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Definite Integral and Summation Equivalence

  1. Jul 13, 2009 #1
    Can someone give me an explanation or possibly a proof that [tex]\int^{a}_{b}f(x)dx= \displaystyle\lim_{m\to\infty}\sum^{m}_{k=1}f(x^{*}_{k})\Delta x[/tex]
     
  2. jcsd
  3. Jul 13, 2009 #2

    HallsofIvy

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    That right hand side is shorthand for some very complicated calculations. It is not just "as m goes to infinity". As the number of points, [itex]x_k^*[/itex] increases, the "partition" of the interval from a to b changes and "[math]\Delta x[/math]" also changes- in fact it might be different for every sub-interval.

    Surely any Calculus text will devote several pages to that proof and can do a much better job that we can do here.
     
  4. Jul 13, 2009 #3
    Oh. I suppose I should have put "as n approaches infinity", given the relationships between n and m, and n and (change in x).

    I am just trying to intuitively understand how the inverse of the derivative differences (definite integral) of a function is equal to the area bounded by the limits of integration a and b, the x-axis, and the curve of the function.
     
  5. Jul 13, 2009 #4
    It's easier to see the relationship between the integral and the derivative if you develop integral calculus before differential calculus (this is also historically the way it was done).


    Here is how we define the integral (taken from Richard Courant's 'Differential and Integral Calculus'):

    ---------------------------------------------------------
    Among the limiting processes of analysis, there are two processes with an especially important role, not only because they arise in many different connections, but chiefly due to their very close reciprocal relationship. Isolated examples of these two limiting processes, differentiation and integration, have even been considered in classical times; but it is the recognition of their complementary nature and the resulting development of a new and methodical mathematical procedure which marks the beginning of the real systematic differential and integral calculus. The credit of initiating this development belongs equally to the two great geniuses of the Seventeenth Century, Newton and Leibnitz, who, as we know to-day, made their discoveries independently of each other. While Newton, in his investigations, may have succeeded in stating his concepts more clearly, Leibnitz's notation and methods of calculation were more highly developed; even to-day, these formal portions of Leibnitz's work form an indispensable element in the theory.

    2.1 The Definite integral

    We first encounter the integral in the problem of measuring the area of a plane region bounded by curved lines. Then, more refined considerations permit us to separate the notion of integral from the naïve intuitive idea of area and to express it analytically in terms of the notion of number only. We shall find this analytical definition of the integral to be of great significance not only because it alone enables us to attain complete clarity in our concepts, but also because its applications extend far beyond the calculation of areas. We shall begin by considering the matter intuitively.

    2.1.1 The integral as an Area: Let there be are given a function f(x), which is continuous and positive in an interval, and two values a and b (a < b) in that interval. We think of the function as being represented by a curve and consider the area of the region which is bounded above by the curve, at the sides by the straight lines x = a and x = b and below by the portion of the x-axis between the points a and b (Fig. 1). That there is a definite meaning to speaking of the area of this region is an assumption inspired by intuition, which we state here

    http://kr.cs.ait.ac.th/~radok/math/mat6/g1.jpg [Broken]

    expressly as a hypothesis. We call this area [tex] F_a^b [/tex] the definite integral of the function f(x) between the limits a and b. When we actually seek to assign a numerical value to this area, we find that we are, in general, unable to measure areas with curved boundaries, but we can measure polygons with straight sides by dividing them into rectangles and triangles. Such a sub-division of our area is usually impossible. It is, however, only a short step to conceive in the following manner the area as the limiting value of a sum of areas of rectangles. We subdivide the part of the x-axis between a and b into n equal parts and erect at each point of sub-division the ordinate up to the curve; the area is thus divided into n strips. We can no more calculate the area of such strips than we could that of the original surface; but if, as shown in Fig. 2, we find first the least and then the greatest value of the function f(x) in each sub-interval and then replace the corresponding strip (1) by a rectangle with height equal to the least value of the function, and (2) by a rectangle with height equal to the greatest value of the function, we obtain two step-shaped figures. (In Fig. 2 above, the first of these is drawn with a solid line, the second with a broken line.) The first step-shaped figure obviously has an area which is at most equal to the area [tex] F_a^b [/tex] which we are trying to determine; the second has an area which is at least as large as Fab. If we denote the sum of the areas of the first set of rectangles by [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g2.jpg [Broken] [Broken] (lower sum) and the sum of the areas of the second set by [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g3.jpg [Broken] [Broken] (upper sum), we find

    http://kr.cs.ait.ac.th/~radok/math/mat6/g4.jpg [Broken]

