Name: ___________________________________ Period: ________ Date: ________________
Exploration Using TI-84’s – Area Under a Curve
DIRECTIONS: To complete this worksheet, you will need to use a TI-84+ graphing
calculator.
Part A: Areas under curves involving regular geometric figures
1. In mathematics, we often wish to determine the area “under a curve.” By this we actually
mean the area between the graph of a function and an axis (usually the x-axis) over a specific
interval. In this exploration, you will discover methods for evaluating this type of area for
various different types of functions.
2. To help understand the meaning of “area under a curve”,
let’s begin with one of the easiest types of functions, a
constant function. Specifically, let’s find the area of the
function, f(x) = 6 or y = 6 over the interval from x = 1 to x
= 3. Use the “Y =” button to enter ‘6’ in for ‘Y1’. Then
using the “WINDOW” button, make your screen look like
that at the right. Finally, use the “GRAPH” button to
generate a graph of this function.
3. Now think of the region bounded by the graph of the
function, y = 6, the x-axis, the vertical line x = 1 and the
vertical line x =3. This region is shown in the figure at the
right.
a) What geometric shape does this region have?
_________________
b) What is the area formula for this shape? _________________
c) What is the length of the base of this figure in units? __________
d) What is the height of this figure in units? _________
e) What is the area of this figure in square units? _______________
4. Now that the concept of “area under a curve (or graph)”
over an interval is clear, let’s try a slightly harder problem.
Use the “Y =” button to enter ‘x’ in for ‘Y1’. Then using
the “WINDOW” button, make x go from 0 to 5 and y go
from 0 to 5. Use the “GRAPH” button to generate a graph
of this function. Let’s find the area under the graph of y =
x between x = 0 and x = 3. See the region shown in the figure at right.
a.
b.
c.
d.
e.
What geometric shape does this region have? _________________
What is the area formula for this shape? _________________
What is the length of the base of this figure in units? __________
What is the height of this figure in units? _________
What is the area of this figure in square units? _______________
5. Using the graph of y = x in problem #4, let’s find the area
under the curve over the interval from x = 1 to x =4. This
area is shown in the figure at the right.
a. What geometric shape does this region have?
_________________
b. What is the area formula for this shape? _________________
c. What are the lengths of the bases of this figure in units? (Hint: think of this figure rotated
90 degrees to the right.) _________________
d. What is the height of this figure in units? _________
e. What is the area of this figure in square units? _______________
6. Let’s do one final problem of this type before moving
on. Use the “Y =” button to enter ‘2x+1’ in for ‘Y1’.
Then using the “WINDOW” button, make x go from
0 to 5 and y go from 0 to 12. Use the “GRAPH”
button to generate a graph of this function. Let’s find
the area under the graph of f(x) = 2x+1 between x = 1
and x = 3. See the region shown in the figure at right.
a. What geometric shape does this region have? _________________
b. What is the area formula for this shape? _________________
c. What are the lengths of the bases of this figure in units? (Hint: what are the values for
f(1) and f(3)?) __________
d. What is the height of this figure in units? _________
e. What is the area of this figure in square units? _______________
7. Before continuing on to more complex functions, let’s review. One side of each region
produced by the area under the curve of f(x) over an interval x = a to x = b is ‘b – a’
along the x-axis. The two vertical sides of the region have lengths equal to f(a) and f(b).
So the area of the trapezoid formed by a linear equation like y = 2x + 1 over an interval x
=1 to x = 3 would be Area = ½ ( f(1) + f(3) ) (b – a) = ½ (3 + 7) (2) = 10 square units.
Part B: Areas under curves of more complex functions.
