Evaluate ∫ |x 2 - 2x|  dx from 1 to 3

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Note : - This is similar to Example 30 of NCERT – Chapter 7 Class 12

Check the answer here https://www.teachoo.com/4811/727/Example-30---Evaluate-integral--1----2--x3---x--dx/category/Examples/

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

Transcript

Question 30 Evaluate โˆซ 3 1 |๐‘ฅ^2โˆ’2๐‘ฅ| dx |๐‘ฅ^2โˆ’2๐‘ฅ|=|๐‘ฅ(๐‘ฅโˆ’2)| =|๐‘ฅ| |๐‘ฅโˆ’2| Thus, ๐‘ฅ=0, ๐‘ฅ=2 Since our integration is from 1 to 3, we ignore x = 0 โˆด |๐‘ฅ^2โˆ’2๐‘ฅ|= {(๐‘ฅร—โˆ’(๐‘ฅโˆ’2) ๐‘–๐‘“ 1โ‰ค๐‘ฅ<2๐‘ฅร—(๐‘ฅโˆ’2) ๐‘–๐‘“ 2โ‰ค๐‘ฅ<3)โ”ค |๐‘ฅ^2โˆ’2๐‘ฅ|= {(โˆ’(๐‘ฅ^2โˆ’2๐‘ฅ) ๐‘–๐‘“ 1โ‰ค๐‘ฅ<2(๐‘ฅ^2โˆ’2๐‘ฅ) ๐‘–๐‘“ 2โ‰ค๐‘ฅ<3)โ”ค Now, โˆซ_1^3 |๐‘ฅ^2โˆ’2๐‘ฅ| dx = โˆ’โˆซ_1^2โ–’(๐‘ฅ^2โˆ’2๐‘ฅ) ๐‘‘๐‘ฅ+โˆซ_2^3โ–’(๐‘ฅ^2โˆ’2๐‘ฅ) ๐‘‘๐‘ฅ = โˆ’โˆซ_1^2โ–’๐‘ฅ^2 ๐‘‘๐‘ฅ+โˆซ_1^2โ–’2๐‘ฅ ๐‘‘๐‘ฅ+โˆซ_2^3โ–’๐‘ฅ^2 ๐‘‘๐‘ฅโˆ’โˆซ_2^3โ–’2๐‘ฅ ๐‘‘๐‘ฅ = โˆ’โˆซ_1^2โ–’๐‘ฅ^2 ๐‘‘๐‘ฅ+โˆซ_2^3โ–’๐‘ฅ^2 ๐‘‘๐‘ฅ+โˆซ_1^2โ–’2๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ_2^3โ–’2๐‘ฅ ๐‘‘๐‘ฅ = โˆ’[๐‘ฅ^3/3]_1^2+[๐‘ฅ^3/3]_2^3+[๐‘ฅ^2 ]_1^2โˆ’[๐‘ฅ^2 ]_2^3 = โˆ’[2^3/3โˆ’1^3/3]+[3^3/3โˆ’2^3/3]+[2^2โˆ’1^2 ]โˆ’[3^2โˆ’2^2 ] = โˆ’[8/3โˆ’1/3]+[27/3โˆ’8/3]+[4โˆ’1]โˆ’[9โˆ’4] = โˆ’[7/3]+[19/3]+[3]โˆ’[5] = 12/3โˆ’2 = 4โˆ’2 = 2

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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.