We have formulas to find areas of shapes like rectangles, triangles, and circles (pi, anyone?).
What if we want to find the area of a less-reasonable shape? Think of sea monkeys. Sure, we all want them. We want to race them and bring them along on space adventures—but how many can we fit in our two-dimensional space ship? We have to use integrals to figure out the area of each sea monkey before heading into orbit.
Let R be the region between the graph y = f (x) = x2+ 1 and the x-axis on the interval [0, 4]:
R is a weird shape, and we don't have any formula that says how to find its area (yet!).
In this section we're going to look at ways to approximate the areas of shapes that are formed, like R, by graphing non-negative functions on specified intervals.
A non-negative function is as it sounds: a function that never outputs a negative y-value.
Sample Problem
The following function is non-negative (hitting zero is allowed):
Sample Problem
The following function is not non-negative:
Back to R—we can approximate the area of R by drawing rectangles that more-or-less cover R, and calculating the total area covered by those rectangles.
There are several different procedures for drawing these rectangles. The most important ones to know are the Left-Hand Sum, the Right-Hand Sum, and the Midpoint Sum.
These are examples of Riemann Sums. There's also a procedure called the Trapezoid Sum, which draws trapezoids instead of rectangles.
The first step in any of these procedures is to chop up the original interval into sub-intervals, usually all of the same size.
On each sub-interval we draw a rectangle whose base is that sub-interval. The height of each rectangle depends on which procedure we're using.
With a Left-Hand Sum (LHS) the height of the rectangle on a sub-interval is the value of the function at the left endpoint of that sub-interval.
We can find the values of the function we need using formulas, tables, or graphs.
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. One way to find these function values is to calculate them using a formula for the function.
Left-Hand Sums with Formulas
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. One way to find these function values is to calculate them using a formula for the function.
Left-Hand Sums with Tables
In order to find a left-hand sum we need to know the value of the function at the left endpoint of each sub-interval. We can take a left-hand sum if we have a table that contains the appropriate function values.
Sample Problem
Some values of the decreasing function f (x) are given by the following table:
- Use a Left-Hand Sum with 2 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [0, 4].
Answer. We don't know what the function f looks like, but we know these points are part of it:
Dividing the interval [0, 4] into 2 equally-sized sub-intervals gives us sub-intervals of length 2.
The height of the rectangle on [0, 2] is f(0) = 20, so the area of this rectangle is
height ⋅ width = 20(2) = 40.
The height of the rectangle on [2, 4] is f (2) = 17, so the area of this rectangle is
height ⋅ width = 17(2) = 34.
Adding the areas of these rectangles, we estimate the area between the graph of f and the x-axis on [0, 4] to be
40 + 34 = 74.
- Use a Left-Hand Sum with 4 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [0, 4].
Answer. Dividing the interval [0, 4] into 4 evenly-sized sub-intervals produces sub-intervals of length 1:
Sub-interval [0, 1]: This rectangle has height f (0) = 20 and width 1, so its area is 20.
Sub-interval [1, 2]: This rectangle has height f (1) = 18 and width 1, so its area is 18.
Sub-interval [2, 3]: This rectangle has height f (2) = 17 and width 1, so its area is 17.
Sub-interval [3, 4]:This rectangle has height f (3) = 11 and width 1, so its area is 11.
Adding the areas of these rectangles, we estimate the area between the graph of f and the x-axis on [0, 4] to be
20 + 18 + 17 + 11 = 66.
- Are the estimates in parts (b) and (c) over- or under-estimates for the area between the function f and the x-axis on the interval [0, 4]?
Answer. We don't know what the function f looks like exactly, but we know it's a decreasing function that passes through these dots:
That means it must look something like this:
and so our estimates in (b) and (c) were both overestimates, because the rectangles covered extra area:
- Could we use the left-hand sum with more than 4 sub-intervals to estimate the area between the function f and the x-axis on the interval [0, 4]?
Answer. No. The table doesn't contain enough data for us to divide the interval [0, 4] into more than 4 sub-intervals. If we tried to use 8 sub-intervals, for example, we would need to know f (0.5), and that value isn't in the table.
Left-Hand Sum with Graphs
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. We can find these values by looking at a graph of the function.
Left-Hand Sum Calculator Shortcuts
When working through left-hand sums, we need to multiply every function value we use by the width of the rectangle. We can use these observations to work more efficiently and make better use of the calculator. Check out the examples for more ideas.
Left-Hand Sum with Math Notation
It's possible to write the process for taking a LHS as a nice tidy formula. Assume that we're using sub-intervals all of the same length and we want to estimate the area between the graph of f (x) and the x-axis on the interval [a, b].
An interval of the form [a, b] has length b – a. This is true whether or not the numbers a and b are positive or negative. For example, the interval [-1, 10] has length 11.
If we wish to divide the interval [a, b] into n equal sub-intervals, each sub-interval will have length
This quantity is often called Δ x:
We split the interval [a, b] into n sub-intervals, each of length 
. We need to know the value of f at each endpoint except at x = b (the right-most endpoint of the original interval).
We need to know what those endpoints are. The first one is x = a. The next endpoint is a + Δx. The next is (a + Δx) + Δx = (a + 2Δx). Then (a + 2Δx) + Δx = (a + 3Δx).
When do we stop? We know that
This means if we start at a and take n steps of size Δx, we'll get to b, which is the end of the original interval. The last endpoint we need for a left-hand sum is the one just before b, which is
a + ( n – 1 )Δx
These endpoints are often labeled like this, so we don't need to write down as much:
To take a LHS we find the value of f at every endpoint but the last, add these values, and multiply by the width of a sub-interval. In this case, the width of a sub-interval is Δx.
[f (x0) + f (x1) + f (x2) + ... + f (xn – 1 )](width) = [f (x0) + f (x1) + f (x2) + ... + f (x n – 1 )]Δx
Using a left-hand sum with n sub-intervals, we estimate the area between the graph of f and the x-axis on the interval [a, b] is
LHS(n) = [f (x0) + f (x1) + f (x2) + ... + f (x n – 1 )]Δx.
If we wanted to be extra fancy, we could use summation notation. Using i to keep track of which endpoint we're on, we can write the left-hand sum as
.
This formula is the same thing as the calculator shortcut. It's a short, tidy way to write down the process for taking a left-hand sum. There are two important things to remember.
- For a left-hand sum, the last endpoint you use is xn – 1 :
- After adding up the values of f at the appropriate endpoints, multiply by the width of a sub-interval:
Left-Hand Sum with Sub-Intervals of Different Lengths
All the left-hand sums we've found so far had sub-intervals that were all the same length. It doesn't need to be this way. There are some situations where we want to use sub-intervals of different lengths.
Sample Problem
Values of the function f are shown in the table below. Use a left-hand sum with the sub-intervals indicated by the data in the table to estimate the area between the graph of f and the x-axis on the interval [1, 8].
Answer. The sub-intervals given in this table aren't all the same. Most of them have length 2, but one has a width of 1.
On sub-interval [1, 3] the height of the rectangle is f (1) = 2 and the width is 2, so the area is
2(2) = 4.
On sub-interval [3, 4] the height of the rectangle is f (3) = 5 and the width is 1, so the area is
5(1) = 5.
On sub-interval [4, 6] the height of the rectangle is f (4) = 3 and the width is 2, so the area is
3(2) = 6.
On sub-interval [6, 8] the height of the rectangle is f (6) = 5 and the width is 2, so the area is
5(2) = 10.
Adding the areas of the rectangles, we estimate the area between f and the x-axis on [1, 8] to be
4 + 5 + 6 + 10 = 25.