ESTIMATING THE AREA UNDER A STRAIGHT LINE
APPLICATION OF SEQUENCES
In this activity, you will explore techniques to estimate the area under a straight line. As you will discover, this method can be
used to estimate the area under any types of curve between two fixed points and can also be extended to find the area between
two curves.
To show how this technique works, we will use the graph y = x + 1 between x = 1 and x = 5. The graph of this function is shown
below.
8
6
y= x+1
4
2
0
0
2
4
6
Notice that the region in consideration, due to its shape, can be divided between a rectangle and a triangle. Use this approach to
find the area between the curve and the x - axis between the points x = 1 and x = 5. The area is _______________________
In general it is hard to break up regions into rectangles and triangles to get an exact area. However, most of the time the area of a
region can be approximated using triangles or rectangles.
You will explore different ways to estimate this area. The techniques discussed can be used to estimate any area in consideration.
Below you find a brief explanation of each technique. The details of each technique will be presented later. All the techniques
rely on subdividing the domain in consideration [1,5] into equal subintervals. The only reason why we use equal subintervals is to
make the computational part easier.
RECTANGLES
Divide the region into rectangles, and add their areas up. The result gives an approximation of the total area.
TRAPEZOIDS
Similar to the "rectangles" approach but instead use trapezoids.
INDIVISIBLES
Calculate the "height" of the function at different points in the domain, and find the average value of those heights. Now construct
a rectangle with width the length of the interval [1,5] and height the average value of the heights you are considering. The area of
this rectangle is an approximation of the area of the region.
PROBLEM 1
Divide the domain [1,5] into two equal subintervals. The end points of these intervals are shown in the graph
8
6
2
AreaUnderStraightLine.nb
PROBLEM 1
Divide the domain [1,5] into two equal subintervals. The end points of these intervals are shown in the graph
8
f HxL = x + 1
6
4
2
0
x1
0
x3
x2
2
4
6
The end points of the intervals are the numbers in the sequence {1, 3, 5}.
This sequence can be defined as xi = 1 + 2 * n, n = 0, 1, 2. Notice that the width of each rectangle is 2. It is denoted as Dx=2.
The heights of the function at these points are the transformation of this sequence under the function. Those values are
8 f Hx1 L, f Hx2 L, f Hx3 L< that can also be defined by f H1 + 2 * nL, n = 0, 1, 2
RECTANGLES
Left Rectangles. The height of each rectangle is the value of the function at the xi value corresponding to the left point of each
subinterval.
8
6
y= x+1
4
R2
2
R1
0
x1
0
x2
2
x3
4
6
Area Expand ≈ area of R1 + area of R2
Area ≈ fHx1 L * Dx + fHx2 L * Dx
≈
≈
2*2 + 4*2
12
This sum is relatively simple to calculate. However, more cumbersome expressions will show up. At that point using technology
will be handy.
TI-89 Instructions to find this sum.
Enter the equation Y1 = x + 1.
Now follow these steps:
a. Go to the home screen
Area ≈ fHx1 L * Dx + fHx2 L * Dx
≈
≈
2*2 + 4*2
AreaUnderStraightLine.nb
12
3
This sum is relatively simple to calculate. However, more cumbersome expressions will show up. At that point using technology
will be handy.
TI-89 Instructions to find this sum.
Enter the equation Y1 = x + 1.
Now follow these steps:
a. Go to the home screen
b. Select F3:Calc
c. From that menu select 4: ∑ ( sum
d. After ∑ ( type Y1(1+2n)*2 , n , 0 , 1). You will see on the screen Ú1n=0 Hy1H1 + 2 nL * 2L
d. Press Enter . You will see 12 on the screen.
TI-83 Instructions.
Follow the instruction at this website. When they ask you to enter the sequence you need to enter the sequence
y1 H1 + 2 nL * 2 = H2 n + 2L * 2
=4n+4
Right Rectangles. The height of each rectangle is the value of the function at the xi value corresponding to the right point of
each subinterval.
8
6
y= x+1
4
2
0
x1
x2
0
x3
2
4
6
Repeat the steps above to calculate the area.
Area ≈
Now verify the information with the calculator.
TRAPEZOIDS
Consider the trapezoids formed by the width of each interval and the end points of the line segments corresponding to the heights
of the function.
