Misc 33 - Chapter 7 Class 12 Integrals (Term 2)

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Misc 33 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≥[email protected]−(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≥[email protected]−(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≥[email protected]−(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