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Chapter 8 Class 12 Application of Integrals

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Last updated at March 16, 2023 by Teachoo

Ex 8.1, 12 Area lying in the first quadrant and bounded by the circle 𝑥2+𝑦2=4 and the lines 𝑥 = 0 and 𝑥 = 2 is (A) π (B) 𝜋2 (C) 𝜋3 (D) 𝜋4 Equation of Given Circle :- 𝑥2+ 𝑦2=4 𝑥2+ 𝑦2= 22 ∴ 𝑟𝑎𝑑𝑖𝑢𝑠 , 𝑟=2 Line 𝑥=0 is y-axis & Line x = 2 passes through point A 2 , 0 So, Required area = Area of shaded region = Area OAB = 02𝑦.𝑑𝑥 We know that, 𝑥2+ 𝑦2=4 𝑦2=4− 𝑥2 ∴ 𝑦=± 4− 𝑥2 As, OBA is in 1st Quadrant ∴ 𝑦= 4− 𝑥2 ∴ Required area = 02𝑦.𝑑𝑥 = 02 4− 𝑥2 𝑑𝑥 = 02 22− 𝑥2 𝑑𝑥 = 𝑥 2 22− 𝑥2+2 sin−1 𝑥2 02 = 22 22− 22+2 sin−1 22 − 02 22− 02−2 sin−1 02 = 0+2 sin−1 1−0 4−2 sin−1 0 = 2 sin−1 1−2 sin−1 0−0 = 2 sin−1 1− sin−1 0 = 2 𝜋2−0 = 2 . 𝜋2 = π ∴ Area Required = π square units Hence, Option (A) is correct