    If we now make the subdivision finer and finer, i.e., let n increase without limit, intuition tells us that the quantities [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g3.jpg [Broken] [Broken] and [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g2.jpg [Broken] [Broken] approach closer and closer to each other and tend to the same limit
    [tex] F_a^b [/tex]. We may therefore consider our integral as the limiting value

    http://kr.cs.ait.ac.th/~radok/math/mat6/g5.jpg [Broken]

    Intuition also tells us the possibility of an immediate generalization. It is by no means necessary that the n sub-intervals should all be of the same length. On the contrary, they may have different lengths provided only that, as n increases, the length of the longest sub-interval tends to 0.


    2.1.2 The Analytical Definition of the Integral: In the above section, we have considered the definite integral as a number given by an area, hence to a certain extent as previously known, and have subsequently represented it as a limiting value. We shall now reverse this procedure. We no longer take the point of view that we know by intuition how an area can be assigned to the region under a continuous curve or, indeed, that this is possible; we shall, on the contrary, begin with sums formed in a purely analytical way, like the upper and lower sums defined previously, and shall then prove that these sums tend to a definite limit. We take this limiting value as the definition of the integral and of the area. We are naturally led to adopt the formal symbols which have been used in the integral calculus since Leibnitz's time.

    Let f(x) be a function which is positive and continuous in the interval [a,b] (of length b - a). We think of the interval as being sub-divided by (n—1) points [tex] x_1, x_2, ··· , x_{n-1} [/tex] into n equal or unequal sub-intervals and, in addition, we let [tex] x_0 = a, x_n = b [/tex]. In each interval, we choose a perfectly arbitrary point, which may be within the interval or at either end; suppose that in the first interval we choose the point [tex] x_1 [/tex] , in the second one the point [tex] x_2, ··· [/tex] and in the n-th interval the point [tex] x_n [/tex]. Instead of the continuous function f(x), we now consider a discontinuous function (step-function) which has the constant value [tex] f(x_1) [/tex] in the first sub-interval, the constant value [tex] f(x_2) [/tex] in the second sub-interval, ··· , the constant value [tex] f(x_n) [/tex] in the n-th sub-interval. As is shown in Fig. 3, the graph of


    http://kr.cs.ait.ac.th/~radok/math/mat6/g6.jpg [Broken]

    this step-function defines a series of rectangles, the sum of the areas of which is given by

    http://kr.cs.ait.ac.th/~radok/math/mat6/g7.jpg [Broken]

    This expression is usually abbreviated by means of the summation sign:

    http://kr.cs.ait.ac.th/~radok/math/mat6/g8.jpg [Broken]

    by introducing the symbol

    http://kr.cs.ait.ac.th/~radok/math/mat6/g9.jpg [Broken]

    we can simplify this formula to

    http://kr.cs.ait.ac.th/~radok/math/mat6/g10.jpg [Broken]

    Our basic assertion may now be stated as follows:

    If we let the number of points of sub-division increase without limit and at the same time let the length of the longest sub-interval tend to 0, then the above sum tends to a limit. This limit is independent of the particular manner in which the points of division [tex] x_1, x_2, ··· , x_{n-1} [/tex] and the intermediate point [tex] x_1, x_2, ··· , x_{n-1} [/tex] are. chosen.

    We shall call this limiting value the definite integral of the function f(x), the integrand, between the limits a and b; as we have already mentioned, we shall consider it as the definition* of the area under the curve y = f(x) for x in the interval [a,b]. Our basic assertion may then be re-worded: If f(x) is continuous in [a,b], its definite integral between the limits a and b exists.

    * Of course, we may also define the notion of are in a purely geometrical way and then prove that such a definition is equivalent to the above limit=definition *

    This theorem on the existence of the definite integral of a continuous function can be proved by purely analytical methods and without appeal to intuition. We shall nevertheless pass it over for the present and return to it later, after the use of the concept of integral has stimulated the reader's interest in constructing for it a firm foundation. For the moment, we shall content ourselves with the fact that the intuitive considerations above have made the theorem appear to be extremely plausible.

