Ex 8.2

Chapter 8 Class 12 Application of Integrals
Serial order wise

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### Transcript

Ex 8.2, 4 Using integration find the area of region bounded by the triangle whose vertices are (β1, 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«_(β1)^1βγπ¦ ππ₯γ π¦β Equation of line AB Equation of line between A(β1, 0) & B(1, 3) is (π¦ β 0)/(π₯ β (β1))=(3 β 0)/(1 β (β1)) 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«_(β1)^1βγπ¦ ππ₯γ = β«1_(β1)^1βγ3/2 (π₯+1) γ dx = 3/2 β«1_(β1)^1β(π₯+1) dx = 3/2 [π₯^2/2+π₯]_(β1)^1 = 3/2 [[1^2/2+1]β[(β1)^2/2+(β1)]] = 3/2 [[3/2]β[(β1)/2]] = 3/2 Γ 2 = 3 Area BDEC Area BDEC = β«_1^3βγπ¦ ππ₯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)=(β1)/2 2(y β 3) = β1(x β 1) 2y β 6 = βx + 1 2y = βx + 7 y = 1/2 (βx + 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«_1^3βγπ¦ ππ₯γ = β«1_1^3βγ1/2 (βπ₯+7)πγx = 1/2 [γβπ₯γ^2/2+7π₯]_1^3 = 1/2 [[γβ3γ^2/2+7(3)]β [γ(β1)/2γ^2+7(1)]] = 1/2 [(β9)/2+21+1/2β7] = 1/2 [(β9 + 1)/2+14] = 1/2 [(β8)/2+14] = 1/2 [β4+14] = 10/2 = 5 Area ACE Area ACE= β«_(β1)^3βγπ¦ ππ₯γ π¦β Equation of line AC Equation of line between A(β1, 0) & C(3, 2) is (π¦ β 0)/(π₯ β (β1))=(2 β 0)/(3 β (β1)) 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«_(β1)^3βγπ¦ ππ₯γ = β«1_(β1)^3βγ1/2 (π₯+1)πγx = 1/2 [π₯^2/2+π₯]_(β1)^3 = 1/2 [3^2/2+3β((β1)^2/2β1)] = 1/2 [9/2+3β(1/2β1)] = 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