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Riemann Sum
Motivation: We start using different approach to evaluate the area under f(x)=x from x=0 and x=1 as shown below:
We all know the area of that triangle is ½ since long time ago. But this time, we are equipped with more tools. And we would like to develop a more general approach to apply on other situation. The interval [0,1] can be divided into n partitions equally. In each partition, it is corresponding to a slender strip with different height. It would be better if we can add up the area of those slender strips by taking n .
But here is a question: what should we use for the height for each strip?
As shown in the diagram, if we take the value of f(x) at point A, the area sum of those strips will be larger than the triangle area that we are going to find in this case. On the other hand, if we choose B, the area sum will be smaller than the triangle area. Anyway, we just denote the area sum by choosing the function value on the right as Rn , and Ln as the area sum by choosing the function value on the left. Thus,
Rn = Ln =
So, Rn = , Ln =
Hence,
Rn = ½ and Ln = ½
The limit does not contradict to what we previously know about the area of the triangle. In general, it is not necessary to partition the interval into equal pieces as long as those pieces are getting thinner and thinner. And there are other approaches other than using the left endpoints or right endpoints as the height in each partition, for example, by taking midpoint of the function in each small interval, or the maximum value or minimum value as the height in each small partition.
The process to sum up those small rectangles is known as Riemann Sum. Finding the limit of Riemann Sum to use it as the area under the curve is known as Riemann Integral.
With different approaches, they should converge to the same limit but with different rates. In other words, some methods will approach to the limit quicker than other methods under some conditions.
Definition ( Lower and Upper Riemann Sum ) Let the set of points x0 , x1, x2, ... , xn be a partition of the interval [a,b] such that
a x0 < x1 < x2 < ... < xn-1 < xn b
In each interval [xi-1, xi] , let mi be the minimum and Mi be the maximum of the function f(x) in the associated interval. ( Strictly speaking, infimum and supremum shall be used to replace minimum and maximum ). Then the Upper Riemann Sum is known as
and the Lower Riemann Sum is
Definition ( Riemann Integral ) If the upper Riemann sum and lower Riemann sum of a function f(x) on an interval [a,b] converge to the same limit, then the limit is defined as Riemann integral. It is denoted as
If the limit of upper Riemann sum is different from the limit of lower Riemann sum, then we say f(x) is not Riemann-integrable .
We might want to ask the criterion for f(x) to be Riemann integrable and what kind of functions are Riemann integrable. We first consider continuous functions. To prove a continuous function is Riemann integrable in a closed interval, It is necessary to introduce the concept “uniformly continuous” .
The definition of the continuity of a function f(x) at a given point x=c is : with a given >0, there exists > 0 such that
| x-c| < | f(x)-f(c)| <
“Uniformly continuous” is not particularly about one point; the idea behind it is: for any two points in the domain of f, if the distance of the two points is less than , then their maps under f will be less than . In other words, it is not dependent on the position of the point, but on the distance of the two points. We write down its definition below to have more clear view.
Definition ( Uniformly Continuous ) A function f(x) is said to be uniformly continuous in an interval I if
for any >0, there exists >0 such that | f(x) – f(y) | < whenever |x-y| < , x, y I
Thus, when you are doing partitioning of an closed interval, if each small interval is small enough and the function is uniformly continuous, we can easily prove that the upper Riemann sum and the lower Riemann sum will approach together as the partition is finer and finer. That is one of the reasons that people want to introduce the concept for “uniformly continuous”. It is obvious that every “uniformly continuous” function is continuous. And we know that the converse is not necessary true. However, the following theorem just brings your attention back to “continuous function” under some condition.
Theorem ( Heine-Cantor Theorem ) If f:[a,b] R is a continuous function, then f is uniformly continuous.
We just state the theorem without proof. The proof would be introduced in more advanced math. It involves another concept “compact set” . And in Rn , every closed set is a compact set.
Based on the result, we are going to prove the following theorem.
Theorem : Every continuous function f(x) on a closed interval [a,b] is Rieman-integrable.
Proof: Every continuous function on a closed interval is uniformly continuous. Thus, for every > 0 , there exists > 0 such that
|f(s)-f(t)| < when |s-t| < , s, t [a,b]
Let’s create a partition {x0 , x1, x2, ... , xn } of the interval [a,b] such that
a x0 < x1 < x2 < ... < xn-1 < xn b and | xi-1 – xi | < , i=1,2,3,4,... n
And let Mi and mi be the maximum and minimum of f(x) in [xi-1, xi] . ( The maximum and minimum exist for a continuous function in a closed interval. Thus, it makes sure that the existence of maximum and minimum in each small interval . )
Then, the lower Riemann sum is
L =
The upper Riemann sum is
U= Mi and mi happen in the interval [xi-1, xi ] | Mi – mi | <
| L – U | = | | < < =
That means: for every >0, if the partition is finer enough, we can have | L – U | <
So, the function f is Riemann-integrable.
Properties of Riemann Integral Let f and g be Riemann integrable functions defined on [a,b]. Then (1) = + for any constants c, d (2) = + if a< c < b (3) | |
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