Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 by Teachoo

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Using integration, find the area in the first quadrant bounded by the curve y = x|x|, the circle x
^{
2
}
+ y
^{
2
}
= 2 and the y-axis

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Question 26
Using integration, find the area in the first quadrant bounded by the curve y = π₯|π₯|, the circle x2 + y2 = 2 and the y-axis
Given
y = π₯|π₯|
y = {β(π₯Γπ₯, π₯β₯0@π₯ Γ(βπ₯), π₯<0)β€
y = {β(π₯^2, π₯β₯0@βπ₯^2, π₯<0)β€
And
x2 + y2 = 2
(x β 0)2 + (y β 0)2 = (β2)^2
So, it is a circle with center (0, 0)
and Radius = β2
Question 26
Using integration, find the area in the first quadrant bounded by the curve y = π₯|π₯|, the circle x2 + y2 = 2 and the y-axis
Given
y = π₯|π₯|
y = {β(π₯Γπ₯, π₯β₯0@π₯ Γ(βπ₯), π₯<0)β€
y = {β(π₯^2, π₯β₯0@βπ₯^2, π₯<0)β€
And
x2 + y2 = 2
(x β 0)2 + (y β 0)2 = (β2)^2
So, it is a circle with center (0, 0)
and Radius = β2
π=γβπγ^π for x < 0
π=γβπγ^π for x < 0
Combining, we form graph
We need to find Area OAB
First we find point B
Point B is point of intersection of circle and parabola
Now,
equation of circle is
π₯^2+π¦^2=2
Putting x2 = y
π¦+π¦^2=2
π¦^2+π¦β2=0
π¦^2+2π¦βπ¦β2=0
π¦(π¦+2)β1(π¦+2)=0
(yβ1)(π¦+2)=0
So, y = 1, y = β2
So, y = 1, y = β2
Since y is in 1st quadrant, it is positive
β΄ y = 1
Now,
y = x2
1 = x2
x2 = 1
x = Β± β1
x = Β± 1
β΄ x = 1, β1
Since x is in 1st quadrant
x = 1
β΄ x = 1, y = 1
So, point B = (1, 1)
Area Required = Area AODB β Area ODB
Area AODB
Area AODB = β«_0^1βγπ¦ ππ₯γ
For circle
x2 + y2 = 2
y2 = 2 β x2
y = Β± β(2βπ₯^2 )
Since AODB is in 1st quadrant
y =β(2βπ₯^2 )
β΄ Area AODB = β«_0^1ββ(2βπ₯^2 ) ππ₯
= β«_0^1ββ((β2)^2βπ₯^2 ) ππ₯ It is of form
β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ
Here, a = β2
= [π₯/2 β((β2)^2βπ₯^2 )+(β2)^2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [π₯/2 β(2βπ₯^2 )+2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [π₯/2 β(2βπ₯^2 )+γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1
= [1/2 β(2β1^2 )+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0/2 β(2β0^2 )+γπ ππγ^(β1)β‘γ 0/β2γ " " ] = [1/2 β(2β1)+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+γπ ππγ^(β1)β‘0 " " ] = [1/2+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+0" " ]
= 1/2+π/4
Area ODB
Area ODB = β«_0^1βγπ¦ ππ₯γ
For parabola
y = x2
β΄ Area ODB = β«_0^1βπ₯^2 ππ₯
= [π₯^3/3]_0^1
=1^3/3β0^3/3 = 1/3
Thus,
Area Required = Area AODB β Area ODB
= 1/2+π/4 β 1/3
= 1/2β1/3+π/4
= π/π+π /π square units

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.