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0-3-x-2-2x-dx-is-there-any-one-to-explain-the-way-to-solve-this-

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Question Number 128834 by bounhome last updated on 10/Jan/21
∫_0 ^3 ∣x^2 −2x∣dx=...? is there any one to explain the way to solve this
∫03∣x2−2x∣dx=…?isthereanyonetoexplainthewaytosolvethis
Answered by bobhans last updated on 10/Jan/21
∫_0 ^( 3) ∣x^2 −2x∣ dx = ∫_0 ^( 2) (2x−x^2 )dx+∫_2 ^( 3) (x^2 −2x) dx
∫03∣x2−2x∣dx=∫02(2x−x2)dx+∫23(x2−2x)dx
Answered by mathmax by abdo last updated on 10/Jan/21
∫_0 ^3 ∣x^2 −2x∣dx =∫_0 ^3 x∣x−2∣dx =∫_0 ^2 x(2−x)dx +∫_2 ^3 x(x−2)dx =∫_0 ^2 (2x−x^2 )dx+∫_2 ^3 (x^2 −2x)dx =[x^2 −(x^3 /3)]_0 ^2 +[(x^3 /3)−x^2 ]_2 ^3 =2^2 −(2^3 /3) +((3^3 /3)−3^2 −(2^3 /3) +2^2 )=4−(8/3) −(8/3) +4 =8−((16)/3)=((24−16)/3) =(8/3)
∫03∣x2−2x∣dx=∫03x∣x−2∣dx=∫02x(2−x)dx+∫23x(x−2)dx=∫02(2x−x2)dx+∫23(x2−2x)dx=[x2−x33]02+[x33−x2]23=22−233+(333−32−233+22)=4−83−83+4=8−163=24−163=83

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