# Example 9 - Chapter 8 Class 12 Application of Integrals (Term 2)

Last updated at Nov. 15, 2019 by Teachoo

Examples

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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 You are here

Example 10 Important Deleted for CBSE Board 2022 Exams

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Example 14 Important Deleted for CBSE Board 2022 Exams

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Chapter 8 Class 12 Application of Integrals (Term 2)

Serial order wise

Last updated at Nov. 15, 2019 by Teachoo

Example 9 Using integration find the area of region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1) Area of ∆ formed by point 1 , 0 , 2 ,2 & 3 , 1 Step 1: Draw the figure Area ABD Area ABD= 12𝑦 𝑑𝑥 𝑦→ equation of line AB Equation of line between A(1, 0) & B(2, 2) is 𝑦 − 0𝑥 − 1= 2 − 02 − 1 𝑦𝑥 − 1= 21 y = 2(x – 1) y = 2x – 2 Area ABD = 12𝑦 𝑑𝑥 = 122 𝑥−1 𝑑𝑥 = 2 𝑥22−𝑥12 =2 222−2− 122−1 =2 2−2− 12+1 =2 12 = 1 Area BDEC Area BDEC = 23𝑦 𝑑𝑥 𝑦→ equation of line BC Equation of line between B(2, 2) & C(3, 1) is 𝑦 − 2𝑥 − 2= 1 − 23 − 2 𝑦 − 2𝑥 − 2= −11 y – 2 = –1(x – 2) y – 2 = –x + 2 y = 4 – x Area BDEC = 23𝑦 𝑑𝑥 = 23 4−𝑥 𝑑𝑥 =4 23𝑑𝑥− 23𝑥𝑑𝑥 =4 𝑥23− 𝑥2223 =4 3−2− 12 32− 22 =4 ×1− 12 9−4 =4− 12 ×5 = 4− 52 = 8 − 52 = 32 Area ACE Area ACE= 13𝑦 𝑑𝑥 𝑦→ equation of line AC Equation of line between A(1, 0) & C(3, 1) is 𝑦 − 0𝑥 − 1= 1 − 03 − 1 𝑦𝑥 − 1= 12 y = 12 (x – 1) Area ACE = 13𝑦 𝑑𝑥 = 13 12 𝑥−1 𝑑𝑥 = 12 13 𝑥−1 𝑑𝑥 = 12 𝑥22−𝑥13 = 12 322−3− 122−1 = 12 92−3− 12+1 = 12 42 =1 Hence Area Required = Area ABD + Area BDEC – Area ACE = 1 + 32−1 = 32