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Thread: riemann integration help

  1. #1
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    riemann integration help

    I need help with the following:
    suppose that f:[0,1]-->R is defined by:
    f(x)=1/n if 1/(n+1) < x <= 1/n
    and 0 if x=0.
    Verify that f is riemann integrable on the interval [0,1]
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  2. #2
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    Quote Originally Posted by binkypoo View Post
    I need help with the following:
    suppose that f:[0,1]-->R is defined by:
    f(x)=1/n if 1/(n+1) < x <= 1/n
    and 0 if x=0.
    Verify that f is riemann integrable on the interval [0,1]

    According to your definition, $\displaystyle f(x)$ is only defined on the set $\displaystyle \left(\frac{1}{n+1},\frac{1}{n}\right)\cup\{0\}$ and not on $\displaystyle [0,1]$ . Check this.

    Tonio
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  3. #3
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    please excuse me, I should have used latex,
    I made x <= 1/n, which reads
    $\displaystyle x \leq 1/n$
    sorry.
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  4. #4
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    Quote Originally Posted by binkypoo View Post
    please excuse me, I should have used latex,
    I made x <= 1/n, which reads
    $\displaystyle x \leq 1/n$
    sorry.

    I don't get it: what is then the function's definition?

    Tonio
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  5. #5
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    $\displaystyle f(x)=\left\{
    \begin{array}{lr}
    \frac{1}{n} & : \frac{1}{n+1}< x\leq \frac{1}{n}\\
    0 & : x=0
    \end{array}\right.$
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  6. #6
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    Quote Originally Posted by tonio View Post
    I don't get it: what is then the function's definition?
    The collection of intervals $\displaystyle \left\{ {\left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]:n \in \mathbb{Z}^ + } \right\}$ partitions $\displaystyle (0,1]$.

    So the function is $\displaystyle f:[0,1]\mapsto [0,1]$ by $\displaystyle f(x) = \left\{ {\begin{array}{rl} 0 & {,x = 0} \\ {\frac{1}{n}} & {,x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]} \\ \end{array} } \right.$
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