Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2 Important

Example 3

Example 4

Example 5 Important

Example 6 Important Deleted for CBSE Board 2023 Exams

Example 7 Important Deleted for CBSE Board 2023 Exams

Example 8 Important Deleted for CBSE Board 2023 Exams

Example 9 Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11

Example 12

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Important You are here

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at March 16, 2023 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