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Integration concepts

  1. Jan 13, 2007 #1
    I am a beginner on integration and I am stuck on some concepts. I hope that someone can explain, thank you so much!:smile:

    1) "Let f be continuous on [a,b].

    Take a partition P={xo,x1,...,xn} of [a,b].

    Then P breaks up [a,b] into n subintervals [xo,x1],[x1,x2],...[x(n-1),xn] of lengths delta x1, delta x2,... delta xn.

    Now pick a point x1* from [xo,x1] and form the product f(x1*)delta x1; pick a point x2* from [x1,x2], and form the product f(x2*)delta x2 and so on, we define the Riemann sum as S*(P)=f(x1*)delta x1 + f(x2*)delta x2 + ... + f(xn*)delta xn

    We define ||P|| as the norm of P, by setting ||P||=max delta xi, where i=1,2,...,n.

    (int_a^b f(x)dx is read the definite integral of f from a to b)
    lim S*(P) = int_a^b f(x)dx "

    I don't understand why the limit of Riemann sum as ||P||->0 will be equal to the definite integral......

    Say if I have delta x1,delta x2,...delta xn, with delta x1 being the largest, now if I only take ||P||->0, i.e. delta x1 ->0, all the other delta x2,...delta xn won't will be affected, then how is it possible that this is the definite integral?

    Graphically, Riemann sum can be pictured as areas of rectangles under the curve, and now I only make the WIDEST rectangle narrower (||P||->0) while keeping all the other rectangles the same widths, now clearly the area of all rectangles & the area under the curve are unequal, then how can it be the definite integral?

    2) Definite integral from a to b is int_a^b f(x)dx

    What is the purpose of writing the "dx" at the end of the integral? Is it possible to have something like int_a^b f(x)dy, or int_a^b f(x)dz?

    3) "A function f is integrable on an interval [a,b] if:
    1. f is increasing on [a,b], or f is decreasing on [a,b]
    2. f is bounded on [a,b] and has at most a finite number of discontinuites."
    There is no key word in my textbook like "and", "or" between 1. and 2., so I am not sure whether both conditions need to be satisfied at the same time or is either one satisfied will be ok??
    Last edited: Jan 13, 2007
  2. jcsd
  3. Jan 14, 2007 #2


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    1) When you take the limit as ||P|| -> 0, what is implied is that it remains the largest sub-interval. Since it is the largest sub-interval, then as its size tends to zero, so must the sizes of the rest of the sub-intervals, and it is in this sense that the Riemann sum limits to the integral.

    2) The dx indicates that you are integrating with respect to x. If you had [itex]\int_a^bf(x)dy[/itex], then you're integrating a function of x with respect to y; since f(x) is independent of the variable y, it is essentially a constant with regards to integration, and the result of such a definite integral would be [itex]f(x)(b-a)[/itex].

    3) I think that perhaps you only need one statement to be true, as I would think a constant function doesn't satisfy statement 1, but is integrable (and of course it satisfies statement 2). Otherwise I don't quite get what statement 1 means by "increasing" or "decreasing".
    Last edited: Jan 14, 2007
  4. Jan 14, 2007 #3
    1) Oh, I see! It makes much more sense if it REMAINS the largest sub-interval. I origianlly thought that it doesn't need to remain the largest

    3) Are you sure about this one? Because from the exercises after this section in my textbook, it asks me to prove something like "if f is bounded on [a,b] and has at most a finite number of discontinuites, then a function f is integrable on an interval [a,b]", this doesn't include condition 1. at the same time, which means 2. alone means f is integrable?

    On the other hand, does f is increasing on [a,b], or f is decreasing on [a,b] ALONE => a function f is integrable on an interval [a,b]?

    Thanks for your help!
  5. Jan 14, 2007 #4

    Gib Z

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    As for your question 2, you could choose to think of it this way. Integration is fancy addition. When we are finding the area under a curve, think of dx as a really really small change in x. So, the area of a small strip, a rectangle, is the height, f(x), times the width, dx, which is f(x) dx. To add up all the tiny tiny strips, we integrate it.
  6. Jan 14, 2007 #5

    matt grime

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    For 3)

    If either condition 1 OR 2 is met, then the function is integrable, or one of many other functions. But just because a function is integrable, it does not mean that either of them is met.

    The proofs of the statements should have made the answer clear.
  7. Jan 14, 2007 #6


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    question 1 i fiund quite fascinating confusion of logic.

    the key as may be explained above, is that you cannot set ||P|| equal to th original deltax1, and then make deltax1 smaller independently of the other deltax's.

    rather ||P|| is a function on partitions P. to evaluate ||P|| on a partition you always have to reconsider all the deltax's and take the largest.

    so if you change the partition until deltax1 is no longer largest then ||P|| cannot anylonger be deltax1.
  8. Jan 14, 2007 #7


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    as to question 2) the f(x)dx tells you what two functions were multiplied togetehr to form the riemann sums.

    i.e. the integral of f(x)dsin(x), would mean you had multiplied f(xi*) bythe difference [sin(xi) - sin(xi-1)] in the riemann sum.

    this lest you also integrate over paths in the plane, where then integrating dx would mean multiply by the difference in x coords of the two endpoints of the subinterval, while dy would mean use instead the diffrerence of their y coords.

    a fucntional answer is also that, when you get ready to learn substitution as a method of antidifrentiation, you will be very glad you are using the d as an aid!
  9. Jan 14, 2007 #8


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    for question 3, as stated above by matt, either of the conditions is sufficient for integrability, and neither is necessary.

    you can easily prove as an exercise that an increasing function is integrable if you try for a bit.

    your book seems better than most by the way. I usually give these criteria (the first of which is apparently due to newton, and thus predates riemanns definition of the integral) but they are seldom found in the books we use.
  10. Jan 14, 2007 #9
    Thanks everyone for explaning!

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