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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.