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Ex 8.2, 5 - Using integration, find area of triangle whose sides y=2x+

Ex 8.2, 5 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.2, 5 - Chapter 8 Class 12 Application of Integrals - Part 3 Ex 8.2, 5 - Chapter 8 Class 12 Application of Integrals - Part 4 Ex 8.2, 5 - Chapter 8 Class 12 Application of Integrals - Part 5 Ex 8.2, 5 - Chapter 8 Class 12 Application of Integrals - Part 6

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Ex 8.2 , 5 Using integration find the area of the triangular region whose sides have the equations 𝑦=2π‘₯+1, 𝑦=3π‘₯+1 and π‘₯=4 Lets Draw the figure & x = 4 Therefore, Required Area = Area ABC Finding point of Intersection B & C For B B is intersection of y = 3x + 1 & x = 4 Putting x = 4 in y = 3x + 1 y = 3(4) + 1 = 13 So, B(4, 13) For C C is intersection of y = 2x + 1 & x = 4 Putting x = 4 in y = 2x + 1 y = 2(4) + 1 = 9 So, C(4, 9) Finding Area Required Area ABC = Area OABD – Area OACD Area OABD Area OABD = ∫1_0^4▒〖𝑦 𝑑π‘₯γ€— Here, y = 3x + 1 Area OABD = ∫1_0^4β–’γ€–(3π‘₯+1) 𝑑π‘₯γ€— = [(3π‘₯^2)/2+π‘₯]_0^4 = [(3γ€–(4)γ€—^2)/2+4βˆ’[(3γ€–(0)γ€—^2)/2+0]] = (3 Γ— 16)/2 + 4 βˆ’ 0 = 24 + 4 = 28 Area OACD Area OACD = ∫1_0^4▒〖𝑦 𝑑π‘₯γ€— Here, y = 2x + 1 Area OACD = ∫1_0^4β–’γ€–(2π‘₯+1) 𝑑π‘₯γ€— = [(2π‘₯^2)/2+π‘₯]_0^4 = [(2γ€–Γ—4γ€—^2)/2+4βˆ’[(2γ€– Γ— 0γ€—^2)/2+0]] = 16 + 4 βˆ’ 0 = 20 Area Required = Area ABDO βˆ’ Area ACDO = 28 βˆ’ 20 = 8 square unit

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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.