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Thread: Riemann integrability

  1. #1
    Member thaopanda's Avatar
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    Riemann integrability

    Decide which of the functions $\displaystyle f_n$ : [0,1] $\displaystyle \rightarrow$ R with n = 0,1,2, defined by setting:

    $\displaystyle f_n(x)$ :=
    $\displaystyle (x^n)sin(\frac{1}{2}$ if $\displaystyle 0 < x \leq 1$
    0 if x = 0

    are Riemann integrable on [0,1].

    more help please
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    Decide which of the functions $\displaystyle f_n$ : [0,1] $\displaystyle \rightarrow$ R with n = 0,1,2, defined by setting:

    $\displaystyle f_n(x)$ :=
    $\displaystyle (x^n)sin(\frac{1}{2}$ if $\displaystyle 0 < x \leq 1$
    0 if x = 0

    are Riemann integrable on [0,1].

    more help please
    Is that $\displaystyle f_n(x)=x^n\sin\left(\frac{1}{2}\right)
    $ ?
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  3. #3
    Member thaopanda's Avatar
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    yeah, sorry, I forgot the other parenthesis
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    yeah, sorry, I forgot the other parenthesis
    Well $\displaystyle \sin(1/2)$ is just a number. Call it $\displaystyle k$. So you need to know for which n is $\displaystyle f_n(x)=kx^n$ integrable on $\displaystyle [0,1]$.

    For $\displaystyle n\geq1$, it's continuous and therefore integrable. For $\displaystyle n=0$, you have one removable discontinuity (at $\displaystyle x=0$), so it's also integrable.
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  5. #5
    Member thaopanda's Avatar
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    if it was $\displaystyle sin(\frac{1}{x})$, how would that change the problem?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by thaopanda View Post
    if it was $\displaystyle sin(\frac{1}{x})$, how would that change the problem?
    $\displaystyle x^n\sin(1/x)$ is continuous for all $\displaystyle n\geq1$. If $\displaystyle n=0$, the function is still continuous on $\displaystyle (0,1]$ and is therefore integrable.
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