We can evaluate Riemann sums on the TI graphing calculators without doing any programming. We use the sum and seq functions to do this.Recall that a Riemann sum is an expression of the form
where the xi* are sample points inside intervals of width ![]()
.
In the examples below, we'll calculate
with
f(x) = 1/(1+x2) and
= 0.01, so that the subintervals are
[0.00, 0.01], [0.01, 0.02], [0.02, 0.03], . . ., [0.99, 1.00] We'll take the sample points to be the midpoints of the subintervals, so that
x1* = 0.005, x2* = 0.015, x3* = 0.025, . . ., x100* = 0.995, Click on one of the links to see this worked out on a
[TI-83], [TI-85], or [TI-89]. On the TI-83
First we store the function 1/(1+x2) as Y1 on the graph page. To do this, we press , clear whatever happens to be in Y1, and enter 1/(1+X2).
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Next we press to return to the Home screen.
We want to evaluate the sum of a sequence of values of Y1, so we need the sum and seq functions. We press and then use the arrow keys to highlight MATH and press to select sum(.
To get the seq function, we again press and then use the arrow keys to highlight OPS. We press to select seq(.
Next we need to enter the function name Y1. To do this, we press , select Y-VARS, and press twice.
At this point, the home screen shows sum(seq(Y1. We want to multiply the value of Y1 by
, which in this case is 0.01. To do so, we just enter *0.01.
To tell the calculator what values of X to use, we complete the command to read
sum(seq(Y1*0.01,X,0.005,0.995,0.01)) This says "Add up the values of Y1*0.01 for X going from 0.005 to 0.995 in increments of 0.01." We press to get the result
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