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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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Find area of x|x|,  circle x^2 + y^2 = 2 and y-axis using integration

Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8 Question 26 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9

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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 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.