8
f HxL = x + 1
6
4
T2
2
0
T1
x1
x2
x3
4
AreaUnderStraightLine.nb
TRAPEZOIDS
Consider the trapezoids formed by the width of each interval and the end points of the line segments corresponding to the heights
of the function.
8
f HxL = x + 1
6
4
T2
2
T1
0
x1
x3
x2
0
2
4
6
Recall that the area of trapezoid of base b and sides of height h and H respectively is (it is a good exercise for derive this formula)
h+H
2
Area =
Using the formula we have that
Area ≈
area T1 + area T2
fHx1 L+fHx2 L
2
≈
*2 +
*b
fHx2 L+fHx3 L
2
≈ fHx1 L + 2 fHx2 L + fHx3 L
*2
≈ 2 + 2*4 + 6
≈
16
INDIVISIBLES
What we call indivisibles are the line segments corresponding to the heights of the function, since they can not be divided any
more.
8
6
f HxL = x + 1
4
2
0
x1
x3
x2
0
2
4
6
Find the average height of the values of the function at the end points of the intervals:
f Hx1 L+f Hx2 L+f Hx3 L
3
= 2+4+6
=4
3
With this average height construct a rectangle as the one shown in the graph.
8
6
f HxL = x + 1
4
Average Height
2
0
x1
0
x3
x2
2
4
6
0
x1
x3
x2
0
2
4
6
Find the average height of the values of the function at the end points of the intervals:
f Hx1 L+f Hx2 L+f Hx3 L
3
= 2+4+6
3
AreaUnderStraightLine.nb
5
=4
With this average height construct a rectangle as the one shown in the graph.
8
6
f HxL = x + 1
4
Average Height
2
0
x1
0
x3
x2
2
4
6
Area of the region is approximately the area of the rectangle which is
Area » Base * height
» Base * Average height of the values of function
» H5 - 1L * 4
» 4*4
» 16
Using the function notation we have:
Area of the region is approximately ≈
≈
f Hx1 L+f Hx2 L+f Hx3 L
3
f Hx1 L+f Hx2 L+f Hx3 L
3
* base of rectangle
*4
You can find this value using the calculator.
ACTIVITY 1
You will now increase the number of subdivisions of the interval [1,5] to approximate the area of the region with each of the
techniques described above. Enter the corresponding expression (you can use summation notation) to estimate the area using each
of the techniques. In the column labeled ≈ enter the numerical value of the approximation
NUMBER OF
RECTANGLES
SUBDIVISIONS
»
TRAPEZOID
» INDIVISIBLES
»
4
8
20
100
n
* For the case "n" you need to find the general pattern for the summation. The actual result of the summation is obtained
using the calculator. This information is needed for activity 2.
ACTIVITY 2
It is expected that as the number of subdivisions n gets larger the approximation to the area is better. In other words the area is the
limiting value of sequence you found in Activity 1 as n ® ¥.
Find the limit for each of the three techniques and convince yourself they are the same.
TECHNOLOGY
The area can be estimated graphically using the graphing calculator. These are the steps to follow:
1. Enter the equation y = x + 1
2. Graph this function on the interval [0,6]. We want to make sure that the interval [1,5] is in the window.
3. Go to F5:MATH (for the TI-89) or CALC (for the TI-83) and select 7:Ù fHxL â x
4. For Lower Limit enter 1 and press Enter
5. For Upper Limit enter 5 and press Enter
6
AreaUnderStraightLine.nb
TECHNOLOGY
The area can be estimated graphically using the graphing calculator. These are the steps to follow:
1. Enter the equation y = x + 1
2. Graph this function on the interval [0,6]. We want to make sure that the interval [1,5] is in the window.
3. Go to F5:MATH (for the TI-89) or CALC (for the TI-83) and select 7:Ù fHxL â x
4. For Lower Limit enter 1 and press Enter
5. For Upper Limit enter 5 and press Enter
6. On the screen you will see Ù fHxL â x = 16. This value should be the value you obtained as the result of the limiting process in
Activity 2.
ACTIVITY 3
One question you may have at this point is why bother with this technique when the actual area under the graph can be calculated
so easily? The reason is that there exist a multitude of graphs for which you cannot calculate the area under its curve as easy as
we did her.
To illustrate this point, choose one of the techniques describe above to estimate the area under the graph y = 2 x2 + 1
between x = 1 and x = 11 by using 20 subdivisions. Explain the technique in details. Also use the calculator to find the area
between the graph and the x-axis as explained above.