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.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


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

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–ˆ(๐‘ฅร—๐‘ฅ, ๐‘ฅโ‰ฅ0@๐‘ฅ ร—(โˆ’๐‘ฅ), ๐‘ฅ<0)โ”ค y = {โ–ˆ(๐‘ฅ^2, ๐‘ฅโ‰ฅ0@โˆ’๐‘ฅ^2, ๐‘ฅ<0)โ”ค And x2 + y2 = 2 (x โ€“ 0)2 + (y โ€“ 0)2 = (โˆš2)^2 So, it is a circle with center (0, 0) and Radius = โˆš2 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–ˆ(๐‘ฅร—๐‘ฅ, ๐‘ฅโ‰ฅ0@๐‘ฅ ร—(โˆ’๐‘ฅ), ๐‘ฅ<0)โ”ค y = {โ–ˆ(๐‘ฅ^2, ๐‘ฅโ‰ฅ0@โˆ’๐‘ฅ^2, ๐‘ฅ<0)โ”ค And x2 + y2 = 2 (x โ€“ 0)2 + (y โ€“ 0)2 = (โˆš2)^2 So, it is a circle with center (0, 0) and Radius = โˆš2 ๐’š=ใ€–โˆ’๐’™ใ€—^๐Ÿ for x < 0 ๐’š=ใ€–โˆ’๐’™ใ€—^๐Ÿ 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‘ฆโˆ’2=0 ๐‘ฆ^2+2๐‘ฆโˆ’๐‘ฆโˆ’2=0 ๐‘ฆ(๐‘ฆ+2)โˆ’1(๐‘ฆ+2)=0 (yโˆ’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 โˆด y = 1 Now, y = x2 1 = x2 x2 = 1 x = ยฑ โˆš1 x = ยฑ 1 โˆด x = 1, โ€“1 Since x is in 1st quadrant x = 1 โˆด x = 1, y = 1 So, point B = (1, 1) Area Required = Area AODB โ€“ Area ODB Area AODB Area AODB = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— For circle x2 + y2 = 2 y2 = 2 โ€“ x2 y = ยฑ โˆš(2โˆ’๐‘ฅ^2 ) Since AODB is in 1st quadrant y =โˆš(2โˆ’๐‘ฅ^2 ) โˆด Area AODB = โˆซ_0^1โ–’โˆš(2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ = โˆซ_0^1โ–’โˆš((โˆš2)^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ It is of form โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 ) ๐‘‘๐‘ฅ=๐‘ฅ/2 โˆš(๐‘Ž^2โˆ’๐‘ฅ^2 )+๐‘Ž^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ๐‘ฅ/๐‘Ž+๐‘ใ€— Here, a = โˆš2 = [๐‘ฅ/2 โˆš((โˆš2)^2โˆ’๐‘ฅ^2 )+(โˆš2)^2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ๐‘ฅ/โˆš2ใ€— " " ]_0^1 = [๐‘ฅ/2 โˆš(2โˆ’๐‘ฅ^2 )+2/2 ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ๐‘ฅ/โˆš2ใ€— " " ]_0^1 = [๐‘ฅ/2 โˆš(2โˆ’๐‘ฅ^2 )+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– ๐‘ฅ/โˆš2ใ€— " " ]_0^1 = [1/2 โˆš(2โˆ’1^2 )+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1/โˆš2ใ€— " " ] โ€“ [0/2 โˆš(2โˆ’0^2 )+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 0/โˆš2ใ€— " " ] = [1/2 โˆš(2โˆ’1)+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1/โˆš2ใ€— " " ] โ€“ [0+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โก0 " " ] = [1/2+ใ€–๐‘ ๐‘–๐‘›ใ€—^(โˆ’1)โกใ€– 1/โˆš2ใ€— " " ] โ€“ [0+0" " ] = 1/2+๐œ‹/4 Area ODB Area ODB = โˆซ_0^1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฅใ€— For parabola y = x2 โˆด Area ODB = โˆซ_0^1โ–’๐‘ฅ^2 ๐‘‘๐‘ฅ = [๐‘ฅ^3/3]_0^1 =1^3/3โˆ’0^3/3 = 1/3 Thus, Area Required = Area AODB โ€“ Area ODB = 1/2+๐œ‹/4 โ€“ 1/3 = 1/2โˆ’1/3+๐œ‹/4 = ๐Ÿ/๐Ÿ”+๐…/๐Ÿ’ square units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.