integrate from 0 to 3 |x^3-3x^2+2x|dx

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integrate from 0 to 3 |x^3-3x^2+2x|dx

Mathematics
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Is that absolute value?
yea
\(x^3-3x^2+2x=x(x^2-3x+2)=x(x-1)(x-2)\) 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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Other answers:

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

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