Riemann Sums

In our last post we discussed what are called Riemann sums. A sum of the form \displaystyle \sum\limits_{i=1}^{n}{f\left( x_{i}^{*} \right)\left( {{x}_{i}}-{{x}_{i-1}} \right)} or the form \sum\limits_{i=1}^{n}{f\left( x_{i}^{*} \right)\Delta x} (with the meanings from the previous post) is called a Riemann sum.

The three most common are these and depend on where the x_{i}^{*} is chosen.

  • Left-Riemann sum, L, uses the left side of each sub-interval, so x_{i}^{*}={{x}_{n-1}}.
  • Right-Riemann sum, R, uses the right side of each sub-interval, so x_{i}^{*}={{x}_{i}}.
  • Midpoint-Riemann sum, M, uses the midpoint of each interval, so x_{i}^{*}=\tfrac{1}{2}\left( {{x}_{i-1}}+{{x}_{i}} \right).

For the AP Exams students should know these and be able to compute them. The actual values are often given in a table, so the long computation of the function values is not necessary.

Another way of approximating the area between the graph and the x-axis is to use trapezoids formed by joining the points at the ends of each sub-interval. The areas can be figured individually and added or the value, T, can be found by averaging the left- and right-Riemann sums, T=\tfrac{1}{2}\left( L+R \right). This trapezoid approximation is usually closer to the true value than the other left- or right sums. However, a trapezoid approximation is not a Riemann sum on the technicality that function values are not being used.

Whenever you are dealing with approximations, you should have some sense of how good they are. All of the approximations discussed will get closer to the true area if more values (more partition points) are used.

If the graph is increasing on the interval, then the left-sum is an underestimate of the actual value and the right-sum is an overestimate.  If the curve is decreasing then the right-sums are underestimates and the left-sums are underestimates. (To see why, draw a sketch.)

If the graph is concave up the trapezoid approximation is an overestimate and the midpoint is an underestimate. If the graph is concave down then trapezoids give an underestimate and the midpoint an overestimate. (To see how these work, draw a sketch. For the midpoint draw the tangent line at the midpoint to the sides of the sub-interval; this trapezoid has the same area as the rectangle drawn at the midpoint of the interval. Why?)

If the graphs are not monotone on the interval or change concavity then all bets are off.

For all of the Riemann sums, including those not mentioned above, as the number of partition points increase (n\to \infty ), or the width of the all the sub-interval decrease (\Delta x\to 0), the limit of a Riemann sum approaches the area between the graph and the x-axis. This will be the subject of the next post.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s