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Ex 8.2, 4 - Using integration find area of triangle - Class 12

Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 2
Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 3 Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 4 Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 5 Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 6 Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 7 Ex 8.2, 4 - Chapter 8 Class 12 Application of Integrals - Part 8

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Ex 8.2, 4 Using integration find the area of region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2) Let points be A(– 1, 0), B(1, 3) and C(3, 2) We mark the points on the diagram Area Δ ABC = Area ABD + Area BDEC – Area ACE Area ABD Area ABD= ∫_(−1)^1▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of line AB Equation of line between A(–1, 0) & B(1, 3) is (𝑦 − 0)/(𝑥 − (−1))=(3 − 0)/(1 − (−1)) 𝑦/(𝑥 + 1)=3/2 y = 3/2 (x + 1) Eq. of line b/w (x1, y1) & (x2, y2) is (𝑦 − 𝑦1)/(𝑥 − 𝑥1)=(𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) Area ABD = ∫_(−1)^1▒〖𝑦 𝑑𝑥〗 = ∫1_(−1)^1▒〖3/2 (𝑥+1) 〗 dx = 3/2 ∫1_(−1)^1▒(𝑥+1) dx = 3/2 [𝑥^2/2+𝑥]_(−1)^1 = 3/2 [[1^2/2+1]−[(−1)^2/2+(−1)]] = 3/2 [[3/2]−[(−1)/2]] = 3/2 × 2 = 3 Area BDEC Area BDEC = ∫_1^3▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of line BC (𝑦 − 3)/(𝑥 − 1)=(2 − 3)/(3 − 1) (𝑦 − 3)/(𝑥 − 1)=(−1)/2 2(y – 3) = –1(x – 1) 2y – 6 = –x + 1 2y = –x + 7 y = 1/2 (–x + 7) Eq. of line b/w (x1, y1) & (x2, y2) is (𝑦 − 𝑦1)/(𝑥 − 𝑥1)=(𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) Area BDEC = ∫_1^3▒〖𝑦 𝑑𝑥〗 = ∫1_1^3▒〖1/2 (−𝑥+7)𝑑〗x = 1/2 [〖−𝑥〗^2/2+7𝑥]_1^3 = 1/2 [[〖−3〗^2/2+7(3)]− [〖(−1)/2〗^2+7(1)]] = 1/2 [(−9)/2+21+1/2−7] = 1/2 [(−9 + 1)/2+14] = 1/2 [(−8)/2+14] = 1/2 [−4+14] = 10/2 = 5 Area ACE Area ACE= ∫_(−1)^3▒〖𝑦 𝑑𝑥〗 𝑦→ Equation of line AC Equation of line between A(–1, 0) & C(3, 2) is (𝑦 − 0)/(𝑥 − (−1))=(2 − 0)/(3 − (−1)) 𝑦/(𝑥 + 1)=2/4 Eq. of line b/w (x1, y1) & (x2, y2) is (𝑦 − 𝑦1)/(𝑥 − 𝑥1)=(𝑦2 − 𝑦1)/(𝑥2 − 𝑥1) y = 1/2 (x + 1) Area ACE = ∫_(−1)^3▒〖𝑦 𝑑𝑥〗 = ∫1_(−1)^3▒〖1/2 (𝑥+1)𝑑〗x = 1/2 [𝑥^2/2+𝑥]_(−1)^3 = 1/2 [3^2/2+3−((−1)^2/2−1)] = 1/2 [9/2+3−(1/2−1)] = 1/2 [9/2+1/2+3+1] = 1/2 [4+4] = 4 Hence, Required area = Area ABD + Area BDEC − Area ACE = 3 + 5 − 4 = 4 square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.