Evaluate ∫ |x 2 - 2x|  dx from 1 to 3

Evaluate Integral |x^2 - 2x| from 1 to 3 - Teachoo - CBSE Class 12 Sam

Question 30 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2
Question 30 - CBSE Class 12 Sample Paper for 2020 Boards - Part 3

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/

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Transcript

Question 30 Evaluate ∫ 3 1 |𝑥^2−2𝑥| dx |𝑥^2−2𝑥|=|𝑥(𝑥−2)| =|𝑥| |𝑥−2| Thus, 𝑥=0, 𝑥=2 Since our integration is from 1 to 3, we ignore x = 0 ∴ |𝑥^2−2𝑥|= {(𝑥×−(𝑥−2) 𝑖𝑓 1≤𝑥<2𝑥×(𝑥−2) 𝑖𝑓 2≤𝑥<3)┤ |𝑥^2−2𝑥|= {(−(𝑥^2−2𝑥) 𝑖𝑓 1≤𝑥<2(𝑥^2−2𝑥) 𝑖𝑓 2≤𝑥<3)┤ Now, ∫_1^3 |𝑥^2−2𝑥| dx = −∫_1^2▒(𝑥^2−2𝑥) 𝑑𝑥+∫_2^3▒(𝑥^2−2𝑥) 𝑑𝑥 = −∫_1^2▒𝑥^2 𝑑𝑥+∫_1^2▒2𝑥 𝑑𝑥+∫_2^3▒𝑥^2 𝑑𝑥−∫_2^3▒2𝑥 𝑑𝑥 = −∫_1^2▒𝑥^2 𝑑𝑥+∫_2^3▒𝑥^2 𝑑𝑥+∫_1^2▒2𝑥 𝑑𝑥−∫_2^3▒2𝑥 𝑑𝑥 = −[𝑥^3/3]_1^2+[𝑥^3/3]_2^3+[𝑥^2 ]_1^2−[𝑥^2 ]_2^3 = −[2^3/3−1^3/3]+[3^3/3−2^3/3]+[2^2−1^2 ]−[3^2−2^2 ] = −[8/3−1/3]+[27/3−8/3]+[4−1]−[9−4] = −[7/3]+[19/3]+[3]−[5] = 12/3−2 = 4−2 = 2

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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.