Using the TI-85 or TI-86 graphing calculator, approximate the value of

using right-hand endpoints, left-hand endpoints, and midpoints with a regular partition of 100 points.

We follow the algorithm given on Riemann Sums:

  1. the function y = f(x) = x2 + 3x
  2. the interval [-1,3]
  3. the positive integer n = 100
we want to find Riemann Sums corresponding to
  1. left-hand endpoints
  2. right-hand endpoints
  3. midpoints
C1. We enter the function on the calculator:

The algorithm is
  1. Subdivide [-1,3] into 100 subintervals of equal length.
  2. The length of each of these subintervals is
    which is (3 -(-1))/100 = 1/25.

  3. We will label the endpoints of the subintervals:

    a0, a1, a2, ..., an
    which, in our example, is

    ai = -1 + i (1/25) = -1 + i/25

4. In each of the subintervals [ai-1, ai], we pick a number xi:

depending upon whether we want to use left-hand endpoints, right-hand endpoints or midpoints, respectively. So

left-hand endpoints: xi = -1 + (i-1)/25
right-hand endpoints: xi = -1 + i/25
midpoints: xi = -1 + (2i-1)/50
C2. Form lists of the xi's:

Left-hand endpoints:

Right-hand endpoints:


5. We then form the Riemann Sum,
C3. After we enter the commands in steps 1. and 2., the calculator automatically makes y1 a list and so we can take the sum:

Screen dumps of calculator:

The following are different ways of calculating the Riemann sums using

Do you know why these methods also work?