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Second Fundamental Theorem of Calculus Question

  1. Jul 3, 2011 #1
    Why do we say that f is a function of t and then take the derivative of the integral with respect to x? This confuses me because x is also the upper limit of integration.
     
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  3. Jul 3, 2011 #2
    Can you please state your version of the second fundamental theorem? My version (the one found on wikipedia) doesn't have your variables in it. It would be easier to explain if I know what we have.
     
  4. Jul 3, 2011 #3
    Ok.
    I have the expression the integral of f(t) with respect to t from a to x.
    The derivative of this with respect to x is f(x).
    This is true for every x chosen.
     
  5. Jul 3, 2011 #4
    I assume you mean something along the lines of
    [tex]\frac{d}{dx}F(x) = \frac{d}{dx}\int_a^x f(t)dt[/tex]

    Which is the http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part".

    When integrating [itex]f(t)[/itex] from [itex]a[/itex] to [itex]x[/itex], you end up plugging [itex]x[/itex] and [itex]a[/itex] in for [itex]t[/itex].
    [tex]\int_a^x f(t)dt = F(x) - F(a)[/tex]
    [tex]\frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx} [F(x) - F(a)][/tex]
    Since [itex]F(a)[/itex] is a constant, you drop it when you differentiate.
    [tex]\frac{d}{dx} [F(x) - F(a)] = \frac{d}{dx} F(x) - \frac{d}{dx} F(a) = \frac{d}{dx} F(x) - 0 = \frac{d}{dx} F(x) = f(x)[/tex]


    When you differentiate, you have to do so with respect to [itex]x[/itex], because [itex]t[/itex] has been integrated out of the formula. Basically, this is just changing from one variable to another, the only difference being wherever you saw [itex]t[/itex], you replace with [itex]x[/itex].

    A specific example of this theorem in action is with [itex]ln\;x[/itex]:

    [tex]ln\;x = \int_1^x \frac{dt}{t}[/tex]

    Does this help?
     
    Last edited by a moderator: Apr 26, 2017
  6. Jul 4, 2011 #5

    HallsofIvy

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    If F is an anti-derivative of f, then
    [tex]\int_a^b f(t)dt= F(b)- F(a)[/tex]
    the "definite integral".

    Similarly,
    [tex]\int_a^x f(t)dt= F(x)- F(a)[/tex]
    so we now have a function of x. The only thing we can differentiate with respect to is x.
     
  7. Jul 5, 2011 #6
    Thank you. After I have posted this thread, I proved to myself that the statement makes sense by using the First Fundamental Theorem (which is what you guys are saying). But now I have another question:
    What if we have f(x) instead of f(t) under the integral and we take the integral with respect to x? Would that make the statement bogus?
     
  8. Jul 5, 2011 #7
    I think it would be (I'm not sure, though).
    [tex]\int_a^x f(t)dt[/tex]
    When you plug into the antiderivative, [itex]F(t)[/itex], you're plugging in the values of [itex]t[/itex] that you're integrating over, from [itex]a[/itex] to [itex]x[/itex]. The way I interpret it (which may be mistaken since I'm only an undergraduate studying math) is that you're plugging in functions [itex]t(x) = a[/itex] to [itex]t(x) = x[/itex], and so when you integrate, you plug in those functions of [itex]x[/itex].
    However, when you're integrating
    [tex]\int_a^x f(x)dx[/tex]
    you're essentially calling [itex]x[/itex] a function of itself, since you'd be plugging in [itex]x(x) = x[/itex].

    Maybe someone that's more knowledgeable on this than myself will comment.
     
  9. Jul 5, 2011 #8

    disregardthat

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    You could use the same sign for the variables, but they are not the same. It isn't "bogus", you just need to know which belong to which operator. Of course, it is more instructive to use different signs for different variables.
     
  10. Jul 5, 2011 #9

    jambaugh

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    Some comments:
    ---------
    It is important to distinguish between a function and the value of a function. Given f is a function then f(x) is a value of that function, and so too is f(t) or f(42) etc. When we define a function we define the action resulting in its value and the variable we use is arbitrary. Thus for example saying f(x) = 2x+3 defines the function f in exactly the same way as saying f(z) = 2z+3 or f(t) = 2t+3. The use of different variable names is immaterial. The use of y=f(x) is a means of defining the graph of the function as a set of points (x,y) satisfying y=f(x) or equivalently the set of points (t,f(t)) (z,f(z))...
    --------
    In the notation for definite integrals (which are values, not functions) the variable of integration is meaningful only inside the the integrand. It is exactly the same as summation notation where the index should only appear inside the summand.
    ---------
    So take a function f and define another function A from it by defining the value of that function using a definite integral involving f:

    [itex]A[/itex] is the function mapping the number [itex]x[/itex] to the value [itex]A(x)[/itex] given by the definite integral:
    [tex]A(x) = \int_a^x f(t)dt[/tex]
    Then the fundamental theorem states that the derivative of A is the function f, i.e. A' = f.

    Note that it is NOT correct to say that "the derivative of A(x) is the function f(x)" which is trying to take the derivative of something which is NOT a function and identifying it with another something which is NOT a function.

    Now we can speak of the derivatives of variable expressions but only when we are thinking of those expressions as functions of a specific independent variable. Thus we can speak of the derivative of f(x) with respect to x, and similarly we can speak of the derivative of f(g(x)) or x f(x) w.r.t. x.
    We are using a function (or functions) to write and expression in x which is itself a function of x.

    [edit] Thus it is OK to say: "The derivative of A(x) with respect to x, is f(x)". [end edit]

    It's a subtle distinction but if ignored it leads to the confusion expressed in the O.P.
     
  11. Jul 5, 2011 #10

    Redbelly98

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    Just want to make a quick point: the following expression:

    [tex]\int_a^x f(t)dt[/tex]

    is not a function of t, and is a function of x. We can't, for example, evaluate the expression at t=5 (or at any other value of t, for that matter). Whereas we can evaluate it when x=5.

    So it is nonsensical to take "the derivative with respect to t". However, it is valid to differentiate it with respect to x.
     
  12. Jul 8, 2011 #11
    OK. Now I understand. Thank you all, especially jambaug.
     
  13. Jul 8, 2011 #12

    lavinia

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    x is a variable so the integral is a function. It is this function that you are differentiating.
     
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