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Questions regarding the Fundamental Theorem of Calculus

  1. Jul 19, 2012 #1
    If you define a function [tex]g(x) = \int_a^x \! f(t) \, \mathrm{d} t[/tex] then from what I currently understand, g(x) gives the value of the area under the curve [tex]y=f(t)[/tex]

    When you differentiate both sides, [tex]g'(x)[/tex] gives the rate of change of the area underneath [tex]y=f(t)[/tex], however, I don't understand intuitively why f(x) is the rate of change of the area underneath its curve.

    Also, when why when you differentiate it (the right hand side), is it [tex]f(x)[/tex] and not [tex]f(t)[/tex]? I can't seem to get my head around the whole respect to x thing - what about t? Why not respect to t? Then if you were to integrate it again, would x become the dummy variable?

    The dummy variable is really confusing!
  2. jcsd
  3. Jul 19, 2012 #2
    Also, why is the condition always differentiable on an OPEN interval, not closed?
  4. Jul 19, 2012 #3
    It is easy to understand intuitively. Say you went an infinitesimal distance dx. Then, dA, the infinitesimal change in area is f(x)dx. Hence, dA/dx becomes f(x) (the derivative of the area with respect to x.)
  5. Jul 19, 2012 #4
    To understand the Second Fundamental Theorem intuitively, let [itex]x_{1}[/itex] and [itex]x_{2}[/itex] be points close together in [itex][a,b][/itex] with [itex]x_{1}<x_{2}.[/itex] Define [itex]\Delta x = x_{2}-x_{1}[/itex] and [itex]\Delta A [/itex] to be the area under the graph of [itex]f[/itex] from [itex]x_{1}[/itex] to [itex]x_{2}[/itex]. (Assume [itex]f[/itex] is nonnegative on [itex][a,b][/itex].) Now pick a point [itex]c[/itex] between [itex]x_{1}[/itex] and [itex]x_{2}[/itex]. If [itex]f[/itex] is continuous from [itex]x_{1}[/itex] to [itex]x_{2}[/itex], then the function values between [itex]x_{1}[/itex] and [itex]x_{2}[/itex] should all be close to [itex]f(c)[/itex]. By considering the area of the rectangle of width [itex]\Delta x[/itex] and height [itex]f(c)[/itex] as an approximation for the area under the graph from [itex]x_{1}[/itex] to [itex]x_{2}[/itex], we obtain the following approximation: [itex]\Delta A \approx f(c)\Delta x .[/itex] If we divide by the change in [itex]x[/itex], we end up with [itex]\Delta A / \Delta x \approx f(c).[/itex]
  6. Jul 19, 2012 #5

    Stephen Tashi

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    Science Advisor

    It doesn't give you "the" area under the curve. It gives you a particular area under the curve. The "a" and the "x" are also used to define which area.

    Would it be intuiitively clear to you if f(x) was a linear function? Would it be intuitively clear that the rate of change of the are under a linear function would only depend on its slope? If so, then think how the derivative is used in finding the formula for a tangent line. The tangent line is an approximation for the function f(x) in the vicinity of the point x where the tangent line is drawn.

    In the original function g we are taking "a" as a constant and "x" as the variable. So the derivative depends on the same variable as the function that is differentiated. But you could call the derivative f(w) or f( Willy_Smith) if you wanted to, if you purpose is merely to name it as a function. In reading mathematics, you have to pay attention the "scope" of variables like "x" and "t".

    If you do computer programming, you should already be familar with the scope of variables. In a computer program, an "x" in one function has nothing to do with an "x" in a different function unless there is some orgnization of the code that forces the "x" to have the same value in both places. Likewise in a mathematics, an "x" at the top of the page need not mean the same thing as an "x" at the bottom of the page, or even as an "x" in the next line or paragraph. If you are trying to convey the idea that the derivative of g(x) evaluated at the argument 5 is equal to the function f evaluated at the argument 5, then it is important to say g'(x) = f(x) and the context indicates that "x" has the same meaning in both the left and right hand sides.

