Example 4 - Chapter 8 Class 12 Application of Integrals
Last updated at Dec. 16, 2024 by Teachoo
Area bounded by curve and horizontal or vertical line
Area bounded by curve and horizontal or vertical line
Last updated at Dec. 16, 2024 by Teachoo
Example 4 Find the area bounded by the curve π¦=cosβ‘π₯ between π₯=0 and π₯=2πArea OAB = β«_0^(π/( 2))βγπ¦ ππ₯γ π¦βcosβ‘π₯ = β«_π^(π /( π))βγπππβ‘π π πγ = [sinβ‘π₯ ]_0^(π/2) =sinβ‘γπ/2βsinβ‘0 γ =1β0 =π Area BCD = β«_(π/( 2))^(3π/( 2))βγπ¦ ππ₯γ = β«_(π /( π))^(ππ /( π))βγπππβ‘π π πγ = [sinβ‘π₯ ]_(π/( 2))^(3π/( 2)) = sin 3π/( 2)βsinβ‘γπ/( 2)γ = β 1 β 1 = β2 Since area cannot be negative Area BCD = 2 Area DEF = β«_(3π/( 2))^2πβγπ¦ ππ₯γ = β«_(ππ /( π))^ππ βγπππβ‘π π πγ = [sinβ‘π₯ ]_(3π/( 2))^2π =sinβ‘2π βsinβ‘γ3π/( 2)γ = 0β(β1) = π Therefore Area Required = Area OAB + Area BCD + Area DEF = 1 + 2 + 1 = 4 square unit