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Question Number 75552 by aliesam last updated on 12/Dec/19

Commented by mathmax by abdo last updated on 12/Dec/19

let I =∫_(−(√3)) ^(3(√3)) ∣(x−2)^2 −4∣dx ⇒I =∫_(−(√3)) ^(3(√3)) ∣x^2 −4x∣ dx =∫_(−(√3)) ^(3(√3)) ∣x∣∣x−4∣dx =∫_(−(√3)) ^0 (−x)(4−x)dx+∫_0 ^(3(√3)) x(x−4)dx =∫_(−(√3)) ^0 (x^2 −4x)dx +∫_0 ^(3(√3)) (x^2 −4x)dx =∫_(−(√3)) ^(3(√3)) (x^2 −4x)dx =[(x^3 /3)−2x^2 ]_(−(√3)) ^(3(√3)) =(((3(√3))^3 )/3)−2(3(√3))^2 −(((−(√3))^3 )/3) +2(−(√3))^2 =27(√3)−52 +(√3)+6 =28(√3)−46

letI=∫−333∣(x−2)2−4∣dx⇒I=∫−333∣x2−4x∣dx=∫−333∣x∣∣x−4∣dx=∫−30(−x)(4−x)dx+∫033x(x−4)dx=∫−30(x2−4x)dx+∫033(x2−4x)dx=∫−333(x2−4x)dx=[x33−2x2]−333=(33)33−2(33)2−(−3)33+2(−3)2=273−52+3+6=283−46

Answered by Kunal12588 last updated on 12/Dec/19

I=∫_(−(√3)) ^( 3(√3)) ∣(x−2)^2 −4∣dx =∫_(−(√3)) ^( 0) [(x−2)^2 −4]dx−∫_0 ^( 4) [(x−2)^2 −4]dx +∫_4 ^(3(√3)) [(x−2)^2 −4]dx =[(1/3)(x−2)^3 −4(x−2)]_(−(√3)) ^0 −[(1/3)(x−2)^3 −4(x−2)]_0 ^4 +[(1/3)(x−2)^3 −4(x−2)]_4 ^(3(√3)) =[(8/3)+8+(1/3)(2+(√3))^3 −(2+(√3))]−[(8/3)−8−(8/3)+8]+[(1/3)(3(√3)−2)^3 −4(3(√3)−2)−(8/3)+8] pls calculate

I=∫−333∣(x−2)2−4∣dx=∫−30[(x−2)2−4]dx−∫04[(x−2)2−4]dx+∫433[(x−2)2−4]dx=[13(x−2)3−4(x−2)]−30−[13(x−2)3−4(x−2)]04+[13(x−2)3−4(x−2)]433=[83+8+13(2+3)3−(2+3)]−[83−8−83+8]+[13(33−2)3−4(33−2)−83+8]plscalculate

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