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


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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.