2 We work in First quadrant with 0โ‰ค๐‘ฅโ‰ค2 So, our figure is Finding point of intersection P & Q Here, P and Q are intersection of parabola and line Solving ๐‘ฆ=๐‘ฅ^2+1 & ๐‘ฆ=๐‘ฅ+1 ๐‘ฅ^2+1=๐‘ฅ+1 ๐‘ฅ^2โˆ’๐‘ฅ+1โˆ’1=0 ๐‘ฅ^2โˆ’๐‘ฅ+0=0 ๐‘ฅ(๐‘ฅโˆ’1)=0 So, ๐‘ฅ=0 , ๐‘ฅ=1 For ๐’™ = 0 ๐‘ฆ=๐‘ฅ+1=0+1=1 So, P(0 , 1) For ๐’™ = 1 ๐‘ฆ=๐‘ฅ+1=1+1=2 So, Q(1 , 2) Finding area Area required = Area OPQRST Area OPQRST = Area OPQT + Area QRST Area OPQT Area OPQT =โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— ๐‘ฆโ†’ equation of Parabola PQ ๐‘ฆ=๐‘ฅ^2+1 โˆด Area OPQT =โˆซ_0^1โ–’(๐‘ฅ^2+1) =[๐‘ฅ^3/3+๐‘ฅ]_0^1 =[1^3/3+1]โˆ’[0^3/3+0] =1/3+1 =4/3 Area QRST Area QRST=โˆซ_1^2โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— Here, ๐‘ฆโ†’ equation of line QP ๐‘ฆ=๐‘ฅ + 1 โˆด Area QRST=โˆซ_1^2โ–’(๐‘ฅ+1) ๐‘‘๐‘ฅ =[๐‘ฅ^2/2+๐‘ฅ]_1^2 =(2^2/2+2)โˆ’(1^2/2+1) =2+2โˆ’(1/2+1) =4โˆ’3/2 =5/2 Thus, Area Required = Area OPQT + Area QPST = 4/3+5/2 = (8 + 15)/6 = ๐Ÿ๐Ÿ‘/๐Ÿ” square units

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.