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Misc 33 - Definite integral |x - 1| + |x - 2| + |x - 3| dx - Definate Integration by properties - P2

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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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 know |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)─ ∴ 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 |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)─ ∴ 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‘₯ Solving 𝐈𝟐 I2=∫_1^4β–’|π‘₯βˆ’2| 𝑑π‘₯ We know |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)─ ∴ 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‘₯ 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)─ ∴ 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 and in i.e., eq. (1) I=9/2 + 5/2 + 5/2 I = πŸπŸ—/𝟐

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