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

Misc 33 - Chapter 7 Class 12 Integrals - Part 2
Misc 33 - Chapter 7 Class 12 Integrals - Part 3
Misc 33 - Chapter 7 Class 12 Integrals - Part 4
Misc 33 - Chapter 7 Class 12 Integrals - Part 5
Misc 33 - Chapter 7 Class 12 Integrals - Part 6
Misc 33 - Chapter 7 Class 12 Integrals - Part 7

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Transcript

Misc 33 Evaluate the definite integral ∫_1^4β–’[|π‘₯βˆ’1|+|π‘₯βˆ’2|+|π‘₯βˆ’3|] 𝑑π‘₯ I=∫_1^4β–’[|π‘₯βˆ’1|+|π‘₯βˆ’2|+|π‘₯βˆ’3|] 𝑑π‘₯ I=∫_1^4β–’|π‘₯βˆ’1| 𝑑π‘₯+∫_1^4β–’|π‘₯βˆ’2| 𝑑π‘₯+∫_1^4β–’|π‘₯βˆ’3| 𝑑π‘₯ Solving 𝐈𝟏 I1=∫_1^4β–’|π‘₯βˆ’1| 𝑑π‘₯ We kow that |π‘₯βˆ’1|= {β–ˆ( (π‘₯βˆ’1) π‘“π‘œπ‘Ÿ π‘₯β‰₯1@βˆ’(π‘₯βˆ’1) π‘“π‘œπ‘Ÿ π‘₯<1)─ Therefore, I1=∫_1^4β–’|π‘₯βˆ’1| 𝑑π‘₯ I1=∫_1^4β–’(π‘₯βˆ’1) 𝑑π‘₯ I1=∫_1^4β–’π‘₯ 𝑑π‘₯βˆ’βˆ«_1^4β–’1 𝑑π‘₯ I1=[π‘₯^2/2]_1^4βˆ’[π‘₯]_1^4 I1=((4)^2 βˆ’ (1)^2)/2 βˆ’ [4βˆ’1] I1=(16 βˆ’ 1)/2 βˆ’ [3] I1=15/2 βˆ’3 I1=(15 βˆ’ 6)/2 I1=9/2 Solving 𝐈𝟐 I2=∫_1^4β–’|π‘₯βˆ’2| 𝑑π‘₯ We know that |π‘₯βˆ’2|= {β–ˆ( (π‘₯βˆ’2) π‘“π‘œπ‘Ÿ π‘₯β‰₯2@βˆ’(π‘₯βˆ’2) π‘“π‘œπ‘Ÿ π‘₯<2)─ Therefore I2=∫_1^4β–’|π‘₯βˆ’2| 𝑑π‘₯ I2=∫_1^2β–’γ€–βˆ’(π‘₯βˆ’2) γ€— 𝑑π‘₯+∫_2^4β–’(π‘₯βˆ’2) 𝑑π‘₯ I2=∫_1^2β–’(βˆ’π‘₯+2) 𝑑π‘₯+∫_2^4β–’(π‘₯βˆ’2) 𝑑π‘₯ I2=∫_1^2β–’γ€–βˆ’π‘₯γ€— 𝑑π‘₯+∫_1^2β–’2 𝑑π‘₯+∫_2^4β–’π‘₯ 𝑑π‘₯βˆ’βˆ«_2^4β–’2 𝑑π‘₯ I2=βˆ’[π‘₯^2/2]_1^2+2[π‘₯]_1^2+[π‘₯^2/2]_2^4βˆ’2[π‘₯]_2^4 I2=βˆ’[(4 βˆ’ 1)/2]+2[2βˆ’1]+[(16 βˆ’ 4)/2]βˆ’2[4βˆ’2] I2=βˆ’[3/2]+2[1]+12/2βˆ’2[2] I2= (βˆ’ 3)/2 + 2+6βˆ’4 I2= (βˆ’3)/2 +8βˆ’4 I2= (βˆ’3)/2 +4 I2= (βˆ’ 3 + 8)/2 I2= 5/2 Solving πˆπŸ‘ I3=∫_1^4β–’|π‘₯βˆ’3| 𝑑π‘₯ We know |π‘₯βˆ’3|= {β–ˆ( (π‘₯βˆ’3) π‘“π‘œπ‘Ÿ π‘₯β‰₯3@βˆ’(π‘₯βˆ’3) π‘“π‘œπ‘Ÿ π‘₯<3)─ Therefore, I3=∫_1^4β–’|π‘₯βˆ’3| 𝑑π‘₯ I3=∫_1^3β–’γ€–βˆ’(π‘₯βˆ’3) γ€— 𝑑π‘₯+∫_3^4β–’(π‘₯βˆ’3) 𝑑π‘₯ I3=∫_1^3β–’(βˆ’π‘₯+3) 𝑑π‘₯+∫_3^4β–’(π‘₯βˆ’3) 𝑑π‘₯ I3=∫_1^3β–’γ€–βˆ’π‘₯γ€— 𝑑π‘₯+∫_1^3β–’3 𝑑π‘₯+∫_3^4β–’π‘₯ 𝑑π‘₯βˆ’βˆ«_3^4β–’3 𝑑π‘₯ I3=βˆ’[π‘₯^2/2]_1^3+3[π‘₯]_1^3+[π‘₯^2/2]_3^4βˆ’3[π‘₯]_3^4 I3=βˆ’[(9 βˆ’ 1)/2]+3[3 βˆ’1]+[(16 βˆ’ 9)/2]βˆ’3[4βˆ’3] I3=(βˆ’ 8)/2 +3[2]+ 7/2 βˆ’ 3[1] I3=βˆ’4 +6+ 7/2 βˆ’ 3 I3=βˆ’7 +6+ 7/2 I3=βˆ’1+ 7/2 I3= (βˆ’2 + 7)/2 I3= 5/2 Putting the values of I1 , I2 , I3 in (1) I=9/2 + 5/2 + 5/2 I = πŸπŸ—/𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.