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

Ex 8.1

Ex 8.1, 1
Important

Ex 8.1, 2

Ex 8.1, 3 (MCQ)

Ex 8.1, 4 (MCQ) Important

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams You are here

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Important Deleted for CBSE Board 2024 Exams

Question 9 Deleted for CBSE Board 2024 Exams

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 3 Find the area of the region bounded by 𝑥2= 4𝑦 , 𝑦 = 2 , 𝑦=4 and the 𝑦-axis in the first quadrant. Here, The curve is 𝑥^2=4𝑦 We have to find area between y = 2 and y = 4 ∴ We have to find area of BCFE Area of BCFE = ∫_2^4▒𝑥 𝑑𝑦 We know that 𝑥^2=4𝑦 Taking square root on both sides 𝑥="±" √4𝑦‖ 𝑥="±" 2√𝑦 Since, BCEF is in 1st Quadrant We take positive value of x ∴ 𝑥=2√𝑦 Area of BCFE = ∫_2^4▒𝑥 𝑑𝑦 = ∫_2^4▒〖2√𝑦〗 𝑑𝑦 = 2∫_2^4▒〖(𝑦)^(1/2) 𝑑𝑦〗 = 2 [𝑦^(3/2)/(3/2)]_2^4 = 2 × 2/3 [𝑦^(3/2) ]_2^4 = 4/3 [(4)^(3/2 )−(2)^(3/2) ] = 4/3 [((4)^(1/2) )^3−((2)^(1/2) )^3 ] = 4/3 [(2)^3−(√2)^3 ] = 4/3 [8 −2√2] = (32 − 8√2)/3 Thus, Area = (𝟑𝟐 − 𝟖√𝟐)/𝟑 square units