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Homework Help: Not really sure how to approach this limit involving integrals

  1. May 21, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex] \lim_{x \to 1} \frac {\int^{x}_{1} \frac{1}{t} dt} {\int^{x}_{1} \frac {1}{(2t+1)}dt} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I am not really sure how I am suppose to approach this problem, I would think that as x approaches 1 the integrals will approach 0 because an integral from n to n is 0. So if that is correct and I am suppose to use l'hopital's rule (since I am still in that section) obviously I am suppose to take the derivatives of those functions but how do I apply the limit after I do that? will it just be:

    [tex] \lim_{x \to 1} \frac {\int^{x}_{1} f'(t) dt}{\int^{x}_{1} g'(t) dt} [/tex]

    or would the integrals not apply anymore?... i really just don't know, if the integral still applies i basically have the same equation though, as it will end up being 0/0, which is no good.

    Supposedly the answer is 3
  2. jcsd
  3. May 21, 2010 #2
    Use a fact that you should know about expressions like this:

    [tex] \frac{d}{dx} \int_a^x f(t) dt = ?? [/tex]
  4. May 21, 2010 #3
    Thanks, I forgot about that portion of my notes, I have never used it till now, so it the equation basically becomes:

    [tex] \lim_{x \to 1} \frac {\frac{1}{x} dx}{ \frac {1}{2x+1} dx} [/tex]

    and with the limit applied the answer is 3.
  5. May 21, 2010 #4
    Yes, although those "dx" are extraneous.

    This is the Fundamental Theorem of Calculus:
    \frac{d}{dx} \int_a^x f(t) dt = f(x)
  6. May 21, 2010 #5
    right but the only portion of FTC i have used this far is F(b)-F(a) so I had forgotten about that part, that's all I was saying
  7. May 21, 2010 #6


    Staff: Mentor

    Or you could just evaluate the two integrals. Seems like that would be a little simpler. (Still need to use L'H rule, though.)
  8. May 21, 2010 #7
    is that not what was done Mark44?


    [tex] \int^{x}_{a} f(t) dt = f(x) [/tex]

    then doing

    [tex] \int^{x}_{1} \frac {1}{t} dt = \frac {1}{x} [/tex]

    is evaluating the integral, no?
  9. May 21, 2010 #8
    No, the derivative of the integral equals f(x), not the integral itself.

    [tex] \int_1^x \frac{1}{t} dt = \ln x - \ln 1 = \ln x [/tex]
    [tex] \int_1^x \frac{1}{2t+1} dt = \frac{1}{2}\ln ({2x+1}) - \ln {3} = \frac{1}{2}\ln \frac{2x+1}{3} [/tex]

    The limits of these are 0 and 0, so then you use L'Hopital's rule, obtaining the original functions over each other, etc. I kind of think the other way is simpler, but YMMV.
  10. May 21, 2010 #9
    ah... so to use

    [tex] \int^{x}_{a} f(t) dt = f(x) dx [/tex]

    you need to find F(x) of f(x) and FTC to actually evaluate the integral?
    Last edited: May 21, 2010
  11. May 21, 2010 #10


    Staff: Mentor

    There should be no dx on the right side - just f(x).
    What I have been suggesting is that you should NOT use the FTC in this problem. You can work this problem in a way that uses the FTC, but that's not what I'm suggesting. Instead, just evaluate each definite integral. hgfalling shows in his post how this is done.
  12. May 21, 2010 #11

    I was not talking about this specific problem anymore, I was speaking, in general about the formula:

    [tex] \int^{x}_{a} f(t) dt = f(x) [/tex]

    for example, the integral in the numerator of my problem is:

    [tex] \int^{x}_{1} \frac {1}{t} dt [/tex]

    using the above formula this would become:

    [tex] \int^{x}_{1} \frac {1}{t} dt = \frac {1}{x}[/tex]

    so using your new function in terms of x, to evaluate the integral you need to find F(x) of f(x) and use FTC. Hgfalling shows this in the post above where I as the question. In doing so the answer of the integral is going to be ln x, and then if you were to use l'hopital's rule, you need to find the derivative, which is 1/x. So actually evaluating the integral seems an unnecessary step in this problem.