    The scope of the "dummy" variable such as [itex] w [/itex] used in definite integrations such as [itex] \int_a^p f(w)dw [/itex] doesn't extend outside the integrand. If you know computer programming, imagine a numerical computation of the integral that is done by a function. The programmer who writes the function needs to introduce some local variable "w" whose value is not visible outside the function. In a computer program, there is no reason to incorporate the name of the dummy variable into the function. For example, a function that computes the above integral might be named "integ(f,a,p)". There is no need to name it "integDw(f,a,p)". In writing mathematics in English there is a reason for naming the dummy variable in an integration. Sometimes there are expressions where several symbols are involved, such as [itex] y \sin(x^2 + z) [/itex] and you want to integrate such an expression assuming some of the symbols are constants and one of the symbols is the variable. The dummy variable notation conveys which symbol is in the expression is treated as the variable.
  7. Jul 19, 2012 #6
    Here's two ways to think intuitively about the first fundamental theorem of calculus. First, let's use limits. Recall that the average value of the function f over the interval [a,b] is given by [itex]\frac{\int^{b}_{a}f(t) dt}{b-a}[/itex]. Now we have [itex]g'(x) = \frac{d}{dx}\int^{x}_{a}f(t) dt = lim_{h\rightarrow 0}\frac{\int^{x+h}_{a}f(t) dt - \int^{x}_{a}f(t) dt}{h} = lim_{h\rightarrow 0}\frac{\int^{x+h}_{x}f(t) dt}{x + h - x}[/itex]. But [itex]\frac{\int^{x+h}_{x}f(t) dt}{x + h - x}[/itex] is just the average value of the function f over the interval [x,x+h]. And as h→0, [x,x+h] becomes the singleton interval [x,x], and the average value of f over the interval [x,x+h] just becomes the value f(x).

    Another way of thinking about it is in terms of infinitesimals. Using the traditional infinitesimal definition of the derivative, we have [itex]g'(x) = \frac{dg}{dx}=\frac{g(x+dx)-g(x)}{dx} = \frac{\int^{x+dx}_{a}f(t) dt - \int^{x}_{a}f(t) dt}{dx}[/itex]
    [itex]= \frac{\int^{x+dx}_{x}f(t) dt}{dx} = \frac{f(x) dx}{dx} = f(x)[/itex], where in the second to last step I used the fact that [itex]\int^{x+dx}_{x}f(t) dt = f(x) dx[/itex] because the area under f between x and x+dx is just the area of the infinitely thin rectangle with height f(x) and width x+dx-x = dx.
  8. Jul 19, 2012 #7
    In fact, the integral of an integrable function f over [a,b] is often written [itex]\int^{b}_{a}f.[/itex] Notice that even in the English description, "the integral of f over [a,b]," I didn't even refer to any variables. Likewise, if the the second endpoint of the interval is a variable, we can write the integral of f over [a,x] as [itex]\int^{x}_{a}f.[/itex] To see what happens if we try to integrate again after differentiating the original integral, consider [itex]\int^{x}_{a}[/itex] as its own function on [a,b] -- call it g (it looks like you're already familiar with this). By the SFT, [itex] \frac{d}{dx} \left[ \int^{x}_{a} f \right] = g'. [/itex] Then,

    [itex] \int^{x}_{a} \frac{d}{dx}\left[ \int^{x}_{a} f\ \right] [/itex]

    [itex]= \int^{x}_{a} g' = g(x)-g(a)[/itex] (by the FFT)

    [itex]= \int^{x}_{a} f - \int^{a}_{a}f[/itex]

    [itex]= \int^{x}_{a} f - 0[/itex]

    [itex]= \int^{x}_{a} f. [/itex]

    So why do we even bother with the dummy variable t in [itex] \int^{b}_{a} f(t)dt?[/itex] One reason, which has already been mentioned, is that when we do iterated integration of a function of several variables, it allows us to keep the variables straight. Another is that an integral with a dummy variable more closely resembles the Riemann sum found in the definition of the integral -- [itex] \int^{b}_{a} f(t)dt = \lim_{n \to \infty} \sum^{n}_{i=1} f(t^{*}_{i})\Delta t .[/itex] Yet another is that statements like [itex] \int^{b}_{a} t^{2}dt = \frac{b^{3}-a^{3}}{3} [/itex] make sense without having to go through the trouble of saying something like "Let [itex]f(t)=t^{2}.[/itex] Then [itex] \int^{b}_{a}f = ... [/itex]"
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