TI-83 Example: Find left and right Riemann sums using 200 subintervals for the function f(x) = 4/(1+x²) on the interval [0,1].

Solution: Since the length of the interval is 1, we have delta x = 1/200.

For the left Riemann sum, we want to add up (1/200) times the sum of 4/(1+x²) evaluated at

This is a perfect job for seq( and sum( on the calculator. The expression we want to evaluate is

To enter sum(, we press [2nd] [LIST] [->] [->] (right-arrow twice, to get the MATH menu) and [5].

To enter seq(, we press [2nd] [LIST] [->] and [5]. We then type in the rest of the expression above, using the ordinary keyboard keys, and finishing off with two right parentheses.

(1/200)*sum(seq( 4/(1+X^2),X,0,19 9/200,1/200)) 3.146588487

When we hit [ENTER], the calculator spends about five seconds thinking things over, and then produces an answer. The screen looks like this:

For the right Riemann sum, we want to use pretty much the same expression, but with values of X going from 1/200 to 2 instead of from 0 to 199/200. The expression we want to evaluate is

4/(1+X^2),X,0,19 9/200,1/200)) 3.146588487 (1/200)*sum(seq( 4/(1+X^2),X,1/20 0,1,1/200)) 3.136588487

To save the trouble of typing in the whole expression again, we press [2nd] [ENTRY] and then use the arrow keys along with [DEL] and [2nd]  [INS] to edit the previous entry. When we press [ENTER] the calculator ponders the problem for a few seconds, and then produces an answer. The screen looks like this:
We conclude that the area A under the curve y=4/(1+x²) between x=0 and x=1 satisfies

Comment: This method uses the TI-83's LIST data type, and TI-83 lists are limited to 999 entries. So for Riemann sums that involve more than 999 terms, we'd have to use another method (probably involving writing a program) or another machine.

TI-85 Example: Find left and right Riemann sums using 2000 subintervals for the function f(x) = 4/(1+x²) on the interval [0,1].

Solution: Since the length of the interval is 1, we have delta x = 1/2000.

For the left Riemann sum, we want to add up (1/2000) times the sum of 4/(1+x²) evaluated at

We use the seq( and sum( functions on the calculator. The expression we want to evaluate is

The sum and seq commands are in the LIST menu, under the sub-menu OPS. To find them, we press [2nd] [LIST] [OPS] [MORE] and [5]. With sum and seq on the screen menu, it's a snap to type in the expression above.

(1/2000)*sum seq(4/(1 +x^2),x,0,1999/2000,1 /2000) 3.14209261192

When we hit [ENTER], the calculator grinds away at the problem for almost a minute, and then produces an answer. Here's the resulting screen (minus the menu-related stuff at the bottom):

For the right Riemann sum, we want to use the same expression, but with values of x going from 1/2000 to 2 instead of from 0 to 1999/2000. Here's what we want to evaluate:

(1/2000)*sum seq(4/(1 +x^2),x,0,1999/2000,1 /2000) 3.14209261192 (1/2000)*sum seq(4/(1 +x^2),x,1/2000,1,1/20 00) 3.14109261192

To save the trouble of typing in the whole expression again, we press [2nd] [ENTRY] and then use the arrow keys along with [DEL] and [2nd]  [INS] to edit the previous entry. When we press [ENTER] the calculator thinks for a while, and then produces an answer. The screen looks like this:
We conclude that the area A under the curve y=4/(1+x²) between x=0 and x=1 satisfies

Comments:

1. The TI-85 doesn't seem to have a hard limit on the length of a list, but it will run out of memory if you try to make use too many subintervals. My TI-85 tends to choke on lists with more than about 2380 elements.

2. The average of the lower and upper bound we have determined should be a pretty good estimate of the area under the curve. What is it?

Excel Example: Find the left and right Riemann sums using 20 subintervals for the function 1/(1+x³) on the interval [1,3].

Solution: We begin by opening a Workbook document in Excel. Usually an untitled workbook comes up automatically when Excel starts.

Since the length of our interval is 2 and we want 20 subintervals, we get delta x = 0.1. In fact, it's not hard to see that we'll want to have the x values

To get these values into the worksheet, we type 1 into cell A1 and 1.1 into cell A2. We then use the mouse to highlight cells A1 and A2.

Next we very carefully pick up the small black square in the lower right corner of the highlight box and drag it down column A to cell A21. Column A magically fills with the numbers 1, 1.1, 1.2, and so on.

To evaluate our function at all these points along the x-axis, we enter the function in cell B1. To do this, we highlight cell B1 and then type

Next we highlight cell B1 and then very carefully pick up the small black square in its lower right corner, and drag it down the column to cell B21. Column B magically fills with the values of our function at the points named in column A.

The left Riemann sum is equal to 0.1 times the sum of the entries in cells B1 through B20. To compute this, we choose an empty cell somewhere, highlight it, and then type

The right Riemann sum is equal to 0.1 times the sum of the entries in cells B2 through B21. To compute this, we choose another empty cell, highlight it, and then type

The left Riemann sum is 0.171305 and the right Riemann sum is 0.148091.