Ex 8.2, 3 - Find area bounded by: y = x2 + 2, y = x, x=0,3

Ex 8.2, 3 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.2, 3 - Chapter 8 Class 12 Application of Integrals - Part 3
Ex 8.2, 3 - Chapter 8 Class 12 Application of Integrals - Part 4
Ex 8.2, 3 - Chapter 8 Class 12 Application of Integrals - Part 5

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Ex 8.2 , 3 Find the area of the region bounded by the curves 𝑦=𝑥2+2, 𝑦=𝑥, 𝑥=0 and 𝑥=3 Here, 𝑦=𝑥2+2 𝑦−2=𝑥^2 𝑥^2=(𝑦−2) So, it is a parabola And, 𝑥=𝑦 is a line x = 3 is a line x = 0 is the y-axis Finding point of intersection B & C Point B Point B is intersection of x = 3 and parabola Putting 𝑥=3 in 𝑥^2=(𝑦−2) 3^2=(𝑦−2) 9 = 𝑦−2 𝑦=11 Hence, B = (3 , 11) Point C Point C is the intersection of x = 3 and x = y Putting 𝑥=3 in 𝑥=𝑦 3=𝑦 i.e. 𝑦=3 Hence C = (3 , 3) Finding Area Area required = Area ABDO – Area OCD Area ABDO Area ABDO = ∫_0^3▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of parabola AB 𝑦=𝑥^2+2 ∴ Area ABDO = ∫_0^3▒〖𝑦 𝑑𝑥〗 = ∫_0^3▒〖(𝑥^2+2) 𝑑𝑥〗 = [𝑥^3/3+2𝑥]_0^3 = [3^3/3+2 ×3−0^3/3] = 9+6 = 15 Area OCD Area OCD = ∫_0^3▒〖𝑦 𝑑𝑥〗 𝑦→ equation of line 𝑦=𝑥 ∴ Area OCD = ∫_0^3▒〖𝑦 𝑑𝑥〗 = ∫_0^3▒〖𝑥 𝑑𝑥〗 = [𝑥^2/2]_0^3 =[3^2/2−0^2/2] = 9/2 Area required = Area ABDO – Area OCD = 15 – 9/2 = 𝟐𝟏/𝟐 square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo