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Anti-derivatives and Elementary Functions

  1. Sep 10, 2009 #1
    I've seen this many times: someone will post a homework question asking them to prove something about an integral [tex] \int_a^b f(t) dt [/tex], where f is some arbitrary, continuous, integrable function. They will write in their proof, [tex] \int_a^b f(t) dt = F(b) - F(a) [/tex] where they're assuming F is the anti-derivative of f. They will get told that they cannot assume that f has an anti-derivative.

    My question now is, if f is continuous and integrable, then doesn't it follow from the FTC that there exists a function F such that F' = f? You may not be able to express F in terms of elementary functions, but F still must exist, right? So why can't the student write in their proof [tex] \int_a^b f(t) dt = F(b) - F(a) [/tex]???
  2. jcsd
  3. Sep 11, 2009 #2
    Eh are you referring to these forums (if so then okay I guess I haven't looked that closely)? I'm not sure if this entirely relates to the situation you are referring to, but I think that many people who take a first calculus course, say AP Calculus, tend to think that the equation you refer to is the definition of an integral. I think I have made this mistake before back when I didn't have a rigorous understanding of riemann integration. But yeah, if your function is continuous (and hence integrable, but we want continuity), then an antiderivative is guaranteed, but just note that F will not be any simpler than a function of the form [itex]\int_{c}^{x}f(t)\,dt,[/itex] if there is no way of expressing the antiderivative in terms of elementary functions.
  4. Sep 11, 2009 #3
    I think what is usually said is that f has no elementary anti-derivative.

    Usually, the case is that OP wants to evaluate an integral exactly by using the FTC. But this doesn't work. Sure enough, the FTC holds, and

    \int_a^b f(t) dt = F(b) - F(a)

    but no progress is made in the problem, because it won't give any help as to how to evaluate F(b) and F(a).
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