1. Now we will look at more complex functions than linear
functions. For instance, consider the function, f(x) = x2,
over the interval from x = 1 to x = 3. See the graph at the
right. This figure looks like a trapezoid, but one side is
curved and not straight. So the area formula for a trapezoid
will not work exactly for this area. Let’s use the area
formula for a trapezoid to get an approximation of this area though.
a. What is f(1)? __________
b. What is f(3)? __________
c. What is the distance along the x-axis? __________
d. Use the area formula for a trapezoid to find an approximate area. _______________
2. We have an approximate area under the curve for the
function above, but let’s try to find a better approximation
for the area. For the sake of simplicity, we will use
rectangles to approximate the area under the curve,
because it is easier to find the areas of rectangles than
trapezoids. To make our approximation better, we will
divide the interval into what are called subintervals. For instance, we could divide the
interval from x =1 to x = 3 (alternatively written as [1,3]) into two subintervals [1,2] and
[2,3]. We will make the subintervals all equal in width for ease of calculation. We will form
two rectangles with widths of 1 unit and heights of f(1) =1 and f(2) = 4. See the figure at
right. The area of the two rectangles are f(1) x 1 = 1 and f(2) x 1 = 4. The total area is 1 + 4 =
5. This area is obviously much lower than the actual area under the curve and is called a
lower sum.
3. Alternatively, we can evaluate the function y = x2 at the
right end of each subinterval rather than the left end as
we did in the previous approximation.
a. What is f(2)? ___________
b. What is f(3)? ___________
c. What is the area of the first subinterval [1,2]?
____________
d. What is the area of the second subinterval [2,3]? _____________
e. What is the total area of both subintervals? ______________
4. The area in problem #3 is obviously larger than the area under the curve and is called an
upper sum? Notice that the actual area under the curve is between the areas of the lower and
upper sums. However, neither of these approximations for the area is as close to the actual
area under the curve as our original trapezoidal approximation. But, if we make more
subintervals, the area should get closer to the actual area under the curve.
5. Let’s use 4 subintervals ([1,1.5],[1.5,2],[2,2.5],[2.5,3])
rather than 2 subintervals and find the lower sum. See
the figure at the right. Answer the following questions.
a. What is the width of each subinterval? _________
b. Evaluate the function f(x) = x2 at the left end of
each subinterval and record all four values in the
space provided. ___________________________
c. Evaluate the area of each of the four rectangles. _____________________________
d. Find the total area of all four rectangles. _________________
e. Now that you have found the lower sum, find the upper sum by using the same process as
above but evaluating the function at the right endpoint of each interval. ______________
f. Notice that the lower sum is still less than the actual area under the curve but is closer
than when two subintervals were used. Likewise, the upper sum is still larger than the
actual area but is closer to it than when two subintervals were used.
6. Next let’s use 10 subintervals ([1,1.2], …,[2.8,3]). The width of each subinterval may be
calculated using the formula (b – a)/n where n is the number of subintervals. To make our
work quicker, let’s use the LIST feature of the calculator.
a. Press the STAT button and select ‘1:Edit’. Enter the left endpoint of each subinterval
into L1 beginning with 1 and ending with 2.8.
b. Use the right arrow key ( ) to move to column L2 and the up arrow key to highlight
‘L2’. Enter the formula ‘L12’ and then press ENTER.
c. Next move to list L3 and highlight ‘L3’. Enter the formula ‘L2*0.2’ to calculate the area
of each rectangle.
d. Finally, move to list L4 and highlight ‘L4’. Press 2ND-STAT then OPS then 6:cumSum(
and enter L3 and end parentheses. Press ENTER.
e. Scroll to the last element in list L4. This is the lower sum. Enter this answer in the space
provided. __________________.
f. Instead of performing parts (d) and (e) above, you could use 2nd-QUIT to exit the LIST
feature and then press 2nd-LIST, MATH, 5-sum(, enter ‘L3’ and ‘)’, and press ENTER.
You should get the same answer as in part (e). Try it!
g. You should be able to perform the same steps as above to find the upper sum, except
enter the right endpoints of each subinterval in list L1. Do this and record the upper sum
in the space provided. ______________. Notice that the lower sum and the upper sum
are closer to each other and to the actual area under the curve.
