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Thread: Help with Integrable Proof

  1. #1
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    Smile Help with Integrable Proof

    Let

    f(x)=
    { 1 if -1 < or equal to x<0;
    {-1 if 0 < or equal to x < or equal to 1.

    Prove that f(x) is integrable on -1 < or equal to x < or equal to 1.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Swamifez View Post
    Let

    f(x)=
    { 1 if -1 < or equal to x<0;
    {-1 if 0 < or equal to x < or equal to 1.

    Prove that f(x) is integrable on -1 < or equal to x < or equal to 1.
    I fail to see how this can be a problem, what do you know about integration?
    (what type of integral are we talking about here).

    If this is a Riemann integral it is obvious that the integral of $\displaystyle f(x)$ is
    defined over $\displaystyle [-1,1]$ from a elementary consideration of the Riemann
    sums of $\displaystyle f(x)$ for partitions of this interval, and that the integral is:

    $\displaystyle \int_{-1}^1 f(x) dx=\int_{-1}^0 f(x) dx + \int_{0}^1 f(x) dx=1-1=0$

    RonL
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  3. #3
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    Quote Originally Posted by Swamifez View Post
    Let

    f(x)=
    { 1 if -1 < or equal to x<0;
    {-1 if 0 < or equal to x < or equal to 1.

    Prove that f(x) is integrable on -1 < or equal to x < or equal to 1.
    I can actually do the limit of the sum to show it exists.
    But the rule is: "A real-function defined on a closed interval is Riemann integrable if and only if it is continous almost everywhere".

    This function is just discontinous on $\displaystyle x=0$. Thus, it is integrable.
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  4. #4
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    Sorry I am posting so much, but these problems are going to be similar to by final. Can someone help me clarify this problem in to one big proof. I also need to pick 5 paritions of this interval and prove it within. Thank you
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  5. #5
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    Okay, for suffiently large $\displaystyle n$.
    You want to find,
    $\displaystyle \int_{-1}^1 f(x)dx$
    I will use right endpoints,
    $\displaystyle \lim_{n\to \infty}\sum_{k=1}^n f(-1+k\Delta x)\Delta x$
    But,
    $\displaystyle \Delta x=\frac{2}{n}$
    Thus,
    $\displaystyle \lim_{n\to \infty}\sum_{k=1}^n f\left(-1+\frac{2k}{n} \right) \frac{2}{n}$
    Note,
    $\displaystyle -1\leq -1+\frac{2k}{n} <0$
    $\displaystyle -n\leq -n+2k<0$
    $\displaystyle 0\leq 2k<n$
    $\displaystyle 0\leq k<n/2$
    $\displaystyle k=0,1,2,...,[n/2]$ (greatest integer).
    Thus, the function is 1 for these values.
    And it is -1 for after these values.
    $\displaystyle \sum_{k=1}^{[n/2]} \frac{2}{n}-\sum_{k=[n/2]+1}^n \frac{2}{n}$
    If $\displaystyle n$ is even we have,
    $\displaystyle \sum_{k=1}^{n/2} \frac{2}{n}-\sum_{k=n/2+1}^n \frac{2}{n}=0$
    Thus, as $\displaystyle n\to \infty$ the limit of the Riemann sum is zero.
    If $\displaystyle n$ is odd we have,
    $\displaystyle \sum_{k=1}^{(n-1)/2}\frac{2}{n}-\sum_{k=(n-1)/2+1}^n \frac{2}{n}=-\frac{2}{n}$
    Thus, as $\displaystyle n\to \infty$ the limit of the Riemann sum is zero.
    Thus, the limit is always zero no matter what.
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