Check sibling questions

Example 15 - Find area: {(x, y) : 0 < y < x2 + 1, 0 < y < x+1

Example 15 - Chapter 8 Class 12 Application of Integrals - Part 2
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 3
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 4
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 5
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 6
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 7
Example 15 - Chapter 8 Class 12 Application of Integrals - Part 8

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Transcript

Example 15 Find the area of the region {(π‘₯, 𝑦) : 0 ≀ 𝑦 ≀ π‘₯2 + 1, 0 ≀ 𝑦 ≀ π‘₯ + 1, 0 ≀ π‘₯ ≀ 2} Here, πŸŽβ‰€π’šβ‰€π’™^𝟐+𝟏 𝑦β‰₯0 So it is above π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 𝑦=π‘₯^2+1 i.e. π‘₯^2=π‘¦βˆ’1 So, it is a parabola πŸŽβ‰€π’šβ‰€π’™+𝟏 𝑦β‰₯0 So it is above π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 𝑦=π‘₯+1 It is a straight line Also πŸŽβ‰€π’™β‰€πŸ Since 𝑦β‰₯0 & 0≀π‘₯≀2 We work in First quadrant with 0≀π‘₯≀2 So, our figure is Finding point of intersection P & Q Here, P and Q are intersection of parabola and line Solving 𝑦=π‘₯^2+1 & 𝑦=π‘₯+1 π‘₯^2+1=π‘₯+1 π‘₯^2βˆ’π‘₯+1βˆ’1=0 π‘₯^2βˆ’π‘₯+0=0 π‘₯(π‘₯βˆ’1)=0 So, π‘₯=0 , π‘₯=1 For 𝒙 = 0 𝑦=π‘₯+1=0+1=1 So, P(0 , 1) For 𝒙 = 1 𝑦=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 =∫_0^1▒〖𝑦 𝑑π‘₯γ€— 𝑦→ equation of Parabola PQ 𝑦=π‘₯^2+1 ∴ Area OPQT =∫_0^1β–’(π‘₯^2+1) =[π‘₯^3/3+π‘₯]_0^1 =[1^3/3+1]βˆ’[0^3/3+0] =1/3+1 =4/3 Area QRST Area QRST=∫_1^2▒〖𝑦 𝑑π‘₯γ€— Here, 𝑦→ equation of line QP 𝑦=π‘₯ + 1 ∴ Area QRST=∫_1^2β–’(π‘₯+1) 𝑑π‘₯ =[π‘₯^2/2+π‘₯]_1^2 =(2^2/2+2)βˆ’(1^2/2+1) =2+2βˆ’(1/2+1) =4βˆ’3/2 =5/2 Thus, Area Required = Area OPQT + Area QPST = 4/3+5/2 = (8 + 15)/6 = πŸπŸ‘/πŸ” square units

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.