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Homework Help: All integrable functions are continuous?

  1. Dec 8, 2005 #1
    I have a couple of true and false questions I'm looking at in order to review for my final next Friday (not tomorrow). First, when someone says that a function is continuous does that mean continuous on its domain or continuous at all reals? For example, tan(x) is continuous on its domain but not continuous at all reals so is tan(x) a continuous function?

    1. All differentiable functions are integrable.
    True because all differentiable functions are continuous and by FTC all continuous functions are integrable.

    2. All integrable functions are continuous.
    This doesn't follow from the FTC, but I'm having trouble thinking of a counter-example. I looked around on the web and saw a couple people say that this is false, but never explain why. Can you integrate piecewise functions? If so then I can think of an easy counter-example. We've never talked about doing so in class.

    3. All integrable functions are differentiable.
    Even though 1 is true this doesn't follow from it. Same difficulty as 2.
  2. jcsd
  3. Dec 8, 2005 #2


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    A continuous function is continuous on its domain. Your intuition is right about 2 (not all integrable functions are continuous). Go back to the definitions to confirm this. By 1 and 2, the answer to 3 follows.
  4. Dec 9, 2005 #3


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    you can integrate piecewise functions by splitting the integral. ie

    f\left( x \right) = \left\{ \begin{array}{l}
    5 \Leftarrow x < 0 \\
    - 5 \Leftarrow x \ge 0 \\
    \end{array} \right. \\
    \int_{ - 1}^1 {f\left( x \right)dx} = \int_{ - 1}^0 {f\left( x \right)dx} + \int_0^1 {f\left( x \right)dx} = \left[ { - 5x} \right]_{ - 1}^0 + \left[ {5x} \right]_0^1 = 0 \\
  5. Dec 9, 2005 #4
    Are all functions that can be differentiated, integratable? It makes sense to think this, but what about something like y = x^x, you can diffentiate implictly, using logs.
    lny = xlnx
    then diffentiate this, but can y = x^x be integrated. I can't see how but I might be wrong.
  6. Dec 9, 2005 #5
    See section 1 in the original post.
  7. Dec 9, 2005 #6

    matt grime

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    Are you talking in the generality of riemann sums?
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