Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

From indefinite integral to fundamental theorem of calculus, how does it all link?

  1. Dec 4, 2004 #1
    Hi there,

    Can someone explain to me what the following are and how each one is used as a tool for the next one:

    1)Indefinite integral
    2)Riemann Sum
    3)Definite Integral
    4)Fundamental Theorem of Calculus(The part which says that the derivative of the integral of f(t)dt from a to x is equal to the value of the function at x, or f(x).

    5)Finally, the part of the fundamental theorem that says the integral of a function form a to b is F(b)-F(a)

    I just need some help with the general link between all of the above 5 and how they relate to each other. The mathematics is meaningless without the concept of each.

    Thank you,
    haribol
     
  2. jcsd
  3. Dec 4, 2004 #2

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    1) The "indefinite integral" is just a fancy name for an anti-derivative of a function.
    2+3) Regard, as for present, these as tightly related concepts:

    Suppose you've got a function f(x) defined on an interval [a,b] and you want to find the area beneath its graph.
    You can approximate this area in two distinct ways:
    a)The upper Riemann sum Ru(N) (with N terms)
    Partition your interval into N sub-intervals.
    Ru(N) is piecewise constant on each interval, and equals there the maximum value/least upper bound of the function on that sub-interval.
    Clearly, the area under f(x) must be less than or equal to the area under Ru(N) (an area which CAN be computed).

    Now, REFINE your partioning, say into 2N intervals, in such a manner that each new sub-interval is completely included in one of the sub-intervals from your previous step.
    Thus, we have that (Area under) Ru(2N)<=(Area under) R(N) (okay with that?)
    But still, the area under f(x) must be less than or equal to the area computed from Ru(2N).

    b) The lower Riemann sum:
    On the same partition as we made for the upper Riemann sum(s), we may define the lower Riemann sum Rl(N), using the minimum values of f(x) on the sub-intervals as our constants.
    Clearly, the refining procedure yields (Area under) Rl(2N)>=(Area under) Rl(N)
    yet the area under f(x) must be greater or equal to the area computed under the lower Riemann sum.

    If now the sequences of upper and lower Riemann sums converges TO THE SAME NUMBER as we let the number of sub-intervals go to infinity, we DEFINE THE DEFINITE INTEGRAL AS THAT NUMBER, and calls it the AREA under f(x) (if, by some perversity, the upper and lower Riemann sums DON'T converge to the same number, we say that f(x) is NOT Riemann-integrable).

    4+5) Calculating an infinite number of terms in order to find a single, puny number is a horrendous bore.
    The fundamental theorem of calculus says we can spare ourselves of that burden, in that THE AREA of f(x) (i.e, the definite integral) over the interval [a,b] can be computed by simply calculating F(b)-F(a), where F is an anti-derivative of f.
     
    Last edited: Dec 4, 2004
  4. Dec 5, 2004 #3
    Thanks for replying arildno,

    My textbook says that as the "norm", which is the largest interval amongst the subintervals, approaches zero of a Riemann sum, then that is a definite integral. I do not understand how the norm approaching zero would affect the other subintervals. Can you please elaborate on this "norm" idea?
     
  5. Dec 5, 2004 #4

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I've not learnt this with the "norm"-concept, but its pretty clear what they mean by it:

    Since the "norm", by definition is the LARGEST sub-interval with respect to a given finite Riemann sum, if the "norm" tends to zero (that is, when we look at the SEQUENCE of Riemann sums) then necessarily the other sub-intervals must also go to zero, since they are, for a particular Riemann sum always less than, or equal to the length of the norm.

    I'm not too sure you've understood correctly what the "definite integral" is, though:
    If the sequence of lower Riemann sums (and we can show that this sequence IS convergent) converges to the SAME NUMBER as the upper Riemann sums converges to, THAT NUMBER is DEFINED as the definite integral (in the Riemann theory of integration).
     
  6. Dec 5, 2004 #5
    Hey,

    So when you say upper and lower Riemann Sums, do you mean taking Riemann Sums with the rectangles inside the graph(lower) and taking Riemann Sums of rectangles outside the graph (upper)?

    If that is so, then I think I understand definite integral as a specific value, a number. Now how about if you need an integral from a lower limit to a variable upper limit?
     
  7. Dec 5, 2004 #6

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You have understood what I meant with the Riemann sums

    Are you thinking of the following example:
    [tex]G(x)=\int_{a}^{x}f(t)dt[/tex]

    If so, think of G's value at a given "x" as being defined as the definite integral of f over the interval "a" to that particular "x".

