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Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y2 = x and the x-axis.

This question is similar to Misc 7 Chapter 8 Class 12 - Applications of Integrals

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Question 12 (Choice 1) Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y2 = x and the x-axis. Let’s first draw the Figure Here, π’šπŸ =𝒙 is a Parabola And, x + y = 2 is a straight line Let A be point of intersection of line and parabola And, Point B is (2, 0) Finding point A Since x + y = 2 y = 2 βˆ’ x Putting y = 2 βˆ’ x in equation of parabola 𝑦^2=π‘₯ (πŸβˆ’π’™)^𝟐=𝒙 4+π‘₯^2βˆ’4π‘₯=π‘₯ π‘₯^2βˆ’4π‘₯βˆ’π‘₯+4=0 𝒙^πŸβˆ’πŸ“π’™+πŸ’=𝟎 π‘₯^2βˆ’4π‘₯βˆ’π‘₯+4=0 π‘₯(π‘₯βˆ’4)βˆ’1(π‘₯βˆ’4)=0 (π‘₯βˆ’4)(π‘₯βˆ’1)=0 So, x = 4, x = 1 Since for point A, x-coordinate will be less than 2 ∴ x = 1 Putting x = 1 in equation of line x + y = 2 1 + y = 2 y = 2 βˆ’ 1 y = 1 So, Coordinates of point A = (1, 1) Finding Area Area Required = Area OAD + Area ADB Area OAD Area OAD = ∫_0^1▒〖𝑦 𝑑π‘₯" " γ€— y β†’ Equation of parabola 𝑦^2 = x 𝑦 = Β± √π‘₯ Since OAD is in 1st quadrant, value of y is positive ∴ π’š = βˆšπ’™ Now, Area OAD = ∫_0^1▒〖𝑦 𝑑π‘₯" " γ€— = ∫_𝟎^πŸβ–’γ€–βˆšπ’™ 𝒅𝒙" " γ€— = [π‘₯^(1/2 + 1)/(1/2 + 1)]_0^1 = [π‘₯^(3/2)/(3/2)]_0^1 = 𝟐/πŸ‘ [𝒙^(πŸ‘/𝟐) ]_𝟎^𝟏 = 2/3 [1^(3/2)βˆ’0^(3/2) ] = 𝟐/πŸ‘ Area ADB Area ADB = ∫1_1^2▒〖𝑦 𝑑π‘₯γ€— y β†’ Equation of line x + y = 2 y = 2 βˆ’ x Therefore, Area ADB = ∫1_𝟏^πŸβ–’(πŸβˆ’π’™)𝒅𝒙 = [2π‘₯βˆ’π‘₯^2/2]_1^2 = [(2(2)βˆ’2^2/2)βˆ’(2(1)βˆ’1^2/2)] = [2π‘₯βˆ’π‘₯^2/2]_1^2 = [(2(2)βˆ’2^2/2)βˆ’(2(1)βˆ’1^2/2)] = [(4βˆ’2)βˆ’(2βˆ’1/2)] = 2βˆ’3/2 = 𝟏/𝟐 Thus, Area Required = Area OAD + Area ADB = 2/3+1/2 = πŸ•/πŸ” 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.