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Fundamental Theorem/Area

  1. Jul 14, 2009 #1
    So, the area under a curve is determined by F(b) - F(a), how does that take into account the changing area between the two extremes, a and b? I mean, since F(b) and F(a) are specific values at the ends of the graph, how is the area in between figured into it, especially if the graph is really wild instead of smooth like, say, x2?
  2. jcsd
  3. Jul 14, 2009 #2
    You put another formula inside the integral, instead of the smooth one, and integrate?

    It does not guess the curve between a and b. You compute the general form of the area formula by integrating, then you compute the area up to point 'b'. Also you compute the area up to point 'a'. Then if you subtract from the first one (F(b)-F(a)!!), you get the difference of areas. I think I am still amazed by it.

    Please keep in mind that integration goes way beyond what we discuss here... But still so beautiful.
  4. Jul 15, 2009 #3
    The function F here is not the function you are integrating, call it f. The integral computes the area formula for the function f, which is then determined over an interval by the values of F at the endpoints. http://books.google.com/books?id=_k...+of+calculus+courant&ei=yG1dSqyAHo3OMomhvY0H" is a good explanation.
    Last edited by a moderator: Apr 24, 2017
  5. Jul 15, 2009 #4
    Thanks for the replies and the link. I'm reading that now, so the answer may be in that. But I will give a simple example. Let's say I have f(x) = x2. Now, I want to know the area under the curve from 0 to 1.

    The way I understand it, I take the integral [tex]\int_{0}^{1} x^2dx[/tex]. So the antiderivative of x2 is [tex]\frac{1}{3}x^3[/tex]. Evaluating at 1 gives me [tex]\frac{1}{3}[/tex] and at 0 gives me 0, so my area is [tex]\frac{1}{3}[/tex]

    But I only computed values for two locations on the graph, and subtracted one from the other. How is the changing graph between the two points taken into account? The graph rises rapidly from 0 to 1, yet I only evaluated those two points. How does the integral "know" about the rest?
  6. Jul 15, 2009 #5


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    Because finding [itex]F(x)= \int_a^x f(t)dt[/itex] requires all the values between a and b!

    What definition of [itex]\int_a^x f(t)dt[/itex] did you learn? One commonly used definition is that it is the area of the region bounded by y= 0, y= f(x), x= a, and the variable vertical line at x. Another is, of course, an "anti-derivative"- the function whose derivative is f(x) and whose value at x= a is 0. It was being able to prove that those two things are the same, the "Fundamental Theorem" you refer to, That made Newton and Leibniz the "founders" of Calculus.
  7. Jul 16, 2009 #6
    You can think about it (very informally) this way. Say you want to find the area under f from a to b. You would take the "average height" of f between a and b and multiply that by (b-a). Imagine any function F whose slope at x is equal to f(x) for any x between a and b (that is, F'(x) = f(x)). Then the "average slope" of F between a and b is equal to the slope of the line drawn between the points (a,F(a)) and (b,F(b)), which is [F(b)-F(a)]/(b-a) (just take rise over run). Since f measures the slope of F, the "average slope" of F would be the "average height" of f. Thus we have area = ("average height" of f)*(b-a) = ("average slope" of F)*(b-a) = {[F(b)-F(a)]/(b-a)}*(b-a) = F(b)-F(a). You can see that you must follow the height of f all the way from a to b in order to know the slope of F from a to b and thus calculate the "average slope" of F from a to b.
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