# Example 13 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Dec. 12, 2019 by Teachoo

Examples

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Example 5 Important

Example 6 Important Deleted for CBSE Board 2022 Exams

Example 7 Important Deleted for CBSE Board 2022 Exams

Example 8 Important Deleted for CBSE Board 2022 Exams

Example 9 Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

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Example 12

Example 13 Important You are here

Example 14 Important Deleted for CBSE Board 2022 Exams

Example 15 Important

Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

Last updated at Dec. 12, 2019 by Teachoo

Example 13 Find the area bounded by the curve π¦=cosβ‘π₯ between π₯=0 and π₯=2π Area Required = Area OAB + Area BCD + Area DEF x = π/2 Area OAB = β«_0^(π/( 2))βγπ¦ ππ₯γ π¦βcosβ‘π₯ = β«_0^(π/( 2))βγcosβ‘π₯ ππ₯γ = [sinβ‘π₯ ]_0^(π/2) =sinβ‘γπ/2βsinβ‘0 γ =1β0 =1 Area BCD = β«_(π/( 2))^(3π/( 2))βγπ¦ ππ₯γ = β«_(π/( 2))^(3π/( 2))βγcosβ‘π₯ ππ₯γ = [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«_(3π/( 2))^2πβγcosβ‘π₯ ππ₯γ = [sinβ‘π₯ ]_(3π/( 2))^2π =sinβ‘2π βsinβ‘γ3π/( 2)γ = 0β(β1) = 1 Therefore Area Required = Area OAB + Area BCD + Area DEF = 1 + 2 + 1 = 4 square unit