Working Towards Riemann Sums

In three previous posts (Nov 30, Dec 3 and Dec 5, 2012) we considered some examples leading up to integration and Riemann sums. Graphically all three can be seen as finding the area of the region between a graph and the x-axis over an interval.  The next thing to do is to abstract that process and see if we can do this in general for any continuous function.

Most all of the textbooks do this in the same way and that’s probably what you should now lead your students through. Start with the simple basic case of a function that is positive, monotone and continuous on an interval [a, b]. You wish to approximate the area of the region between the graph and the x-axis with vertical sides at x = a and x = b. (The actual minimum conditions are only that the function be bounded and continuous at all but a finite number of points. But that can wait.)

Follow the steps in your textbook.

The things you should emphasize first with numbers and then eventually with symbols are these.

  1. The interval is divided into n sub-intervals each of equal length \Delta x=\frac{b-a}{n}.
  2. The x-coordinates of the endpoints are written in terms of a and \Delta x. This is called a partition of the interval. The x-coordinates are a={{x}_{0}}, {{x}_{1}}=a+(1)\Delta x, {{x}_{2}}=a+(2)\Delta x, {{x}_{3}}=a+(3)\Delta x up to {{x}_{n}}=a+(n)\Delta x=b.
  3. Decide on a scheme to evaluate the function a one point in each of the sub-intervals defined by the partition. This is usually done at the left or right endpoint, but in fact may be anywhere in the sub-interval. The notation for this “any point” is f\left( x_{i}^{*} \right), which may be a little complicated. If you decide on using the right end then for the ith sub-interval [{{x}_{i-1}},{{x}_{i}}] the value is f\left( {{x}_{i}} \right), for the left side the value is f\left( {{x}_{i-1}} \right). This is the vertical distance between the graph and the x-axis.
  4. Multiply this by the width of the sub-interval, this gives f\left( {{x}_{i}} \right)\Delta x. Do this for each sub-interval and add the results to get your approximation. For the right side approximation this looks like \sum\limits_{i=1}^{n}{f\left( {{x}_{i}} \right)\Delta x}.

Have your students compute this number for several functions with n in the 3 – 6 range. Your book will have plenty of examples. The idea is to learn and understand the procedure.

Some comments about each step above in case the kids ask. (Hope that they do.) Referring to the numbers above:

  1. The sub-intervals do not have to be the same length. When we get to taking limits of the sum above, the actual requirement is that whatever way we partition the interval the largest sub-interval (called the norm of the partition) approaches zero in length as n increases. For actually doing the computation, equal length sub-intervals are easiest.
  2. There will be a different \Delta x for each interval: \Delta {{x}_{i}}={{x}_{i}}-{{x}_{i-1}}.
  3. You may pick any point in a sub-interval, and you do not even have to pick the same way in each sub-interval. For computation, the right side is usually the easiest, with the left side not far behind. For the theory involved, picking the largest and/or smallest value in each sub-interval is used. This value may not be at an endpoint.
  4. Regardless of any of the above you will still multiply the function value by the width of the interval. The notation is more complicated. The sum is now written \sum\limits_{i=1}^{n}{f\left( x_{i}^{*} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)}

The important thing here is not the notation, but the process of partitioning the interval, calculating the function values multiplied by the width of the interval, and adding these products.

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