CBSE Class 12 Sample Paper for 2018 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

### 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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### Transcript

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₯Γπ₯, π₯β₯[email protected]π₯ Γ(βπ₯), π₯<0)β€ y = {β(π₯^2, π₯β₯[email protected]βπ₯^2, π₯<0)β€ And x2 + y2 = 2 (x β 0)2 + (y β 0)2 = (β2)^2 So, it is a circle with center (0, 0) and Radius = β2 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₯Γπ₯, π₯β₯[email protected]π₯ Γ(βπ₯), π₯<0)β€ y = {β(π₯^2, π₯β₯[email protected]βπ₯^2, π₯<0)β€ And x2 + y2 = 2 (x β 0)2 + (y β 0)2 = (β2)^2 So, it is a circle with center (0, 0) and Radius = β2 π=γβπγ^π for x < 0 π=γβπγ^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¦β2=0 π¦^2+2π¦βπ¦β2=0 π¦(π¦+2)β1(π¦+2)=0 (yβ1)(π¦+2)=0 So, y = 1, y = β2 So, y = 1, y = β2 Since y is in 1st quadrant, it is positive β΄ y = 1 Now, y = x2 1 = x2 x2 = 1 x = Β± β1 x = Β± 1 β΄ x = 1, β1 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«_0^1βγπ¦ ππ₯γ For circle x2 + y2 = 2 y2 = 2 β x2 y = Β± β(2βπ₯^2 ) Since AODB is in 1st quadrant y =β(2βπ₯^2 ) β΄ Area AODB = β«_0^1ββ(2βπ₯^2 ) ππ₯ = β«_0^1ββ((β2)^2βπ₯^2 ) ππ₯ It is of form β(π^2βπ₯^2 ) ππ₯=π₯/2 β(π^2βπ₯^2 )+π^2/2 γπ ππγ^(β1)β‘γ π₯/π+πγ Here, a = β2 = [π₯/2 β((β2)^2βπ₯^2 )+(β2)^2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1 = [π₯/2 β(2βπ₯^2 )+2/2 γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1 = [π₯/2 β(2βπ₯^2 )+γπ ππγ^(β1)β‘γ π₯/β2γ " " ]_0^1 = [1/2 β(2β1^2 )+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0/2 β(2β0^2 )+γπ ππγ^(β1)β‘γ 0/β2γ " " ] = [1/2 β(2β1)+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+γπ ππγ^(β1)β‘0 " " ] = [1/2+γπ ππγ^(β1)β‘γ 1/β2γ " " ] β [0+0" " ] = 1/2+π/4 Area ODB Area ODB = β«_0^1βγπ¦ ππ₯γ For parabola y = x2 β΄ Area ODB = β«_0^1βπ₯^2 ππ₯ = [π₯^3/3]_0^1 =1^3/3β0^3/3 = 1/3 Thus, Area Required = Area AODB β Area ODB = 1/2+π/4 β 1/3 = 1/2β1/3+π/4 = π/π+π/π square units