Check sibling questions

      Slide37.JPG

Slide38.JPG
Slide39.JPG Slide40.JPG Slide41.JPG Slide42.JPG Slide43.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 31 Evaluate the definite integral ∫_1^4▒[|𝑥−1|+|𝑥−2|+|𝑥−3|] 𝑑𝑥 I=∫_1^4▒[|𝑥−1|+|𝑥−2|+|𝑥−3|] 𝑑𝑥 I=∫_1^4▒|𝑥−1| 𝑑𝑥+∫_1^4▒|𝑥−2| 𝑑𝑥+∫_1^4▒|𝑥−3| 𝑑𝑥 Solving 𝐈𝟏 I1=∫_1^4▒|𝑥−1| 𝑑𝑥 We kow that |𝑥−1|= {█( (𝑥−1) 𝑓𝑜𝑟 𝑥≥1@−(𝑥−1) 𝑓𝑜𝑟 𝑥<1)┤ Therefore, I1=∫_1^4▒|𝑥−1| 𝑑𝑥 I1=∫_1^4▒(𝑥−1) 𝑑𝑥 I1=∫_1^4▒𝑥 𝑑𝑥−∫_1^4▒1 𝑑𝑥 I1=[𝑥^2/2]_1^4−[𝑥]_1^4 I1=((4)^2 − (1)^2)/2 − [4−1] I1=(16 − 1)/2 − [3] I1=15/2 −3 I1=(15 − 6)/2 I1=9/2 Solving 𝐈𝟐 I2=∫_1^4▒|𝑥−2| 𝑑𝑥 We know that |𝑥−2|= {█( (𝑥−2) 𝑓𝑜𝑟 𝑥≥2@−(𝑥−2) 𝑓𝑜𝑟 𝑥<2)┤ Therefore I2=∫_1^4▒|𝑥−2| 𝑑𝑥 I2=∫_1^2▒〖−(𝑥−2) 〗 𝑑𝑥+∫_2^4▒(𝑥−2) 𝑑𝑥 I2=∫_1^2▒(−𝑥+2) 𝑑𝑥+∫_2^4▒(𝑥−2) 𝑑𝑥 I2=∫_1^2▒〖−𝑥〗 𝑑𝑥+∫_1^2▒2 𝑑𝑥+∫_2^4▒𝑥 𝑑𝑥−∫_2^4▒2 𝑑𝑥 I2=−[𝑥^2/2]_1^2+2[𝑥]_1^2+[𝑥^2/2]_2^4−2[𝑥]_2^4 I2=−[(4 − 1)/2]+2[2−1]+[(16 − 4)/2]−2[4−2] I2=−[3/2]+2[1]+12/2−2[2] I2= (− 3)/2 + 2+6−4 I2= (−3)/2 +8−4 I2= (−3)/2 +4 I2= (− 3 + 8)/2 I2= 5/2 Solving 𝐈𝟑 I3=∫_1^4▒|𝑥−3| 𝑑𝑥 We know |𝑥−3|= {█( (𝑥−3) 𝑓𝑜𝑟 𝑥≥3@−(𝑥−3) 𝑓𝑜𝑟 𝑥<3)┤ Therefore, I3=∫_1^4▒|𝑥−3| 𝑑𝑥 I3=∫_1^3▒〖−(𝑥−3) 〗 𝑑𝑥+∫_3^4▒(𝑥−3) 𝑑𝑥 I3=∫_1^3▒(−𝑥+3) 𝑑𝑥+∫_3^4▒(𝑥−3) 𝑑𝑥 I3=∫_1^3▒〖−𝑥〗 𝑑𝑥+∫_1^3▒3 𝑑𝑥+∫_3^4▒𝑥 𝑑𝑥−∫_3^4▒3 𝑑𝑥 I3=−[𝑥^2/2]_1^3+3[𝑥]_1^3+[𝑥^2/2]_3^4−3[𝑥]_3^4 I3=−[(9 − 1)/2]+3[3 −1]+[(16 − 9)/2]−3[4−3] I3=(− 8)/2 +3[2]+ 7/2 − 3[1] I3=−4 +6+ 7/2 − 3 I3=−7 +6+ 7/2 I3=−1+ 7/2 I3= (−2 + 7)/2 I3= 5/2 Putting the values of I1 , I2 , I3 in (1) I=9/2 + 5/2 + 5/2 I = 𝟏𝟗/𝟐

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.