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Math Help - Definite integral problem

  1. #1
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    Definite integral problem

    So I have the integral \int - \frac{3y^2}{2} over 0 to 4.

    Every computer software evaluates this as -32 which is the correct answer but I keep getting -24... what am I doing wrong? Any help is appreciated!
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  2. #2
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    Show some work so we can tell you where you went wrong and what you need to do
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  3. #3
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    Original integral \int - \frac{3y^2}{2} over 0 to 4.


    Well I did -\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24

    and 0 makes the whole second fraction 0 so we can ignore it... but apparently I'm wrong
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  4. #4
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    Quote Originally Posted by CleanSanchez View Post
    Original integral \int - \frac{3y^2}{2} over 0 to 4.


    Well I did -\frac {(3)(4^2)}{2} = -\frac {(3)(16)}{2} = -\frac {48}{2} = -24

    and 0 makes the whole second fraction 0 so we can ignore it... but apparently I'm wrong
    Did you not forget to integrate? :P
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  5. #5
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    Ah yes, I was getting ahead of myself and thats not where my mistake was. It's part of a larger double integral problem and the bounds of x are 0 - 2 not 0 -4. Reversing double integrals can get confusing!
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