Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide42.JPG

Slide43.JPG
Slide44.JPG Slide45.JPG Slide46.JPG Slide47.JPG Slide48.JPG

  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 33 Evaluate the definite integral โˆซ_1^4โ–’[|๐‘ฅโˆ’1|+|๐‘ฅโˆ’2|+|๐‘ฅโˆ’3|] ๐‘‘๐‘ฅ I=โˆซ_1^4โ–’[|๐‘ฅโˆ’1|+|๐‘ฅโˆ’2|+|๐‘ฅโˆ’3|] ๐‘‘๐‘ฅ I=โˆซ_1^4โ–’|๐‘ฅโˆ’1| ๐‘‘๐‘ฅ+โˆซ_1^4โ–’|๐‘ฅโˆ’2| ๐‘‘๐‘ฅ+โˆซ_1^4โ–’|๐‘ฅโˆ’3| ๐‘‘๐‘ฅ Solving ๐ˆ๐Ÿ I1=โˆซ_1^4โ–’|๐‘ฅโˆ’1| ๐‘‘๐‘ฅ We kow that |๐‘ฅโˆ’1|= {โ–ˆ( (๐‘ฅโˆ’1) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅโ‰ฅ1@โˆ’(๐‘ฅโˆ’1) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<1)โ”ค Therefore, I1=โˆซ_1^4โ–’|๐‘ฅโˆ’1| ๐‘‘๐‘ฅ I1=โˆซ_1^4โ–’(๐‘ฅโˆ’1) ๐‘‘๐‘ฅ I1=โˆซ_1^4โ–’๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ_1^4โ–’1 ๐‘‘๐‘ฅ I1=[๐‘ฅ^2/2]_1^4โˆ’[๐‘ฅ]_1^4 I1=((4)^2 โˆ’ (1)^2)/2 โˆ’ [4โˆ’1] I1=(16 โˆ’ 1)/2 โˆ’ [3] I1=15/2 โˆ’3 I1=(15 โˆ’ 6)/2 I1=9/2 Solving ๐ˆ๐Ÿ I2=โˆซ_1^4โ–’|๐‘ฅโˆ’2| ๐‘‘๐‘ฅ We know that |๐‘ฅโˆ’2|= {โ–ˆ( (๐‘ฅโˆ’2) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅโ‰ฅ2@โˆ’(๐‘ฅโˆ’2) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<2)โ”ค Therefore I2=โˆซ_1^4โ–’|๐‘ฅโˆ’2| ๐‘‘๐‘ฅ I2=โˆซ_1^2โ–’ใ€–โˆ’(๐‘ฅโˆ’2) ใ€— ๐‘‘๐‘ฅ+โˆซ_2^4โ–’(๐‘ฅโˆ’2) ๐‘‘๐‘ฅ I2=โˆซ_1^2โ–’(โˆ’๐‘ฅ+2) ๐‘‘๐‘ฅ+โˆซ_2^4โ–’(๐‘ฅโˆ’2) ๐‘‘๐‘ฅ I2=โˆซ_1^2โ–’ใ€–โˆ’๐‘ฅใ€— ๐‘‘๐‘ฅ+โˆซ_1^2โ–’2 ๐‘‘๐‘ฅ+โˆซ_2^4โ–’๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ_2^4โ–’2 ๐‘‘๐‘ฅ I2=โˆ’[๐‘ฅ^2/2]_1^2+2[๐‘ฅ]_1^2+[๐‘ฅ^2/2]_2^4โˆ’2[๐‘ฅ]_2^4 I2=โˆ’[(4 โˆ’ 1)/2]+2[2โˆ’1]+[(16 โˆ’ 4)/2]โˆ’2[4โˆ’2] I2=โˆ’[3/2]+2[1]+12/2โˆ’2[2] I2= (โˆ’ 3)/2 + 2+6โˆ’4 I2= (โˆ’3)/2 +8โˆ’4 I2= (โˆ’3)/2 +4 I2= (โˆ’ 3 + 8)/2 I2= 5/2 Solving ๐ˆ๐Ÿ‘ I3=โˆซ_1^4โ–’|๐‘ฅโˆ’3| ๐‘‘๐‘ฅ We know |๐‘ฅโˆ’3|= {โ–ˆ( (๐‘ฅโˆ’3) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅโ‰ฅ3@โˆ’(๐‘ฅโˆ’3) ๐‘“๐‘œ๐‘Ÿ ๐‘ฅ<3)โ”ค Therefore, I3=โˆซ_1^4โ–’|๐‘ฅโˆ’3| ๐‘‘๐‘ฅ I3=โˆซ_1^3โ–’ใ€–โˆ’(๐‘ฅโˆ’3) ใ€— ๐‘‘๐‘ฅ+โˆซ_3^4โ–’(๐‘ฅโˆ’3) ๐‘‘๐‘ฅ I3=โˆซ_1^3โ–’(โˆ’๐‘ฅ+3) ๐‘‘๐‘ฅ+โˆซ_3^4โ–’(๐‘ฅโˆ’3) ๐‘‘๐‘ฅ I3=โˆซ_1^3โ–’ใ€–โˆ’๐‘ฅใ€— ๐‘‘๐‘ฅ+โˆซ_1^3โ–’3 ๐‘‘๐‘ฅ+โˆซ_3^4โ–’๐‘ฅ ๐‘‘๐‘ฅโˆ’โˆซ_3^4โ–’3 ๐‘‘๐‘ฅ I3=โˆ’[๐‘ฅ^2/2]_1^3+3[๐‘ฅ]_1^3+[๐‘ฅ^2/2]_3^4โˆ’3[๐‘ฅ]_3^4 I3=โˆ’[(9 โˆ’ 1)/2]+3[3 โˆ’1]+[(16 โˆ’ 9)/2]โˆ’3[4โˆ’3] I3=(โˆ’ 8)/2 +3[2]+ 7/2 โˆ’ 3[1] I3=โˆ’4 +6+ 7/2 โˆ’ 3 I3=โˆ’7 +6+ 7/2 I3=โˆ’1+ 7/2 I3= (โˆ’2 + 7)/2 I3= 5/2 Putting the values of I1 , I2 , I3 in (1) I=9/2 + 5/2 + 5/2 I = ๐Ÿ๐Ÿ—/๐Ÿ

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.