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.

### Transcript

Question 26 Using integration, find the area in the first quadrant bounded by the curve y = ๐ฅ|๐ฅ|, the circle x2 + y2 = 2 and the y-axis Given y = ๐ฅ|๐ฅ| y = {โ(๐ฅร๐ฅ, ๐ฅโฅ0@๐ฅ ร(โ๐ฅ), ๐ฅ<0)โค y = {โ(๐ฅ^2, ๐ฅโฅ0@โ๐ฅ^2, ๐ฅ<0)โค And x2 + y2 = 2 (x โ 0)2 + (y โ 0)2 = (โ2)^2 So, it is a circle with center (0, 0) and Radius = โ2 Question 26 Using integration, find the area in the first quadrant bounded by the curve y = ๐ฅ|๐ฅ|, the circle x2 + y2 = 2 and the y-axis Given y = ๐ฅ|๐ฅ| y = {โ(๐ฅร๐ฅ, ๐ฅโฅ0@๐ฅ ร(โ๐ฅ), ๐ฅ<0)โค y = {โ(๐ฅ^2, ๐ฅโฅ0@โ๐ฅ^2, ๐ฅ<0)โค And x2 + y2 = 2 (x โ 0)2 + (y โ 0)2 = (โ2)^2 So, it is a circle with center (0, 0) and Radius = โ2 ๐=ใโ๐ใ^๐ for x < 0 ๐=ใโ๐ใ^๐ 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 ๐ฅ^2+๐ฆ^2=2 Putting x2 = y ๐ฆ+๐ฆ^2=2 ๐ฆ^2+๐ฆโ2=0 ๐ฆ^2+2๐ฆโ๐ฆโ2=0 ๐ฆ(๐ฆ+2)โ1(๐ฆ+2)=0 (yโ1)(๐ฆ+2)=0 So, y = 1, y = โ2 So, y = 1, y = โ2 Since y is in 1st quadrant, it is positive โด y = 1 Now, y = x2 1 = x2 x2 = 1 x = ยฑ โ1 x = ยฑ 1 โด x = 1, โ1 Since x is in 1st quadrant x = 1 โด x = 1, y = 1 So, point B = (1, 1) Area Required = Area AODB โ Area ODB Area AODB Area AODB = โซ_0^1โใ๐ฆ ๐๐ฅใ For circle x2 + y2 = 2 y2 = 2 โ x2 y = ยฑ โ(2โ๐ฅ^2 ) Since AODB is in 1st quadrant y =โ(2โ๐ฅ^2 ) โด Area AODB = โซ_0^1โโ(2โ๐ฅ^2 ) ๐๐ฅ = โซ_0^1โโ((โ2)^2โ๐ฅ^2 ) ๐๐ฅ It is of form โ(๐^2โ๐ฅ^2 ) ๐๐ฅ=๐ฅ/2 โ(๐^2โ๐ฅ^2 )+๐^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/๐+๐ใ Here, a = โ2 = [๐ฅ/2 โ((โ2)^2โ๐ฅ^2 )+(โ2)^2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1 = [๐ฅ/2 โ(2โ๐ฅ^2 )+2/2 ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1 = [๐ฅ/2 โ(2โ๐ฅ^2 )+ใ๐ ๐๐ใ^(โ1)โกใ ๐ฅ/โ2ใ " " ]_0^1 = [1/2 โ(2โ1^2 )+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0/2 โ(2โ0^2 )+ใ๐ ๐๐ใ^(โ1)โกใ 0/โ2ใ " " ] = [1/2 โ(2โ1)+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0+ใ๐ ๐๐ใ^(โ1)โก0 " " ] = [1/2+ใ๐ ๐๐ใ^(โ1)โกใ 1/โ2ใ " " ] โ [0+0" " ] = 1/2+๐/4 Area ODB Area ODB = โซ_0^1โใ๐ฆ ๐๐ฅใ For parabola y = x2 โด Area ODB = โซ_0^1โ๐ฅ^2 ๐๐ฅ = [๐ฅ^3/3]_0^1 =1^3/3โ0^3/3 = 1/3 Thus, Area Required = Area AODB โ Area ODB = 1/2+๐/4 โ 1/3 = 1/2โ1/3+๐/4 = ๐/๐+๐/๐ square units

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.