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Ex 8.2

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Chapter 8 Class 12 Application of Integrals

Serial order wise

Last updated at March 16, 2023 by Teachoo

Ex 8.2, 4 Using integration find the area of region bounded by the triangle whose vertices are (β1, 0), (1, 3) and (3, 2) Let points be A(β 1, 0), B(1, 3) and C(3, 2) We mark the points on the diagram Area Ξ ABC = Area ABD + Area BDEC β Area ACE Area ABD Area ABD= β«_(β1)^1βγπ¦ ππ₯γ π¦β Equation of line AB Equation of line between A(β1, 0) & B(1, 3) is (π¦ β 0)/(π₯ β (β1))=(3 β 0)/(1 β (β1)) π¦/(π₯ + 1)=3/2 y = 3/2 (x + 1) Eq. of line b/w (x1, y1) & (x2, y2) is (π¦ β π¦1)/(π₯ β π₯1)=(π¦2 β π¦1)/(π₯2 β π₯1) Area ABD = β«_(β1)^1βγπ¦ ππ₯γ = β«1_(β1)^1βγ3/2 (π₯+1) γ dx = 3/2 β«1_(β1)^1β(π₯+1) dx = 3/2 [π₯^2/2+π₯]_(β1)^1 = 3/2 [[1^2/2+1]β[(β1)^2/2+(β1)]] = 3/2 [[3/2]β[(β1)/2]] = 3/2 Γ 2 = 3 Area BDEC Area BDEC = β«_1^3βγπ¦ ππ₯γ π¦β Equation of line BC (π¦ β 3)/(π₯ β 1)=(2 β 3)/(3 β 1) (π¦ β 3)/(π₯ β 1)=(β1)/2 2(y β 3) = β1(x β 1) 2y β 6 = βx + 1 2y = βx + 7 y = 1/2 (βx + 7) Eq. of line b/w (x1, y1) & (x2, y2) is (π¦ β π¦1)/(π₯ β π₯1)=(π¦2 β π¦1)/(π₯2 β π₯1) Area BDEC = β«_1^3βγπ¦ ππ₯γ = β«1_1^3βγ1/2 (βπ₯+7)πγx = 1/2 [γβπ₯γ^2/2+7π₯]_1^3 = 1/2 [[γβ3γ^2/2+7(3)]β [γ(β1)/2γ^2+7(1)]] = 1/2 [(β9)/2+21+1/2β7] = 1/2 [(β9 + 1)/2+14] = 1/2 [(β8)/2+14] = 1/2 [β4+14] = 10/2 = 5 Area ACE Area ACE= β«_(β1)^3βγπ¦ ππ₯γ π¦β Equation of line AC Equation of line between A(β1, 0) & C(3, 2) is (π¦ β 0)/(π₯ β (β1))=(2 β 0)/(3 β (β1)) π¦/(π₯ + 1)=2/4 Eq. of line b/w (x1, y1) & (x2, y2) is (π¦ β π¦1)/(π₯ β π₯1)=(π¦2 β π¦1)/(π₯2 β π₯1) y = 1/2 (x + 1) Area ACE = β«_(β1)^3βγπ¦ ππ₯γ = β«1_(β1)^3βγ1/2 (π₯+1)πγx = 1/2 [π₯^2/2+π₯]_(β1)^3 = 1/2 [3^2/2+3β((β1)^2/2β1)] = 1/2 [9/2+3β(1/2β1)] = 1/2 [9/2+1/2+3+1] = 1/2 [4+4] = 4 Hence, Required area = Area ABD + Area BDEC β Area ACE = 3 + 5 β 4 = 4 square units