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Ex 8.2
Ex 8.2, 2 Deleted for CBSE Board 2023 Exams
Ex 8.2, 3 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 4 Important Deleted for CBSE Board 2023 Exams
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Ex 8.2, 6 (MCQ) Deleted for CBSE Board 2023 Exams
Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2023 Exams
Last updated at March 16, 2023 by Teachoo
Ex 8.2 , 5 Using integration find the area of the triangular region whose sides have the equations π¦=2π₯+1, π¦=3π₯+1 and π₯=4 Lets Draw the figure & x = 4 Therefore, Required Area = Area ABC Finding point of Intersection B & C For B B is intersection of y = 3x + 1 & x = 4 Putting x = 4 in y = 3x + 1 y = 3(4) + 1 = 13 So, B(4, 13) For C C is intersection of y = 2x + 1 & x = 4 Putting x = 4 in y = 2x + 1 y = 2(4) + 1 = 9 So, C(4, 9) Finding Area Required Area ABC = Area OABD β Area OACD Area OABD Area OABD = β«1_0^4βγπ¦ ππ₯γ Here, y = 3x + 1 Area OABD = β«1_0^4βγ(3π₯+1) ππ₯γ = [(3π₯^2)/2+π₯]_0^4 = [(3γ(4)γ^2)/2+4β[(3γ(0)γ^2)/2+0]] = (3 Γ 16)/2 + 4 β 0 = 24 + 4 = 28 Area OACD Area OACD = β«1_0^4βγπ¦ ππ₯γ Here, y = 2x + 1 Area OACD = β«1_0^4βγ(2π₯+1) ππ₯γ = [(2π₯^2)/2+π₯]_0^4 = [(2γΓ4γ^2)/2+4β[(2γ Γ 0γ^2)/2+0]] = 16 + 4 β 0 = 20 Area Required = Area ABDO β Area ACDO = 28 β 20 = 8 square unit