anonymous
  • anonymous
integrate from 0 to 3 |x^3-3x^2+2x|dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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watchmath
  • watchmath
Is that absolute value?
anonymous
  • anonymous
yea
watchmath
  • watchmath
\(x^3-3x^2+2x=x(x^2-3x+2)=x(x-1)(x-2)\) So the integral is equal to \[\int_0^1 x^3-3x^2+2x\,dx+\int_1^2-x^3+3x^2-2x\,dx+\int_2^3x^3-3x^2+2x\,dx\]

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anonymous
  • anonymous
why did you inverse the sign in the 2nd summation term??
watchmath
  • watchmath
Because over \([1,2]\) the function \(f(x)=x(x-1)(x-2)\) is negative. So \(|f(x)|=-f(x)\) on \([1,2]\).
anonymous
  • anonymous
i see... do we have to include 1 and 2?? because you used square brackets to show them
watchmath
  • watchmath
Well, in the integration it won't make any difference whether we have open interval or closed interval.
anonymous
  • anonymous
yea thats true.. thanx again
anonymous
  • anonymous
please check my other question

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