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Thread: For b>0, deduce that f(x)=x^3 is integrable on [0,b]

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    For b>0, deduce that f(x)=x^3 is integrable on [0,b]

    This question is a peculiar one, to say the least.

    For $\displaystyle b>0$, let $\displaystyle P$ be the partition of $\displaystyle [0,b]$ into $\displaystyle n$ equal subintervals. Calculate both $\displaystyle S_P$ and $\displaystyle s_P$ for $\displaystyle f(x)=x^3$ and prove that
    $\displaystyle J(=glb(S_P))\leq 1/4b^4(1+1/n)^2$ and that $\displaystyle I(=lub(s_P))\geq 1/4b^4(1-1/n)^2$
    for any positive $\displaystyle n$.
    Deduce that $\displaystyle f(x)$ is integrable on $\displaystyle [0,b]$ and $\displaystyle \int_0^b x^3dx=1/4b^4$.
    NOTE: $\displaystyle \sum_{k=1}^n k^3=1/4n^2(n+1)^2$

    EDIT: I found some help on the question, though his answer is a little long. I'll post a link below to his work.
    http://answers.yahoo.com/question/in...8081532AALd5im

    If he made any mistakes, be sure to point them out.
    Last edited by Runty; Nov 18th 2010 at 09:39 AM. Reason: Got the question answered; link provided
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