






Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Area between 2 curves
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams You are here
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 (MCQ) Deleted for CBSE Board 2024 Exams
Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2024 Exams
Area between 2 curves
Last updated at May 29, 2023 by Teachoo
Question 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