Section 5.1 Notes Page 1
5.1 Area and Estimating with Finite Sums
This chapter deals with finding the area under curves, which is what integrals will do.
EXAMPLE: Find the area under the curve y  1  x 2 between 0 and 1 using Lower Sums, Upper Sums, and the
Midpoint Rule. Use four rectangles.
This area cannot be found be using a geometric formula.
Therefore we need to estimate it. We are going to find the area of
small rectangular strips to estimate the sum. Of course the more
strips you have then the better estimate. In this example I will use
four rectangles to estimate the area under the curve y  1  x 2 .
There are three ways that we can approach this. It involves using
upper sums, lower sums, and the Midpoint Rule. In each case I
will use four rectangles, however I will draw then in different
places:
Upper Sums
In upper sums, the rectangles height is determined by where a corner of the box
hits the curve, and the rectangles are draw above the curve as shown. We are
splitting the area up into four regions. To find the distance, we take 1 and divide
it by 4 to get 0.25 as the width of each box. In the picture the left side of the box
determines the height, so we need to know the y-values at x = 0, 0.25, 0.5, and 0.75.
The corresponding y values are shown in the figure. Now we want to find the area
of each rectangle and add it together. I will use the formula A = LW. In our case,
the width of each box is 0.25.
A = 1(0.25) + 0.9375(0.25) + 0.75(0.25) + 0.4375(0.25) = 0.78125.
Lower Sums
In lower sums, the rectangles height is determined by where a corner of the box
hits the curve, and the rectangles are draw below the curve as shown. We are
splitting the area up into four regions. Once again the width of each rectangle is 0.25.
In the picture the right side of the box determines the height, so we need to know the
y-values at x = 0, 0.25, 0.5, and 0.75. The corresponding y values are shown in the
figure. Now we want to find the area of each rectangle and add it together. I will use
the formula A = LW. In our case, the width of each box is 0.25.
A = 0.9375(0.25) + 0.75(0.25) + 0.4375(0.25) + 0(0.25) = 0.53125.
Notice that we do have a fourth rectangle, but it has an area of zero since the height is 0.
Section 5.1 Notes Page 2
Midpoint Rule
In the midpoint rule, the rectangles height is determined by where the midpoint of
the box hits the curve, and the rectangles are draw on curve as shown. We are
splitting the area up into four regions. Once again the width of each rectangle is
0.25. In the picture the midpoint of the box determines the height, so we need to
know the half-way point between each interval. To find this, for the first interval,
we do (0 + 0.25)/2 = 0.125. Then we do (0.25 + 0. 5)/2 = 0.375, and so forth. You
will get the x values below the figure. Then we take each of these midpoint values
and put them into our equation, y  1  x 2 . The corresponding y values are shown
in the figure. Now we want to find the area of each rectangle and add it together.
I will use the formula A = LW. In our case, the width of each box is 0.25.
A = 0.9844(0.25) + 0.8594(0.25) + 0.6094(0.25) + 0.2344(0.25) = 0.6719.
Exact Area
In the next section we will see how to get the exact value under the curve by using infinite boxes. If you were
to do this, then you would find that the exact area under this curve is 2/3. Notice that of the three processes, the
midpoint rule is the best approximation to the actual area since you are not overshooting or undershooting by
too much.
EXAMPLE: Use upper and lower sums to approximate the area under the curve y  x  2 and above the xaxis on the interval [0, 2]. Use 8 subintervals as shown below:
If we take 2 and divide it by 8 then the width of each rectangle is 0.25. Now we
can label each x coordinate . We start with 0 and keep adding 0.25 until we reach
2. Each coordinate is: 0.25, 0.5, 0.75, 1, 1.25, 1.5, 1.75. Now let’s make a table
of values:
x 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2.0
y 2 2.5 2.71 2.87 3 3.12 3.22 3.32 3.41
We will start with the lower sum. The first rectangle will have a height of 2 according to the picture. The
second height is 2.5 when x = 0.25. Each sequential height is read off of the table above. Now we can calculate
the area. Notice we do not have a height of 2 since we are doing lower sums.
A = 2(0.25) + 2.5(0.25) + 2.71(0.25) + 2.87(0.25)+ 3(0.25) + 3.12(0.25)+ 3.22(0.25) + 3.32(0.25) = 5.67.
Now we will do the upper sum. Now the first rectangle has a height of 2.5.
A = 2.5(0.25) + 2.71(0.25) + 2.87(0.25)+ 3(0.25) + 3.12(0.25)+ 3.22(0.25) + 3.32(0.25) + 3.41(0.25) = 6.04.
The exact area should be between 5.67 and 6.04.
Section 5.1 Notes Page 3
EXAMPLE: Use upper and lower sums to approximate the area under the curve y  1  x 2 and above the xaxis on the interval [0, 1]. Use 5 subintervals as shown below:
If we take 1 and divide it by 5 then the width of each rectangle is 0.2. Now we
can label each x coordinate . We start with 0 and keep adding 0.2 until we reach
1. Each coordinate is: 0.2, 0.4, 0.6, 0.8, 1. Now let’s make a table of values:
x 0 0.2 0.4 0.6 0.8 1
y 1 0.98 0.92 0.8 0.6 0
We will start with the lower sum. The first rectangle will have a height of 0.98 according to the picture. The
second height is 0.92 when x = 0.4. Each sequential height is read off of the table above. Now we can calculate
the area.
A = 0.98(0.2) + 0.92(0.2) + 0.8(0.2) + 0.6(0.2) + 0(0.2) = 0.66
Now we will do the upper sum. Now the first rectangle has a height of 1.
A = 1(0.2) + 0.98(0.2) + 0.92(0.2) + 0.8(0.2) + 0.6(0.2) = 0.86.
EXAMPLE: Using the Midpoint rule, estimate the area under the curve f ( x)  x 3 between x = 0 and x = 2
using first two and then four rectangles.
Two Rectangles: To find the width of the rectangles, take the change in x and divide it by the number of
rectangles. In our case the change in x is two and the number of rectangles is two, so the width of the rectangles
one. So we have two intervals, [0,1] and [1,2]. We need to find the midpoint of each interval, which will be at
0.5 and 1.5. Then we put 0.5 and 1.5 into our equation to find the height. We find that f (0.5)  (0.5) 3  0.125
and f (1.5)  (1.5) 3  3.375 . To find our area, we will use A = LW. In this case the width of each rectangle is
one. Here is our area:
A = 0.125(1) + 3.375(1) = 3.5
:
Four Rectangles: To find the width of the rectangles, take the change in x and divide it by the number of
rectangles. In our case the change in x is two and the number of rectangles is four, so the width of the rectangles
0.5. So we have four intervals, [0,0.5], [0.5,1], [1,1.5], and [1.5,2]. We need to find the midpoint of each
interval, which will be at 0.25, 0.75, 1.25, and 1.75. Then we put these into our equation to find the height. We
find that f (0.25)  0.1563 , f (0.75)  0.4219 , f (1.25)  1.9531 , f (1.75)  5.3594 . Now we find the area of
each rectangle and add them together. The width of each rectangle is 0.5.
A = 0.1563(0.5) + 0.4219(0.5) + 1.9531(0.5) + 5.3594(0.5) = 3.9452
Note: The exact area under the curve f ( x)  x 3 from 0 to 2 is 4. Notice that using more rectangles brings us
closer to the actual area.