Question 12 (Choice 1) - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at May 29, 2023 by Teachoo

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.

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,
𝒚𝟐 =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 − x
Putting y = 2 − x in equation of parabola
𝑦^2=𝑥
(𝟐−𝒙)^𝟐=𝒙
4+𝑥^2−4𝑥=𝑥
𝑥^2−4𝑥−𝑥+4=0
𝒙^𝟐−𝟓𝒙+𝟒=𝟎
𝑥^2−4𝑥−𝑥+4=0
𝑥(𝑥−4)−1(𝑥−4)=0
(𝑥−4)(𝑥−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 − 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 𝑑𝑥" " 〗
y → Equation of parabola
𝑦^2 = x
𝑦 = ± √𝑥
Since OAD is in 1st quadrant,
value of y is positive
∴ 𝒚 = √𝒙
Now,
Area OAD = ∫_0^1▒〖𝑦 𝑑𝑥" " 〗
= ∫_𝟎^𝟏▒〖√𝒙 𝒅𝒙" " 〗
= [𝑥^(1/2 + 1)/(1/2 + 1)]_0^1
= [𝑥^(3/2)/(3/2)]_0^1
= 𝟐/𝟑 [𝒙^(𝟑/𝟐) ]_𝟎^𝟏
= 2/3 [1^(3/2)−0^(3/2) ]
= 𝟐/𝟑
Area ADB
Area ADB = ∫1_1^2▒〖𝑦 𝑑𝑥〗
y → Equation of line
x + y = 2
y = 2 − x
Therefore,
Area ADB = ∫1_𝟏^𝟐▒(𝟐−𝒙)𝒅𝒙
= [2𝑥−𝑥^2/2]_1^2
= [(2(2)−2^2/2)−(2(1)−1^2/2)]
= [2𝑥−𝑥^2/2]_1^2
= [(2(2)−2^2/2)−(2(1)−1^2/2)]
= [(4−2)−(2−1/2)]
= 2−3/2
= 𝟏/𝟐
Thus,
Area Required = Area OAD + Area ADB
= 2/3+1/2
= 𝟕/𝟔 square units

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.

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