7. Finally, let’s try using 100 subintervals. Even with the LIST feature on the calculator, this
would take a long time. So, we will try to condense the process by developing a formula that
combines the steps in lists L2, L3, and L4 above. Let’s call the width of each subinterval Δx
where Δx = (b – a)/n = (3 – 1)/100 = 0.02. We will calculate the lower sum first, so the height
of each subinterval will be f(xi) = f(1 + (i-1)* Δx) where ‘i’ is the number of each
subinterval from 1 up to n. So, the heights will be f(1) = 1, f(1.02) = 1.0404, …, f(2.98) =
8.8804. The areas of the subintervals will then be f(xi) Δx = xi2 (0.02).
8. To produce the values (xi) for the left endpoints of the subintervals and the areas of the
individual rectangles, we will use another feature of the graphing calculator called sequences.
The ‘seq(‘ operation involves 5 parameters: function rule ( areas= f(xi) Δx ), the variable (xi),
the beginning value of x, the ending value of x, and the increment (Δx). So, if we wanted to
use ten subintervals, we would press 2nd-LIST, OPS, 5:seq(. Then we would enter the
parameters separated by commas. Specifically, for 10 subintervals this would look like this:
seq(x2 * 0.2, x,1,2.8,0.2). Try entering this and write the list of areas produced in the
sequence in the space provided. _______________________________________________
To get the sum of the areas, use the sum( feature discussed earlier as follows: sum(2ndANS).. Record your answer for this sum in the blank provided. How does it compare to your
answer in question #6(e)? ____________________________________________________
9. Use the seq and sum functions to calculate the lower and upper sums for 100 subintervals.
Hint: use questions #7 and #8 to help you write the correct formulas.
a. Formula for lower sum = __________________________________________________
b. Area for lower sum = _______________
c. Formula for upper sum = __________________________________________________
d. Area for upper sum = _______________
10. If you correctly calculated the lower and upper sums, you should have gotten 8.5868 for the
lower sum and 8.7468 for the upper sum. These areas are close together, and the actual area
is somewhere between them. If we could use 1000, 10000, or even more subintervals, we
could get even closer to the actual area under the curve. But the areas would still be
approximations. To get the exact area under the curve, we will use the concept of limits that
we used earlier to define the derivative function. In this case we will take the limit of the
rectangular areas as the number of subintervals (n) approaches infinity or as the size of the
subintervals (Δx) approaches zero. This is written as follows:
𝑛
lim ∑(𝑓(𝑥𝑖) ∗ Δx)
𝑛→ ∞
𝑖=1
Further, we will refer to this limit as the definite integral of the function f over the interval
[a,b] and will symbolize this as follows:
𝑏
∫ 𝑓(𝑥)𝑑𝑥
𝑎
11. The TI-84+ calculator is able to evaluate definite integrals of functions by using the sum of a
very large number of areas of very small subintervals. Although this is not the exact area, the
area calculated is correct within the precision possible on the calculator. To use this feature
of the calculator, enter the function whose area under the curve is to be found by using the ‘Y
=’ button. Select an appropriate window to graph the interval over which the area is to be
found. Then press ‘2nd-CALC’ button and select ‘7: ‘. When the graph shows up, type in the
number for the lower limit and press ENTER, then type in the number for the upper limit
and press ENTER. Use this feature to calculate the area under the graph of y = x2 over the
interval [1,3]. Enter the answer in the space provided. How does this compare to the lower
and upper sums you calculated in question #9?
_________________________________________________________________________
12. One last method to calculate a definite integral is possible using the TI-84+ calculator. Press
the MATH button and then 9:fnInt(. Enter the function to find the area under (e.g., x2), the
variable to integrate by (x), the left endpoint of the interval, and the right endpoint of the
interval all separated by commas with a right parenthesis at the end. Use this function to
determine the definite integral of f(x) = x2 over the interval [1,3]. How does this answer
compare to your answer for question #11?
_________________________________________________________________________
13. As we continue our exploration of integrals, we will learn several analytical methods for
determining definite integrals without using the graphing calculators.
Illinois Learning Standards – Illinois Mathematics Assessment Framework:
8.11.02 – Represent mathematical relationships using symbolic algebra.
8.11.04 – Determine a specific term, a finite sum, or a rule that generates terms of a
pattern.
8.11.06 – Evaluate variable expressions and functions.
8.11.13 – Represent quantitative relationships graphically, and interpret the meaning of the
graph or a specific part of the graph as it relates to the situation represented by the graph.