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
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Question 1 (Choice 2)
Question 2 Important
Question 3
Question 4 Important
Question 5
Question 6 Important
Question 7 Important
Question 8 (Choice 1)
Question 8 (Choice 2)
Question 9 Important
Question 10 (Choice 1)
Question 10 (Choice 2)
Question 11 Important
Question 12 (Choice 1) You are here
Question 12 (Choice 2) Important
Question 13 Important
Question 14 - Case Based Important
CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)
Last updated at Jan. 25, 2022 by Teachoo
This question is similar to Misc 7 Chapter 8 Class 12 - Applications of Integrals
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. Letβs first draw the Figure Here, ππ =π 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 β x Putting y = 2 β x in equation of parabola π¦^2=π₯ (πβπ)^π=π 4+π₯^2β4π₯=π₯ π₯^2β4π₯βπ₯+4=0 π^πβππ+π=π π₯^2β4π₯βπ₯+4=0 π₯(π₯β4)β1(π₯β4)=0 (π₯β4)(π₯β1)=0 So, x = 4, x = 1 Since for point A, x-coordinate will be less than 2 β΄ x = 1 Putting x = 1 in equation of line x + y = 2 1 + y = 2 y = 2 β 1 y = 1 So, Coordinates of point A = (1, 1) Finding Area Area Required = Area OAD + Area ADB Area OAD Area OAD = β«_0^1βγπ¦ ππ₯" " γ y β Equation of parabola π¦^2 = x π¦ = Β± βπ₯ Since OAD is in 1st quadrant, value of y is positive β΄ π = βπ Now, Area OAD = β«_0^1βγπ¦ ππ₯" " γ = β«_π^πβγβπ π π" " γ = [π₯^(1/2 + 1)/(1/2 + 1)]_0^1 = [π₯^(3/2)/(3/2)]_0^1 = π/π [π^(π/π) ]_π^π = 2/3 [1^(3/2)β0^(3/2) ] = π/π Area ADB Area ADB = β«1_1^2βγπ¦ ππ₯γ y β Equation of line x + y = 2 y = 2 β x Therefore, Area ADB = β«1_π^πβ(πβπ)π π = [2π₯βπ₯^2/2]_1^2 = [(2(2)β2^2/2)β(2(1)β1^2/2)] = [2π₯βπ₯^2/2]_1^2 = [(2(2)β2^2/2)β(2(1)β1^2/2)] = [(4β2)β(2β1/2)] = 2β3/2 = π/π Thus, Area Required = Area OAD + Area ADB = 2/3+1/2 = π/π square units