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Thread: Calculate the average over the given interval.

  1. #1
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    Calculate the average over the given interval.

    I do not understand the final step taken to solve this problem (It was an example given in class) I would really appreciate it if someone would be able to explain the simplification process for the final step?

    f
    (x) = xn
    for
    n ≥ 0, [0, 2]

    {\frac{1}{2-0} }\int\limits_{0}^{2}$x^{n}${dx} = {\frac{1}{2n + 2}$x^{n+1}$ \Bigr\rvert\biggr \limits_{0}^{2}

    I understand that one must solve using the given limits, which would give me a value of:

    \biggl[{\frac{1}{2n + 2}$2^{n+1}$ \biggr] - \biggl[{\frac{1}{2n + 2}$0^{n+1}$ \biggr]

    However, I am not following the what happens in order to have the final solution of:

    {$2^{n}$\frac{1}{n+1}}



    I end up with my final solution being:
    {\frac{1}{2n + 2}$2^{n+1}$

    Any sort of help is greatly appreciated! Thank you
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  2. #2
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    Re: Calculate the average over the given interval.

    Quote Originally Posted by cathrinbleu View Post
    I do not understand the final step taken to solve this problem (It was an example given in class) I would really appreciate it if someone would be able to explain the simplification process for the final step?

    f(x) = xn
    for
    n ≥ 0, [0, 2]

    {\frac{1}{2-0} }\int\limits_{0}^{2}$x^{n}${dx} = {\frac{1}{2n + 2}$x^{n+1}$ \Bigr\rvert\biggr \limits_{0}^{2}

    I understand that one must solve using the given limits, which would give me a value of:

    \biggl[{\frac{1}{2n + 2}$2^{n+1}$ \biggr] - \biggl[{\frac{1}{2n + 2}$0^{n+1}$ \biggr]

    However, I am not following the what happens in order to have the final solution of:

    {$2^{n}$\frac{1}{n+1}}


    I end up with my final solution being: {\frac{1}{2n + 2}$2^{n+1}$
    {\frac{1}{2n + 2}2^{n+1}={\frac{1}{2(n + 1)}2(2^{n})=~?
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  3. #3
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    Re: Calculate the average over the given interval.

    I suppose I am not following where you take the 2 out of $2^{n+1}$ in order to get 2($2^{n})$

    I imagine it's quite simple, yet for some reason it's lost on me....how frustrating
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  4. #4
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    Re: Calculate the average over the given interval.

    Quote Originally Posted by cathrinbleu View Post
    I suppose I am not following where you take the 2 out of $2^{n+1}$ in order to get 2($2^{n})$
    Do you add exponents?

    2(2^n)=2^1(2^n)=2^{n+1}
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  5. #5
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    Re: Calculate the average over the given interval.

    Why...yes you do! Wow...

    Thank you!
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