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Example 15 Important You are here

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at Dec. 12, 2019 by Teachoo

Example 15 Find the area of the region {(π₯, π¦) : 0 β€ π¦ β€ π₯2 + 1, 0 β€ π¦ β€ π₯ + 1, 0 β€ π₯ β€ 2} Here, πβ€πβ€π^π+π π¦β₯0 So it is above π₯βππ₯ππ π¦=π₯^2+1 i.e. π₯^2=π¦β1 So, it is a parabola πβ€πβ€π+π π¦β₯0 So it is above π₯βππ₯ππ π¦=π₯+1 It is a straight line Also πβ€πβ€π Since π¦β₯0 & 0β€π₯β€2 We work in First quadrant with 0β€π₯β€2 So, our figure is Finding point of intersection P & Q Here, P and Q are intersection of parabola and line Solving π¦=π₯^2+1 & π¦=π₯+1 π₯^2+1=π₯+1 π₯^2βπ₯+1β1=0 π₯^2βπ₯+0=0 π₯(π₯β1)=0 So, π₯=0 , π₯=1 For π = 0 π¦=π₯+1=0+1=1 So, P(0 , 1) For π = 1 π¦=π₯+1=1+1=2 So, Q(1 , 2) Finding area Area required = Area OPQRST Area OPQRST = Area OPQT + Area QRST Area OPQT Area OPQT =β«_0^1βγπ¦ ππ₯γ π¦β equation of Parabola PQ π¦=π₯^2+1 β΄ Area OPQT =β«_0^1β(π₯^2+1) =[π₯^3/3+π₯]_0^1 =[1^3/3+1]β[0^3/3+0] =1/3+1 =4/3 Area QRST Area QRST=β«_1^2βγπ¦ ππ₯γ Here, π¦β equation of line QP π¦=π₯ + 1 β΄ Area QRST=β«_1^2β(π₯+1) ππ₯ =[π₯^2/2+π₯]_1^2 =(2^2/2+2)β(1^2/2+1) =2+2β(1/2+1) =4β3/2 =5/2 Thus, Area Required = Area OPQT + Area QPST = 4/3+5/2 = (8 + 15)/6 = ππ/π square units