    The above definition of the integral as the limit of a sum led Leibnitz to express the integral by the symbol http://kr.cs.ait.ac.th/~radok/math/mat6/g11.jpg [Broken].



    ---------------------------------------------------------


    Now that you know the definition of a definite integral, we can move towards showing that differentiation is the inverse operation of differentiation. Without proof, this is the mean-value theorem of integral calculus:

    If f(x) is continuous in the interval [a,b], then [tex] \int_a^b f(x) dx = (b-a)f(\xi) [/tex] where [tex] \xi [/tex] is some intermediate value in the interval [a,b].

    Now we define the indefinite integral:

    [tex] \phi(x) = \int_a^x f(u) du [/tex] where x is any point in the domain of f.

    The following is taken from Richard Courant's Introduction to Calculus and Analysis (virtually the same book as his Differential and Integral Calculus):

    -------------------------------------------------------

    Statement of the Fundamental Theorem of Calculus (Part One):

    The indefinite integral [tex] \phi(x) [/tex] of a continuous function f(x) always possesses a derivative [tex] \phi'(x) [/tex], and moreover [tex] \phi'(x) = \phi(x) [/tex].

    That is, differentiation of the indefinite integral of a continuous function always reproduces the integrand

    [tex] \frac{d}{dx} \int_a^x f(u) du = f(x) [/tex].

    This inverse character of the operations of differentiation and integration is the basic fact of calculus. The proof is an immediate consequence of the mean value theorem of integral calculus. According to that theorem we have for any values x and x + h of the domain of f

    [tex] \phi(x+h) - \phi(x) = \int_x^{x+h} f(u) du = h f(\xi) [/tex]

    where [tex] \xi [/tex] is some value in the interval with end points x and x + h. For h tending to zero the value [tex] \xi [/tex] must tend to x, because [tex] x < \xi < x+h [/tex].

    So,

    [tex] \lim_{h \rightarrow 0} \frac{ \phi(x+h) - \phi(x)}{h} = \lim_{h \rightarrow 0} f(\xi) = f(x) [/tex]

    since f is continuous. Hence [tex] \phi'(x) = f(x) [/tex] as stated by the theorem.

    ------------------------
    We see now from the Fundamental Theorem that an indefinite integral F(x) is the same as a primitive of continuous function f(x) or anti-derivative of f(x). (F(x) is a primitive of f(x) if F'(x) = f(x) ). Notice that we say 'an' indefinite integral, and not 'the' definite integral, because the derivative of F(x) is equal to the derivative of F(x) + c, for a real number c (since the derivative of a constant is 0), which is equal to f(x). Thus there are an infinite amount of anti-derivatives for an integrable function.

    The following is taken from Courant's Differential and Integral calculus:

    ----------------------------------------


    Suppose that we know any one primitive function http://kr.cs.ait.ac.th/~radok/math/mat6/g175.jpg [Broken] for the function f(x) and that we wish to evaluate the definite integral http://kr.cs.ait.ac.th/~radok/math/mat6/g177.jpg [Broken]. We know that the indefinite integral

    [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g179.jpg [Broken] [Broken]

    being also a primitive of f(x), can only differ from F(x) by an additive constant. Consequently,

    http://kr.cs.ait.ac.th/~radok/math/mat6/g180.jpg [Broken]

    and the additive constant c is at once determined, if we recollect that the indefinite integral
    [PLAIN]http://kr.cs.ait.ac.th/~radok/math/mat6/g179.jpg [Broken] [Broken], must take the value 0 when x = a. We thus obtain

    http://kr.cs.ait.ac.th/~radok/math/mat6/g181.jpg [Broken]

    whence c = - F(a) and F(x)=F(x) - F(a). In particular, we have for the value x=b

    http://kr.cs.ait.ac.th/~radok/math/mat6/g182.jpg [Broken]

    which yields the important rule:

    If F(x) is any primitive of the function f(x), the definite integral of f(x) between the limits a and b is equal to the difference F(b) - F(a).

    If we use the relation F '(x) = f(x), this may be written in the form

    http://kr.cs.ait.ac.th/~radok/math/mat6/g183.jpg [Broken]

    ----------------------------------------------
     
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