Chapter 7 Class 12 Integrals
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Misc 31 - Chapter 7 Class 12 Integrals

Last updated at December 16, 2024 by

Ā  Ā  Ā  Definite integral |x - 1| + |x - 2| + |x - 3| dx from 1 to 4 - Teachoo - Miscellaneous

part 2 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 3 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 4 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals
part 5 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals part 6 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals part 7 - Misc 31 - Miscellaneous - Serial order wise - Chapter 7 Class 12 Integrals

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Misc 31 Evaluate the definite integral āˆ«_1^4ā–’[|š‘„āˆ’1|+|š‘„āˆ’2|+|š‘„āˆ’3|] š‘‘š‘„ I=āˆ«_1^4ā–’[|š‘„āˆ’1|+|š‘„āˆ’2|+|š‘„āˆ’3|] š‘‘š‘„ I=āˆ«_1^4ā–’|š‘„āˆ’1| š‘‘š‘„+āˆ«_1^4ā–’|š‘„āˆ’2| š‘‘š‘„+āˆ«_1^4ā–’|š‘„āˆ’3| š‘‘š‘„ Solving šˆšŸ I1=āˆ«_1^4ā–’|š‘„āˆ’1| š‘‘š‘„ We kow that |š‘„āˆ’1|= {ā–ˆ( (š‘„āˆ’1) š‘“š‘œš‘Ÿ š‘„ā‰„1@āˆ’(š‘„āˆ’1) š‘“š‘œš‘Ÿ š‘„<1)ā”¤ Therefore, I1=āˆ«_1^4ā–’|š‘„āˆ’1| š‘‘š‘„ I1=āˆ«_1^4ā–’(š‘„āˆ’1) š‘‘š‘„ I1=āˆ«_1^4ā–’š‘„ š‘‘š‘„āˆ’āˆ«_1^4ā–’1 š‘‘š‘„ I1=[š‘„^2/2]_1^4āˆ’[š‘„]_1^4 I1=((4)^2 āˆ’ (1)^2)/2 āˆ’ [4āˆ’1] I1=(16 āˆ’ 1)/2 āˆ’ [3] I1=15/2 āˆ’3 I1=(15 āˆ’ 6)/2 I1=9/2 Solving šˆšŸ I2=āˆ«_1^4ā–’|š‘„āˆ’2| š‘‘š‘„ We know that |š‘„āˆ’2|= {ā–ˆ( (š‘„āˆ’2) š‘“š‘œš‘Ÿ š‘„ā‰„2@āˆ’(š‘„āˆ’2) š‘“š‘œš‘Ÿ š‘„<2)ā”¤ Therefore I2=āˆ«_1^4ā–’|š‘„āˆ’2| š‘‘š‘„ I2=āˆ«_1^2ā–’怖āˆ’(š‘„āˆ’2) 怗 š‘‘š‘„+āˆ«_2^4ā–’(š‘„āˆ’2) š‘‘š‘„ I2=āˆ«_1^2ā–’(āˆ’š‘„+2) š‘‘š‘„+āˆ«_2^4ā–’(š‘„āˆ’2) š‘‘š‘„ I2=āˆ«_1^2ā–’怖āˆ’š‘„怗 š‘‘š‘„+āˆ«_1^2ā–’2 š‘‘š‘„+āˆ«_2^4ā–’š‘„ š‘‘š‘„āˆ’āˆ«_2^4ā–’2 š‘‘š‘„ I2=āˆ’[š‘„^2/2]_1^2+2[š‘„]_1^2+[š‘„^2/2]_2^4āˆ’2[š‘„]_2^4 I2=āˆ’[(4 āˆ’ 1)/2]+2[2āˆ’1]+[(16 āˆ’ 4)/2]āˆ’2[4āˆ’2] I2=āˆ’[3/2]+2[1]+12/2āˆ’2[2] I2= (āˆ’ 3)/2 + 2+6āˆ’4 I2= (āˆ’3)/2 +8āˆ’4 I2= (āˆ’3)/2 +4 I2= (āˆ’ 3 + 8)/2 I2= 5/2 Solving šˆšŸ‘ I3=āˆ«_1^4ā–’|š‘„āˆ’3| š‘‘š‘„ We know |š‘„āˆ’3|= {ā–ˆ( (š‘„āˆ’3) š‘“š‘œš‘Ÿ š‘„ā‰„3@āˆ’(š‘„āˆ’3) š‘“š‘œš‘Ÿ š‘„<3)ā”¤ Therefore, I3=āˆ«_1^4ā–’|š‘„āˆ’3| š‘‘š‘„ I3=āˆ«_1^3ā–’怖āˆ’(š‘„āˆ’3) 怗 š‘‘š‘„+āˆ«_3^4ā–’(š‘„āˆ’3) š‘‘š‘„ I3=āˆ«_1^3ā–’(āˆ’š‘„+3) š‘‘š‘„+āˆ«_3^4ā–’(š‘„āˆ’3) š‘‘š‘„ I3=āˆ«_1^3ā–’怖āˆ’š‘„怗 š‘‘š‘„+āˆ«_1^3ā–’3 š‘‘š‘„+āˆ«_3^4ā–’š‘„ š‘‘š‘„āˆ’āˆ«_3^4ā–’3 š‘‘š‘„ I3=āˆ’[š‘„^2/2]_1^3+3[š‘„]_1^3+[š‘„^2/2]_3^4āˆ’3[š‘„]_3^4 I3=āˆ’[(9 āˆ’ 1)/2]+3[3 āˆ’1]+[(16 āˆ’ 9)/2]āˆ’3[4āˆ’3] I3=(āˆ’ 8)/2 +3[2]+ 7/2 āˆ’ 3[1] I3=āˆ’4 +6+ 7/2 āˆ’ 3 I3=āˆ’7 +6+ 7/2 I3=āˆ’1+ 7/2 I3= (āˆ’2 + 7)/2 I3= 5/2 Putting the values of I1 , I2 , I3 in (1) I=9/2 + 5/2 + 5/2 I = šŸšŸ—/šŸ

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo