# Question 5 - Examples - Chapter 8 Class 12 Application of Integrals

Last updated at April 16, 2024 by Teachoo

Examples

Example 1

Example 2 Important

Example 3

Example 4 Important

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Question 4 Important Deleted for CBSE Board 2025 Exams

Question 5 Important Deleted for CBSE Board 2025 Exams You are here

Question 6 Important Deleted for CBSE Board 2025 Exams

Question 7 Deleted for CBSE Board 2025 Exams

Question 8 Important Deleted for CBSE Board 2025 Exams

Question 9 Deleted for CBSE Board 2025 Exams

Question 10 Important Deleted for CBSE Board 2025 Exams

Question 11 Important Deleted for CBSE Board 2025 Exams

Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 5 Find the area lying above x-axis and included between the circle ๐ฅ2 +๐ฆ2=8๐ฅ and inside of the parabola ๐ฆ2=4๐ฅ Since equation of circle is of form (๐ฅโ๐)^2+(๐ฆโ๐)^2=๐^2 , We convert our equation ๐ฅ^2+๐ฆ^2=8๐ฅ ๐ฅ^2โ8๐ฅ+๐ฆ^2=0 ๐ฅ^2โ2 ร4 ร๐ฅ+๐ฆ^2=0 ๐ฅ^2โ2 ร4 ร๐ฅ+4^2โ4^2+๐ฆ^2=0 (๐ฅโ4)^2+๐ฆ^2=4^2 So, Circle has center (4 , 0) & Radius =4 We need to find Area OPQC Point P is point of intersection of circle and parabola Finding Point P Equation of circle is ๐ฅ^2+๐ฆ^2=8๐ฅ Putting ๐ฆ^2=4๐ฅ ๐ฅ^2+4๐ฅ=8๐ฅ ๐ฅ^2=8๐ฅโ4๐ฅ ๐ฅ^2=4๐ฅ ๐ฅ^2โ4๐ฅ=0 ๐ฅ(๐ฅโ4)=0 So, ๐ฅ=0 & ๐ฅ=4 For ๐ = 0 ๐ฆ^2=4๐ฅ=4 ร 0=0 ๐ฆ=0 So, point is (0, 0) For ๐ = 4 ๐ฆ^2=4๐ฅ=4 ร4=4^2 ๐ฆ=4 So, point is (4, 4) So, ๐ฅ=0 & ๐ฅ=4 Since point P is in 1st quadrant, Coordinates of P = (4, 4) Note that ๐ฅ-coordinate same as that of center (4, 0) โด P lies above point C So, we need to change the figure New figure Area Required Area Required = Area OPC + Area PCQ Area OPC Area OPC = โซ_0^4โใ๐ฆ ๐๐ฅใ Here, y โ Equation of parabola y2 = 4x y = ยฑ โ4๐ฅ y = ยฑ 2โ๐ฅ Since OPC is in 1st quadrant, value of y is positive y = 2โ๐ฅ โด Area OPC = โซ_0^4โใ2โ๐ฅใ ๐๐ฅ = 2 โซ_0^4โ๐ฅ^(1/2) ๐๐ฅ = 2 [๐ฅ^(1/2 + 1)/(1/2 + 1)]_0^4 = 2 [๐ฅ^(3/2)/(3/2)]_0^4 = 2 ร 2/3 [(4)^(3/2)โ(0)^(3/2) ] = 4/3 [8โ0] = 32/3 Area PCQ Area PCQ = โซ_4^8โใ๐ฆ ๐๐ฅใ Here, y โ Equation of circle x2 + y2 = 8x y2 = 8x โ x2 y = ยฑ โ(8๐ฅโ๐ฅ^2 ) Since PCQ is in 1st quadrant, value of y is positive y = โ(8๐ฅโ๐ฅ^2 ) โด Area PCQ = โซ_4^8โโ(8๐ฅโ๐ฅ^2 ) ๐๐ฅ = โซ_4^8โโ(โ(๐ฅ^2โ8๐ฅ)) ๐๐ฅ = โซ_4^8โโ(โ(๐ฅ^2โ8๐ฅ+16โ16)) ๐๐ฅ = โซ_4^8โโ(โ(๐ฅ^2โ8๐ฅ+16)โ(โ16)) ๐๐ฅ = โซ_4^8โโ(16โ(๐ฅ^2โ8๐ฅ+16)) ๐๐ฅ = โซ_4^8โโ(16โ(๐ฅโ4)^2 ) ๐๐ฅ = โซ_4^8โโ(4^2โ(๐ฅโ4)^2 ) ๐๐ฅ = [((๐ฅ โ 4))/2 โ(4^2โใ(๐ฅโ4)ใ^2 )+4^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ((๐ฅ โ 4))/4ใ " " ]_4^8 It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ Here, a = 4, x = x โ 4 = [((๐ฅ โ 4))/2 โ(16โ(๐ฅ^2โ8๐ฅ+4^2))+16/2 ใ๐ ๐๐ใ^(โ1)โกใ ((๐ฅ โ 4))/4ใ " " ]_4^8 = [((๐ฅ โ 4))/2 โ(โ(๐ฅ^2โ8๐ฅ))+8 ใ๐ ๐๐ใ^(โ1)โกใ ((๐ฅ โ 4))/4ใ " " ]_4^8 = [((8 โ 4))/2 โ(โ(8^2โ8(8)))+8 ใ๐ ๐๐ใ^(โ1)โกใ ((8 โ 4))/4ใ ] โ [((4 โ 4))/2 โ(โ(4^2โ8(4)))+8 ใ๐ ๐๐ใ^(โ1)โกใ ((4 โ 4))/4ใ ] = [4/2 โ0+8 ใ๐ ๐๐ใ^(โ1)โกใ 1ใ ] โ [0+8 ใ๐ ๐๐ใ^(โ1)โกใ 0ใ ] = 8 ใ๐ ๐๐ใ^(โ1)โกใ 1ใ โ 8 ใ๐ ๐๐ใ^(โ1)โกใ 0ใ = 8(๐/2) โ 8 ร 0 = 4๐ As ใ๐ ๐๐ใ^(โ1)โกใ 1ใ = ๐/2 & ใ๐ ๐๐ใ^(โ1)โกใ 0ใ = 0 Thus, Area Required = Area OPC + Area PCQ = 32/3 + 4๐ = ๐/๐ (8 + 3๐ ) square units