Results 1 to 4 of 4

Thread: Real Analysis: Riemann Integrable

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    65

    Real Analysis: Riemann Integrable

    Here is the question:



    How do I do this problem?? Thank you!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by Phyxius117 View Post
    Here is the question:



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

    Is $\displaystyle k$ also a natural number?

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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 $\displaystyle \varepsilon<1$. Consider then the intervals


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


    Notice then that if


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


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


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


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


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


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


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


    where we've used the obvious fact that $\displaystyle U(P,f)=U(P,x^3)$ for any $\displaystyle P\in\mathcal{P}$ and since $\displaystyle x^3\in\mathcal{R}([0,2])$ that $\displaystyle \displaystyle \sup_{P\in\mathcal{P}}U(P,x^3)=\int_0^2 x^3\text{ }dx$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,718
    Thanks
    3003
    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].
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. f & g Riemann integrable, show fg is integrable
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Feb 12th 2011, 09:19 PM
  2. Analysis, riemann and lebesque integrable
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 27th 2010, 03:01 PM
  3. Real Analysis: Prove BV function bounded and integrable
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Jan 17th 2010, 02:31 PM
  4. Real Analysis - Riemann Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 10th 2009, 07:06 PM
  5. real analysis riemann integrable
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Dec 3rd 2008, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum