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I'm currently using a textbook that covers Calculus and Analytic Geometry, however, the author seems to have confused me with notations and definitions. He first proves that the area under the curve within the interval $[a,b]$ can be calculated by allowing $n$ to approach infinity in

$\Sigma_{k = 1}^ n f(c_k) \Delta x$, ... (1)

where $c_k$ is the minimum of the function at the $k$th interval, and $\Delta x$ is the length of each interval (so that $\Delta x = \frac{b – a}{n}$).

He first defines (1) to be the "Area under the graph". Later in the text, he calls (1) specifically the "lower bound", because it approaches the actual graph from below. And so $\Sigma_{k = 1}^ n f(c_k) \Delta x$, where $c_k$ is the maximum of the function at the $k$th interval is called the "upper bound", because it approaches the actual graph from below. A question arises for me: although the upper bound and lower bound are equal as $n$ approaches infinity, is the notion of "Area under the graph" specifically defined as the lower bound? However, this question is only subsidiary to my next. Later, the author begins considering arbitrary intervals instead of fixed ones. He defines the "norm" to be the length of the greatest interval. He proves that $\Sigma_{k = 1}^ n f(c_k) \Delta x_k$, (where $c_k$ is a randomly chosen point in the interval $\Delta x_k$) is equal to the upper and lower bound as the norm approaches $0$. He calls this expression the Riemann Integral (named adter Georg Friedrich Bernhard Riemann), and denotes it by $\int_a^b f(x) dx$. He then says that the area we defined earlier in (1) is the Riemann Integral over the interval $[a,b]$. He then says that the Integral $\int_a^b f(x) dx$ is called the Definite Integral.

I was completely thrown off. So are the "area under the graph", the Riemann Integral, the Definite Integral all identical concepts? Is there a difference between the terminologies? I seek thorough guidance. Thank you in advcalance.

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  • $\begingroup$ The author seems to give a roundabout approach to define the definite integral. All definitions are the same (until you get to measure theory). $\endgroup$ Commented Jul 2, 2023 at 21:47
  • $\begingroup$ Are you sure? Because I thought perhaps they differ in the technique they use to find the area under the graph? $\endgroup$
    – Camelot823
    Commented Jul 2, 2023 at 21:51
  • $\begingroup$ "although the upper bound and lower bound are equal as $n$ approaches infinity". This is false in general. There exists functions for which they are different. $\endgroup$
    – jjagmath
    Commented Jul 2, 2023 at 21:52
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    $\begingroup$ Although the approaches are different, they give the same result for any integrable function. $\endgroup$ Commented Jul 2, 2023 at 22:17
  • $\begingroup$ @herbsteinberg it would be more than wonderful if you could tell me which approach the Riemann integral refers to, and which approach the definite integral refers to $\endgroup$
    – Camelot823
    Commented Jul 3, 2023 at 1:16

1 Answer 1

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In a very vague sense, the definite integral $\int_a^b f(x)\ dx$ can be "defined" as being the area under the graph of $y = f(x)$ between $x = a$ and $x = b$.

The Riemann integral is one method of deriving a value for the definite integral. We take a partition of the interval $[a, b]$ as the points $\{x_0 = a, x_1, \ldots, x_n = b\}$ and a selection of one point in each segment $t_i \in [x_i, x_{i + 1}]$ and calculate the sum

$$\sum_{i = 0}^{n - 1} f(t_i) (x_{i + 1} - x_i)$$

and this is the Riemann sum of $f$ on the given partition with the chosen sample of points.

Then, we take the limit of the Riemann sum as the partition gets arbitrarily fine, i.e. as $\max (x_{i+1} - x_i) \rightarrow 0$. If this limit is well-defined - i.e. if the limit converges to a single value regardless of how we take the partition and the sample points, then the value of that limit is what we define as the Riemann integral of the function on the interval $[a, b]$, and we say that $f$ is Riemann integrable on that interval.

This is a fairly horrible thing to work with in most situations, so we usually make some simplifications. One such simplification is to work with the lower and upper Darboux sums (which are often also called the lower and upper Riemann sums), where we look at the cases where $f(t_i)$ takes either its minimum or maximum value in each segment. If the lower and upper Darboux sums both converge to the same value as we refine the partition, then in fact they converge to the Riemann integral.

We can then prove a few basic results, including:

  1. If $f$ consists of straight line segments, then the Riemann integral of $f$ is consistent with calculating the area under $f$ geometrically as a combination of rectangles and triangles.

  2. If $f$ is continuous on $[a, b]$ then it is Riemann integrable on $[a, b]$.

  3. The Riemann integral follows various "intuitive" notions of area, for example if you split an interval up then the integral over the whole interval is equal to the sum of its parts, if you add two integrable functions together the integral is the sum of their components, if you scale a function vertically the integral scales similarly, and so on.

  4. If $f$ is sufficiently well-behaved on $[a, b]$ (and I won't go into the specifics of what that means here) and $F$ is a function on $[a, b]$ whose derivative $F'(x) = f(x)$, then $\int_a^b f(x)\ dx = F(b) - F(a)$ (this is the Fundamental Theorem of Calculus).

So in other words for many practical purposes the concepts of "definite integral", "Riemann integral", and "limit of the lower and upper Riemann sums" are indistinguishable. There are various pathological cases of functions that aren't Riemann integrable, and there are different integrals (such as the Lebesgue integral) that cover some of that territory while giving the same results as the Riemann integral in cases where it exists, but for your purposes it's easiest to not worry about those just yet.

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