    The fundamental theorem of calculus PROVES that G(x) (considered as a function over x's) is an anti-derivative to f(x).
     
    Last edited: Dec 5, 2004
  8. Dec 5, 2004 #7
    Hi again,

    Yes thats exactly the function. So far what I've understood about the fundamental theorem of calculus is something like this. Say you have a curve in interval [a,b] and x is somewhere in between. So,

    The area from "a" to a value "x" is given by [tex]G(x)=\int_{a}^{x}f(t)dt[/tex]

    Assuming h is some small increment, the area from "a" to "x+h" is given by:

    [tex]G(x+h)=\int_{a}^{x+h}f(t)dt[/tex]

    So [tex]G(x+h)-G(x)[/tex] will give me the area under the curve from "x" to "x+h".
    Now as h becomes smaller and smaller, the rectangle becomes thinner and thinner and almost with only one dimension, which is the value of f at x or f(x). So therefore we get G'(x)=f(x).

    Is this what they mean? Or is there some other way I should think of this theorem?
     
  9. Dec 5, 2004 #8

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yeah, that's basically it.
    I'll add a few details on this:
    1) You should regard the FUNCTION VALUE of G at "x" as the definite integral over f from [a,x].
    That is, the function G maps the x's into a set of definite integrals over f.
    Both G(x+h) and G(x) are therefore to be understood as definite integrals.

    2) On basis of the fact that Riemann sums are additive, we can reach the conclusion that the definite integral must be additive as well:
    [tex]\int_{a}^{c}f(t)dt=\int_{a}^{b}f(t)dt+\int_{b}^{c}f(t)dt[/tex]

    This provable feature of the definite integral is crucial in the proof of the fundamental theorem of calculus.

    3) On basis of the property of additivity of integrals, we can note at first the following property of a G-function.
    Suppose we have defined G(x) as:
    [tex]G(x)=\int_{0}^{x}f(t)dt[/tex]
    (using zero as my lower limit is for illustration purposes only)
    Then, we have, for example two function evaluations at "a" and "b" respectively:
    [tex]G(b)=\int_{0}^{b}f(t)dt[/tex]
    [tex]G(a)=\int_{0}^{a}f(t)dt[/tex]
    Then, by the additivite property of the definite integral, we have:
    [tex]G(b)-G(a)=\int_{a}^{b}f(t)dt[/tex]
    That is, the area under f(x) over the interval [a,b] is seen to be calculable from an APPARENTLY simple difference of G's values at "a" and "b".

    Make note of the word "APPARENTLY" here!
    We haven't, as yet, found any way to calculate the function values of G(x) aside from computing the limit of a infinite sequence of Riemann sums!!

    4) This is PRECISELY where the fundamental theorem of calculus comes to our aid.
    Armed with that (and a few details), we find a COMPLETELY DIFFERENT WAY of calculating the values of G(b)-G(a):
    By SHOWING that G(x) is an ANTI-DERIVATIVE of f(x), tFToC gives us the tool to compute the area under f through a comparably ridiculously simple procedure.

    With a few details fixed, the theorem says to us:
    Pick any antiderivative of f(x).
    Evaluate the difference between your anti-derivative's values at "b" and "a".
    That number is the definite integral of f(x) on the interval [a,b]
     
    Last edited: Dec 5, 2004
  10. Dec 5, 2004 #9
    Thanks for the nice explanation arildno,

    You know when you have something like

    [tex]\int_{a}^{b}f(t)dt[/tex]

    is "f(t)dt" a first order differential?
     
  11. Dec 5, 2004 #10

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I would STRONGLY advise you at present not to worry about "differentials"!
    If you want to, anyway, you should regard f(t)dt as the infinitesemal area of a column of width "dt" (centered about "t") and height "f(t)"
     
  12. Dec 5, 2004 #11
    Is it alright to say that,

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


    means the sum of all infinitesemal area of a column of width "dt" and height "f(t)" between the interval [a,b] of the area function G(x)?

    Also, when we say G'(x), do we mean the rate at which the area under the curve f(t) changes as x changes?
     
  13. Dec 5, 2004 #12

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Yes, the whole area can be seen as the sum of all the tiny areas.
    Yes.
     
  14. Dec 5, 2004 #13
    Ok, so when we say the rate at which the area changes, does that mean the rate at which the height of each of infinite infinitesemal rectangles which contribute to the total area. If that is so, then the height of each of those rectangles depends on the rate of change of the function f with respect to time. Does it make sense, what I've just said?

    Thanks for your help and time arildno
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: From indefinite integral to fundamental theorem of calculus, how does it all link?
Loading...