Question 26 (OR 1 st question)
Find the area bounded by the curves y = √x, 2y + 3 = x and x axis
Last updated at Oct. 1, 2019 by Teachoo
Question 26 (OR 1 st question)
Find the area bounded by the curves y = √x, 2y + 3 = x and x axis
Transcript
Question 26 (OR 1st question) Find the area bounded by the curves y = βπ₯, 2y + 3 = x and x axis Given equation of curves y = βπ₯ 2y + 3 = x Here, y = βπ₯ y2 = x So, it is a parabola, with only positive values of y Drawing figure Drawing line 2y + 3 = x on the graph Finding point of intersection of line and curve y = βπ₯ Putting x = 2y + 3 from equation of line y = β(2π¦+3) Squaring both sides y2 = (β(2π¦+3))^2 y2 = 2y + 3 y2 β 2y β 3 = 0 y2 β 3y + y β 3 = 0 y(y β 3) + 1(y β 3) = 0 (y β 3) (y + 1) = 0 So, y = 3, y = β1 Since y cannot be negative y = 3 Since y cannot be negative y = 3 Now, putting y = 3 in lineβs equation 2y + 3 = x 2(3) + 3 = x 6 + 3 = x 9 = x x = 9 So, point is (9, 3) Now, letβs find the area Area required Area required = Area OAC β Area ABC Area OAC Area OAC = β«1_0^9βγπ¦ ππ₯γ y β Equation of curve y = βπ₯ Therefore, Area OAC = β«1_0^9βγβπ₯ ππ₯γ = β«1_0^9βγπ₯^(1/2) ππ₯γ = [π₯^(3/2)/(3/2)]_0^9 = 2/3 [9^(3/2)β0^(3/2) ] = 2/3 Γ 9^(3/2) = 2/3 Γ 3^((2 Γ 3/2) ) = 2/3 Γ 3^3 = 2 Γ 32 = 18 Area ABC Area ABC = β«1_3^9βγπ¦ ππ₯γ y β Equation of line 2y + 3 = x 2y = x β 3 y = π₯/2 β 3/2 Area ABC = β«1_3^9βγ(π₯/2β3/2) ππ₯γ = β«1_3^9βγπ₯/2 ππ₯γ β β«1_3^9βγ3/2 ππ₯γ = [π₯^2/(2 Γ 2)]_3^9 β 3/2 [π₯]_3^9 = [π₯^2/4]_3^9 β 3/2 [π₯]_3^9 = [9^2/4β3^2/4] β 3/2 [9β3] = [(81 β 9)/4] β 3/2 [6] = 18 β 9 = 9 Therefore, Area required = Area OAC β Area ABC = 18 β 9 = 9 square units
CBSE Class 12 Sample Paper for 2019 Boards
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CBSE Class 12 Sample Paper for 2019 Boards
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