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Finding the integral of an unknown function

  1. Feb 14, 2005 #1
    I need the help of all you math folks out there...I'm working on a bonus assignment in my calculus 2 class. Here's the problem:

    Suppose the curve y=f(x) passes through the origin and the point (1,1). Find the value of the integral of f'(x) dx from 0 to 1.

    I thought this question just wanted a restatement of the fundamental theorem of calculus. But since it's a bonus problem, I know that there has to be something tricky here. It does ask for a "value".

    Obviously this curve can take any shape, so there's no general formula for the area.

    Does anyone have any suggestions?
  2. jcsd
  3. Feb 14, 2005 #2
    Hi pattie,
    Write down what you know about f mathematically, and write down your integral. The deceptively simple result is just one of the many results that illustrate the power of calculus.
    The only prerequisite assumption missing from your statement is that f(x) is continuous.
    Last edited: Feb 14, 2005
  4. Feb 14, 2005 #3
    The book doesn't even specify it's continuous...but assuming it is I have the integral of f'(x)dx from 0 to 1 is [f(1)-f(0)]. Is it really that simple?
  5. Feb 14, 2005 #4
    Have you proven/read a proof of the fundamental theorem of calculus ? :smile: Unless you spot a flaw in the proof, it really is that simple.
  6. Feb 14, 2005 #5


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    well i guess you can assume that f' is integrable, since they said find its integral. then the MVT implies that an integrable function which is a derivative, can be comouted as you say, i.e. the FTC holds. continuity is not needed.

    this stronger proof of FTC, without continuity, like most other things i know about calc can be found in courant.
  7. Feb 15, 2005 #6


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    If they say to compute the integral of f', it's obvious to assume f is differentiable and hence that f is continuous.
  8. Feb 15, 2005 #7

    matt grime

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    The continuity of f isn't the issue. It is the continuity of df/dx that is "missing".
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