Question 26 (OR 1 st question)

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

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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 lines 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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.