Area bounded by curve and horizontal or vertical line

Chapter 8 Class 12 Application of Integrals
Concept wise

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### Transcript

Example 4 Find the area bounded by the curve π¦=cosβ‘π₯ between π₯=0 and π₯=2πArea OAB = β«_0^(π/( 2))βγπ¦ ππ₯γ π¦βcosβ‘π₯ = β«_π^(π/( π))βγπππβ‘π ππγ = [sinβ‘π₯ ]_0^(π/2) =sinβ‘γπ/2βsinβ‘0 γ =1β0 =π Area BCD = β«_(π/( 2))^(3π/( 2))βγπ¦ ππ₯γ = β«_(π/( π))^(ππ/( π))βγπππβ‘π ππγ = [sinβ‘π₯ ]_(π/( 2))^(3π/( 2)) = sin 3π/( 2)βsinβ‘γπ/( 2)γ = β 1 β 1 = β2 Since area cannot be negative Area BCD = 2 Area DEF = β«_(3π/( 2))^2πβγπ¦ ππ₯γ = β«_(ππ/( π))^ππβγπππβ‘π ππγ = [sinβ‘π₯ ]_(3π/( 2))^2π =sinβ‘2π βsinβ‘γ3π/( 2)γ = 0β(β1) = π Therefore Area Required = Area OAB + Area BCD + Area DEF = 1 + 2 + 1 = 4 square unit