Problem:
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.

Visualization:
We follow the algorithm given on Riemann Sums:

AlgorithmCalculator
Given:
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
where 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: Midpoints: 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:

• Left-hand endpoints: • Right-hand endpoints: • Midpoints: The following are different ways of calculating the Riemann sums using

• Left-hand endpoints: • Midpoints: Do you know why these methods also work?