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anonymous

  • 5 years ago

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

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  1. watchmath
    • 5 years ago
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    Is that absolute value?

  2. anonymous
    • 5 years ago
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    yea

  3. watchmath
    • 5 years ago
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    \(x^3-3x^2+2x=x(x^2-3x+2)=x(x-1)(x-2)\) So the integral is equal to \[\int_0^1 x^3-3x^2+2x\,dx+\int_1^2-x^3+3x^2-2x\,dx+\int_2^3x^3-3x^2+2x\,dx\]

  4. anonymous
    • 5 years ago
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    why did you inverse the sign in the 2nd summation term??

  5. watchmath
    • 5 years ago
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    Because over \([1,2]\) the function \(f(x)=x(x-1)(x-2)\) is negative. So \(|f(x)|=-f(x)\) on \([1,2]\).

  6. anonymous
    • 5 years ago
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    i see... do we have to include 1 and 2?? because you used square brackets to show them

  7. watchmath
    • 5 years ago
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    Well, in the integration it won't make any difference whether we have open interval or closed interval.

  8. anonymous
    • 5 years ago
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    yea thats true.. thanx again

  9. anonymous
    • 5 years ago
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    please check my other question

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