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Question 26 (OR 1 st question)

Find the area bounded by the curves y = √x, 2y + 3 = x and x axis

Find the area bounded by curves y = √x, 2y + 3 = x and x-axis

Question 26 (Or 1st) - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
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Question 26 (OR 1st question) Find the area bounded by the curves y = √π‘₯, 2y + 3 = x and x axis Given equation of curves y = √π‘₯ 2y + 3 = x Here, y = √π‘₯ 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 = √π‘₯ Putting x = 2y + 3 from equation of line y = √(2𝑦+3) Squaring both sides y2 = (√(2𝑦+3))^2 y2 = 2y + 3 y2 – 2y – 3 = 0 y2 – 3y + y – 3 = 0 y(y – 3) + 1(y – 3) = 0 (y – 3) (y + 1) = 0 So, y = 3, y = –1 Since y cannot be negative y = 3 Since y cannot be negative y = 3 Now, putting y = 3 in line’s equation 2y + 3 = x 2(3) + 3 = x 6 + 3 = x 9 = x x = 9 So, point is (9, 3) Now, let’s find the area Area required Area required = Area OAC – Area ABC Area OAC Area OAC = ∫1_0^9▒〖𝑦 𝑑π‘₯γ€— y β†’ Equation of curve y = √π‘₯ Therefore, Area OAC = ∫1_0^9β–’γ€–βˆšπ‘₯ 𝑑π‘₯γ€— = ∫1_0^9β–’γ€–π‘₯^(1/2) 𝑑π‘₯γ€— = [π‘₯^(3/2)/(3/2)]_0^9 = 2/3 [9^(3/2)βˆ’0^(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 = ∫1_3^9▒〖𝑦 𝑑π‘₯γ€— y β†’ Equation of line 2y + 3 = x 2y = x – 3 y = π‘₯/2 – 3/2 Area ABC = ∫1_3^9β–’γ€–(π‘₯/2βˆ’3/2) 𝑑π‘₯γ€— = ∫1_3^9β–’γ€–π‘₯/2 𝑑π‘₯γ€— – ∫1_3^9β–’γ€–3/2 𝑑π‘₯γ€— = [π‘₯^2/(2 Γ— 2)]_3^9 – 3/2 [π‘₯]_3^9 = [π‘₯^2/4]_3^9 – 3/2 [π‘₯]_3^9 = [9^2/4βˆ’3^2/4] – 3/2 [9βˆ’3] = [(81 βˆ’ 9)/4] – 3/2 [6] = 18 – 9 = 9 Therefore, Area required = Area OAC – Area ABC = 18 – 9 = 9 square units

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.