Riemann Sum Example Problem

Riemann Sums and Area

The curve in red in the pictures below is the graph of y = x² + 1.

The goal is to approximate and then compute exactly the area between the x-axis and the graph of y = x² + 1 with x between 0 and 2. The region being described is shown at the right in red. Initially I will use four approximating rectangles. As demonstrated in class, I will do it three times. I will first use a left approximation, then a right approximation, and then a midpoint approximation.

Left

In the picture above the width of each rectangle will be 1/2 and the length (height) will be the value of the function evaluated at the left end of the subinterval corresponding to that rectangle. In this case the area of the region will be approximated by

Right

In the picture above the width of each rectangle will be 1/2 and the length (height) will be the value of the function evaluated at the right end of the subinterval corresponding to that rectangle. In this case the area of the region will be approximated by

The left approach would give a low estimate and the right approach would give a high estimate.

A pretty good estimate could be computed by taking the average of the left and right approximations.

A good estimate of the area of the region would appear to be the area approximation derived from the midpoint approach taken on the right. At least this "looks" pretty accurate with the "top" of each approximating rectangle partially above and partially below the red graph of the function f(x) = x² + 1.

Midpoint

In the picture above the width of each rectangle will be 1/2 and the length (height) will be the value of the function evaluated at the midpoint of the subinterval corresponding to that rectangle. In this case the area of the region will be approximated by

Let's try using more approximating rectangles and focus first on the right approximation.

Quicktime Animation: right Riemann sum

At this point we can certainly conclude that the area of the region cannot be more than the limit computed above.

It also cannot be less than than a limit we could compute relating to a left approximation method. Let's see what that would be.

Quicktime Animation: left Riemann sum

Just for fun let's look further at the midpoint method.

Quicktime Animation: midpoint Riemann sum

What if we do not have a summation formula that applies? For example, how could we approximate the area under the graph of y = sin(x) and above the x-axis with x between 0 and pi.

We will soon discover that the exact area of the region is 2 square units. Using the summation feature of a TI-89 and n = 1000 yields 1.99999835507. In the picture above on the right n = 10 and the area approximation is 1.983523538 using a TI-89 or Maple. The approximation for n = 2000 is 1.99999958876 (TI-89). The accuracy of these "right" approximations is increased by the fact that the function is increasing over the first half of the interval and decreasing over the second half of the interval. Thus some of the approximating rectangles are too large and some are too small and the errors tend to balance themselves. Quicktime animation using the midpoint method.

One More Example--Again Using a TI to Approximate the Sum. The approximated area for n = 10 is 16.26332364.

Here is a Maple worksheet for computing the sum above.

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Lane Vosbury, Mathematics, Seminole State College email: vosburyl@seminolestate.edu