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Math Help - Real Analysis: Riemann Integrable

  1. #1
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    Real Analysis: Riemann Integrable

    Here is the question:



    How do I do this problem?? Thank you!!
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  2. #2
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    Quote Originally Posted by Phyxius117 View Post
    Here is the question:



    How do I do this problem?? Thank you!!

    Is k also a natural number?

    Is n a fixed natural number, or does {k\over n} represent all rational numbers on the interval?

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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Phyxius117 View Post
    Here is the question:



    How do I do this problem?? Thank you!!
    There is a very general theorem which does away with this, but assuming that you don't know this theorem here's what I would do:

    Start by considering \varepsilon<1. Consider then the intervals


    I_k=\begin{cases}\left[\frac{k}{n}-\frac{\varepsilon}{64n},\frac{k}{n}+\frac{\varepsi  lon}{64n}\right] & \mbox{if}\quad k=1,\cdots,2n-1\\ \left[2-\frac{\varepsilon}{64n},2] & \mbox{if}\quad k=2n\end{cases}


    Notice then that if


    J_k=\begin{cases}[0,\frac{1}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=1\\ [\frac{k-1}{n}+\frac{\varepsilon}{64n},\frac{k}{n}-\frac{\varepsilon}{64n}] & \mbox{if}\quad k=2,\cdots,2n\end{cases}


    Then [0,2]=I_1\cup\cdots\cup I_{2n}\cup J_1\cup\cdots\cup J_{2n}. Note though that f_{\mid J_k}=x^3,\quad k=1,\cdots,2n and so by continuity of x^3 on J_1,\cdots,J_{2n} we may find partitions P_1,\cdots,P_{2n} such that U(P_k,f_{\mid J_k})-L(P_k,f_{\mid J_k})<\frac{\varepsilon}{4n}. Note next that P=P_1\cup\cdots\cup P_{2n} is a partition for [0,2] and


    \displaystyle U(P,f)-L(P,f)=\underbrace{\sum_{k=1}^{2n}\left(U(P_k,f_{\  mid J_k})-L(P_k,f_{\mid J_k})\right)}_{\mathbf{(1)}}+\underbrace{\frac{\va  repsilon}{32n}\sum_{k=1}^{n}\left(\sup_{x\in I_k}f(x)-\inf_{x\in I_k}f(x)\right)}_{\mathbf{(2)}}


    Note though that by construction \mathbf{(1)}=2n\cdot \frac{\varepsilon}{4n}=\frac{\varepsilon}{2} and ,since \displaystyle \inf_{x\in I_k}f(x)=0,


    \displaystyle \mathbf{(2)}=\frac{\varepsilon}{32n}\sum_{k=1}^{n}  \sup_{x\in I_k}f(x)\leqslant \frac{\varepsilon}{32n}\sum_{k=1}^{2n}8=\frac{\var  epsilon}{2}


    And thus U(P,f)-L(P,f)=\mathbf{(1)}+\mathbf{(2)}<\frac{\varepsilon  }{2}+\frac{\varepsilon}{2}=\varepsilon and since \varepsilon (after taking the min of 1 and \varepsilon or any \varepsilon) was arbitrary it follows that[ math]f[/tex] is integrable. Thus,


    \displaystyle \int_0^2 f(x)\text{ }dx=\overline{\int_0^2}f(x)dx=\sup_{P\in\mathcal{P  }}U(P,f)=\sup_{p\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx=4


    where we've used the obvious fact that U(P,f)=U(P,x^3) for any P\in\mathcal{P} and since x^3\in\mathcal{R}([0,2]) that \displaystyle \sup_{P\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx.
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  4. #4
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    Assuming that n is a fixed natural number and k must be an integer, then k/n is in [0, 2] only for k= 0, 1, 2, ..., 2n. That is, f(x) is continuous every where except on the finite set {0, 1/n , 2/n, ..., 2}. Every finite set has Lebesque measure 0 so f is continuous except on a set of measure 0 and so is Riemann integrable. its integral is exactly [itex]\int_0^2 x^3dx= 4[/itex].
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