1. Chapter 8 Class 12 Application of Integrals
2. Serial order wise

Transcript

Ex 8.2 , 7 Area lying between the curves 𝑦2 = 4𝑥 and 𝑦=2𝑥 is (A) 𝟐﷮𝟑﷯ (B) 𝟏﷮𝟑﷯ (C) 𝟏﷮𝟒﷯ (D) 𝟑﷮𝟒﷯ Step 1: Drawing figure Parabola is 𝑦2 =4x Also, 𝑦=2𝑥 passes through (0, 0) & (1, 2) Point (1, 2) lies in parabola y2 = 4x Hence, intersecting point A = (1, 2) Area required Area required = Area OBAD – Area OAD Area OBAD Area OBAD = 0﷮1﷮𝑦 𝑑𝑥﷯ y → Equation of parabola 𝑦﷮2﷯ = 4x 𝑦 = ﷮4x﷯ 𝑦 = 2 ﷮x﷯ Therefore, Area OBAD = 0﷮1﷮2 ﷮x﷯ 𝑑𝑥﷯ = 2 0﷮1﷮ 𝑥﷮ 1﷮2﷯﷯ 𝑑𝑥﷯ = 2 𝑥﷮ 1﷮2﷯+1﷯﷮ 1﷮2﷯+1﷯﷯﷮0﷮1﷯ = 2 𝑥﷮ 3﷮2﷯﷯﷮ 3﷮2﷯﷯﷯﷮0﷮1﷯ = 2 × 2﷮3﷯ 𝑥﷮ 3﷮2﷯﷯﷯﷮0﷮1﷯ = 4﷮3﷯ 1﷮ 3﷮2﷯﷯− 0﷮ 3﷮2﷯﷯﷯ = 4﷮3﷯ Area OAD Area OAD = 0﷮1﷮𝑦 𝑑𝑥﷯ y → Equation of line y = 2x Therefore, Area OAD = 0﷮1﷮2𝑥 𝑑𝑥﷯ = 2 0﷮1﷮𝑥 𝑑𝑥﷯ = 2 𝑥﷮2﷯﷮2﷯﷯﷮0﷮1﷯ = 2 × 1﷮2﷯ 𝑥﷮2﷯﷯﷮0﷮1﷯ = 1﷮2﷯− 0﷮2﷯﷯ = 1 Area required = Area OBAD − Area OAD = 4﷮3﷯ − 1 = 1﷮3﷯ So, B is correct option

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