Ex 8.1, 10 - Find area bounded by x2 = 4y and line x = 4y - 2 - Ex 8.1

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  1. Chapter 8 Class 12 Application of Integrals
  2. Serial order wise
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Ex 8.1, 10 Find the area bounded by the curve 𝑥2=4𝑦 and the line 𝑥=4𝑦 – 2 Let AB represent the line 𝑥=4𝑦−2 AOB represent the curve 𝑥﷮2﷯=4𝑦 First we find Points A and B Points A & B are the intersection of curve and line We know that, 𝑥=4𝑦−2 Putting in equation of curve , we get 𝑥﷮2﷯=4𝑦 4𝑦−2﷯﷮2﷯=4𝑦 16 𝑦﷮2﷯+4−16𝑦=4𝑦 16 𝑦﷮2﷯−16𝑦−4𝑦+4=0 16 𝑦﷮2﷯−20𝑦+4=0 4 4 𝑦﷮2﷯−5𝑦+1﷯=0 4 𝑦﷮2﷯−5𝑦+1=0 4 𝑦﷮2﷯−4𝑦−𝑦+1=0 4𝑦 𝑦−1﷯−1 𝑦−1﷯=0 4𝑦−1﷯ 𝑦−1﷯=0 So, y = 1﷮4﷯ , y = 1 As Point A is in 2nd Quadrant ∴ A = −1 , 1﷮4﷯﷯ & Point B is in 1st Quadrant ∴ B = 2 , 1﷯ Finding required area Required Area = Area APBQ – Area APOQBA = −1﷮2﷮𝑦1𝑑𝑥﷯− −1﷮2﷮𝑦2𝑑𝑥﷯ Required Area = −1﷮2﷮ 𝑥 + 2﷮4﷯﷯𝑑𝑥− −1﷮2﷮ 𝑥﷮2﷯﷮4﷯﷯𝑑𝑥 = 1﷮4﷯ −1﷮2﷮ 𝑥+2﷯𝑑𝑥﷯ − 1﷮4﷯ −1﷮2﷮ 𝑥﷮2﷯ 𝑑𝑥﷯ = 1﷮4﷯ 𝑥﷮2﷯﷮2﷯+2𝑥﷯﷮−1﷮2﷯− 1﷮4﷯ 𝑥﷮3﷯﷮3﷯﷯﷮−1﷮2﷯ = 1﷮4﷯ 2﷯﷮2﷯ − −1﷯﷮2﷯﷮2﷯+2 2− −1﷯﷯﷯− 1﷮4﷯ 2﷯﷮3﷯ − −1﷯﷮3﷯﷮3﷯﷯ = 1﷮4﷯ 4 − 1﷮2﷯+2 3﷯﷯− 1﷮4﷯ 8 + 1﷮3﷯﷯ = 1﷮4﷯ 3﷮2﷯+6﷯− 1﷮4﷯ 9﷮3﷯﷯ = 1﷮4﷯ 3﷮2﷯+6−3﷯ = 1﷮4﷯ 3﷮2﷯+3﷯ = 1﷮4﷯ 9﷮2﷯﷯ = 9﷮8﷯ ∴ Required Area = 𝟗﷮𝟖﷯ Square units

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