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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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Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.