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Thread: Integration

  1. #1
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    Integration

    I can't figure out why when integrating

    $\displaystyle

    -4{x}^\frac{3}{2}

    $

    you get

    $\displaystyle

    -\displaystyle{\frac{8}{3}{x}}^\frac{3}{2}

    $

    and not

    $\displaystyle
    -8{x}^\frac{3}{2}

    $

    Why does it become a fraction?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Envy View Post
    I can't figure out why when integrating

    $\displaystyle

    -4{x} \frac{3}{2}

    $

    you get

    $\displaystyle

    -\displaystyle{\frac{8}{3}{x}}
    \frac{3}{2}

    $

    and not

    $\displaystyle
    -8{x} \frac{3}{2}

    $

    Why does it become a fraction?
    The rule is:

    $\displaystyle \int k x^r\ dx= \frac{k}{r+1}x^{r+1} +C$

    In your case $\displaystyle k=-4$, $\displaystyle r=3/2$ and so:

    $\displaystyle \int -4 x^{3/2}\ dx= -\ \frac{4}{(5/2)}x^{5/2} +C=-\ \frac{8}{5}x^{5/2} + C$

    CB
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  3. #3
    MHF Contributor
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    Quote Originally Posted by Envy View Post
    I can't figure out why when integrating

    $\displaystyle

    -4{x}^\frac{3}{2}

    $

    you get

    $\displaystyle

    -\displaystyle{\frac{8}{3}{x}}^\frac{3}{2}

    $

    and not

    $\displaystyle
    -8{x}^\frac{3}{2}

    $

    Why does it become a fraction?
    You shouldn't get either of those answers you've listed...

    If $\displaystyle f(x) = ax^n$ then it's antiderivative is $\displaystyle F(x) = \int{ax^n\,dx} = \frac{ax^{n + 1}}{n + 1}+C$.

    In other words, you add 1 to your power, and then divide by the new power.


    In your case you have $\displaystyle -4x^{\frac{3}{2}}$.


    Add 1 to the power and you get $\displaystyle \frac{5}{2}$.


    Divide by your new power and you get

    $\displaystyle -4\div\frac{5}{2}$

    $\displaystyle = -4\times\frac{2}{5}$

    $\displaystyle = -\frac{8}{5}$.



    So your antiderivative is $\displaystyle -\frac{8x^{\frac{5}{2}}}{5} + C$ where C is a constant.
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post


    $\displaystyle -\ \frac{4}{(5/2)}x^{5/2} +C=-\ \frac{8}{5}x^{5/2} + C$
    Thanks, I forgot to flip the fraction and then times it

    Quote Originally Posted by Prove It View Post
    You shouldn't get either of those answers you've listed...
    Totally correct, I wrote down the wrong quesion as $\displaystyle -4{x}^\frac{3}{2}$ which was my partially intergrated answer using $\displaystyle {ax^{n + 1}}$, the actual question was $\displaystyle -4{x}^\frac{1}{2}$!
    Last edited by Envy; Jun 3rd 2009 at 03:05 AM.
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