Misc 31 - Chapter 7 Class 12 Integrals

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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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