    Also why is the dx in:

    [tex] \int^{x}_{a} f(t) dt = f(x) dx [/tex]

    not correct? I copied my notes straight out of the book and that's the form they used, I am just curious to that point.
  13. May 22, 2010 #12


    Staff: Mentor

    This is incorrect. The integral on the left side is a function of x, but what you are showing as f on both sides generally represents different functions.

    For example,
    [tex] \int^{x}_{a} t^2 dt = (1/3)x^3 - (1/3)a^3 [/tex]

    Notice that if the function in the integrand is f(t) = t2, the function that appears on the right is NOT f(x), which would be x2. What appears on the right is the antiderivative of f, evaluated at x and a. The antiderivative of f is often written as F, which is different from f.

    That is not correct. The value of the definite integral is ln|x| - ln(1) = ln|x|. The natural log function is the antiderivate of the function f(t) = 1/t.
    It is incorrect for two reasons.
    1. As I already noted, the correct version of the formula just above would be
    [tex] \int^{x}_{a} f(t) dt = F(x) - F(a)[/tex]
    where F is an antiderivative of f. Another way to say this is that F'(x) = f(x).
    2. The dt in the integral identifies the variable to be used in integration, in a similar way that the dx in d/dx identifies the independent variable with respect to which we are differentiating. Leaving the dx on the right side in what you have above is just as incorrect as writing that d/dx(x3) = 3x2 dx.

    It seems that you are confused about two important things and are mixing them up. One if the definite integral and the other is the Fundamental Theorem of Calculus.

    I won't give an example of the definite integral, since I already gave one in this post. The FTC has two parts, the first of which states that differentiation and antidifferentiation are essentially inverse operations. It is often presented this way:

    [tex] \frac{d}{dx}\int^{x}_{a} f(t) dt = f(x)[/tex]

    What I think you are missing is that differentiation operator in the FTC.

    You said that you copied this out of the book
    Go back and take another look. I don't believe that there are any calculus books out there that would make such a mistake.
  14. May 22, 2010 #13
    Oh wow yeah, I totally skipped the differential marker in my notes, it reads:

    [tex] d(\int^{x}_{a} f(t)dt) = f(x)dx [/tex]

    sorry to cause all that controversy, with the different notation (or rather the part i skipped over) it makes sense now.
  15. May 22, 2010 #14


    Staff: Mentor

    OK, now it's right. That's the differential form.

    The derivative form is
    [tex] \frac{d}{dx}\int^{x}_{a} f(t) dt = f(x)[/tex]
  16. May 22, 2010 #15
    ok, so, forgive me if this is obvious or something I should know but whats the difference? They both read the same to me, is it just different notation for the same thing?
  17. May 22, 2010 #16


    Staff: Mentor

    No, they're different by a factor of dx. The FTC (first part) is usually given in terms of the derivative of the definite integral (the way I wrote it). What you have is the differential of the definite integral (the way you wrote it from your text). I've seen quite a few calculus texts, but I've never seen the FTC presented in terms of differentials.

    Here's the relationship between the differential of a function and the derivative of the same function:

    df = (df/dx) * dx
  18. May 22, 2010 #17
    Well first off, I am using, what I consider to be, a none standard text. It is the free book put out by Jerome H. Keisler entitled "Elementary Calculus: An Infinitesimal Approach", I find that it differs quite widely from standard calculus texts, for one example limits are taught in the 5th chapter (even after integrals, which is the fourth) when a standard calculus book starts with limits.

    So, once again, forgive me for being naive but its not quite apparent to me what the actual difference is, how do the answers differ between the two equations? As I understand it dx is infinitely small and just a marker for which variable you are differentiating with respect to and does not actually have anything to do with the answer.

    Maybe your relationship equation just hasn't sunk in yet....
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