Imagine you are hanging out with two of your friends. Suddenly one takes out a chocolate bar from their bag! They are willing to share it with you and your other friend.
A chocolate bar - pixabay.com
You think of the "fairest" way to divide it, and conclude to give 1/3 of the chocolate bar to each person. However, since this was out of the blue, you do not have the means for dividing the bar in an exact way. You just need each person to have roughly a third of the bar and everyone will be happy!
The chocolate bar divided in three - pixabay.com
Sometimes you just need an answer that is close enough in a practical way to solve our problems. This very idea can be used in many situations, in particular when approximating areas.
What is the meaning of approximating an area?
Sometimes a problem is really hard to solve, but you do not need the exact answer to make practical use of it.
Suppose you need to fence a garden. If you want to find the exact shape of your garden you would need to be extremely skilled with the use of a measuring tape!
Rather than spending time getting the best possible measurements, you just need a rough shape of your garden to go to the store and buy the amount of wire you need. You might have a little leftover, but it does not matter, you solved your problem!
Just like the garden example, some figures have extravagant shapes, with no formula at all for finding their area.
Fig. 1. Figure with an extravagant, non-conventional shape.
If you want to find the area of a shape like the above picture, you will need to find an approximation instead. You can use the grid shown in the picture to make the approximation! Remember that in integration you are finding the area under a curve, which can be approximated in many different ways.
Approximating areas under curves
There are many formulas for finding the area of different geometric figures. In elementary school, we all were taught how to find the area of a triangle, a rectangle, a circle, and many more formulas. It is now time to find the area of more complex shapes: the area below curves.
Fig. 2. Area below a parabola.
But how can you even start? This figure is not like any of the geometrical shapes we are all familiar with. You can start relating it to more basic figures, like rectangles.
Fig. 3. Approximation of the area of the parabola using rectangles.
This, of course, does not match the area of the figure, but it will give you a means of approximation.
Formula for Approximating Areas
Rather than a formula, you can approximate an area following different methods. The first thing you need to do to approximate the area below a curve is to divide up the region of interest. For this purpose, you use what is known as a partition.
Let \( [a,b] \) be an interval. A partition is a set of \(N+1\) values \( P= \{x_i\} \) for \( i=0, 1, 2, ..., N \) such that all \(x_i\) are in the given interval and \( x_0 < x_1 < \dots < x_N.\) The first value is \(a=x_0\) and the last one is \(b=x_N.\)
The values from the partition divide the interval into \(N\) subintervals, and \(N+1\) partition points. The partition is usually chosen so that the subintervals have the same length, but according to the definition they don't have to be the same length!
Sometimes, the partition values are referred to as points, so \(x_0\) and \(x_N\) are known as the endpoints. Why does the index of \(x_i\) starts at \(0\)? Let's look at it with a quick example.
Suppose you are given the interval \( [2,4] \) and you want to divide it into two subintervals. This means that your partition must include three points.
Fig. 4. A partition of three points dividing an interval into two subintervals.
What if you want 4 subintervals? You need to use 5 points, and so on.
Fig. 5. A partition of five points dividing an interval into four subintervals.
Note that you need one more point in the partition than the number of subintervals. This extra point is \(x_0.\)
In the above example, you divided the interval into subintervals of the same length. This is the most common scenario, and you can easily find the length of each subinterval by doing a simple division.
Let \( [a,b] \) be an interval, and let \(P=\{x_i\} \) for \(i=0, 1, ..., N\) be a partition of the interval. The length of each equally-spaced subinterval formed by the partition is given by
$$\Delta x = \frac{b-a}{N}.$$
Remember that N is the number of subintervals. It is not the number of values of the partition!
The above is a formula for finding the length of each equally-spaced subinterval, but is not a formula for finding the required approximation. There are different ways of doing it. Let's address each of them one at a time!
Approximating areas using rectangles
You can now divide the area below a curve using a partition as a guide. But you still need to relate it to simpler figures. The first choice is to use rectangles, as the formula for finding the area of a rectangle is very simple.
Here is a quick reminder: The area of a rectangle is its width times its height!
But how to place the rectangles? There are many ways of doing it. The most common are the left-endpoint approximation and the right-endpoint approximation.
Fig. 6. Left-endpoint approximation of the area below a parabola.
Fig. 7. Right-endpoint approximation of the area below a parabola.
Furthermore, the approximation can be done by using a different amount of rectangles. You will get better approximations by using more rectangles!
Left-Endpoint Approximation
In the left-endpoint approximation, you need to use the leftmost value of \(x\) for each calculation. That is, you use
$$x_0, x_1, x_2, \dots, x_{N-1}$$
to calculate the height of each rectangle, which would be
$$f(x_0), f(x_1), f(x_2), \dots, f(x_{N-1}).$$
Fig. 8. Left-endpoints with its corresponding rectangles.
You then add the area of each rectangle and name it \(L_N,\) where \(N\) is the number of subintervals formed by the partition.
The left-endpoint approximation \(L_N\) of an area divided into \(N\) equally-spaced subintervals is given by:
$$L_N=\sum_{i=0}^{N-1}\Delta x f(x_i).$$
This is easier done than said, so let's see an example!
Approximate the area below the function
$$f(x)=-\frac{1}{2}x^2+\frac{3}{2}x+2$$
in the interval \( [0,4] \) by dividing it into four equally-spaced subintervals. Use a left-endpoint approximation.
Answer:
A visual aid is always very useful. Graphing the function and drawing the rectangles is a good first step.
Fig. 9. Left-endpoint approximation for the area below the function using 4 subintervals.
You now need to find the width and the height of each rectangle for the approximation. Use the formula for the length of each subinterval, which gives you
$$\begin{align}\Delta x &= \frac{b-a}{N} \\[0.5em] &= \frac{4-0}{4} \\[0.5em] &= 1.\end{align}$$
To find the height of each rectangle you will need to evaluate the function at the different points of the partition. Start by evaluating the function at \(x_0=0\)
$$\begin{align}f(0) &=-\frac{1}{2}(0)^2+\frac{3}{2}(0)+2 \\[0.5em] &= 2.\end{align}$$
Find the height of the rest of the rectangles similarly.
\( f(x) \) | \( -\frac{1}{2}x^2+\frac{3}{2}x+2 \) | \(h\) |
| \( f(0) \) | \( -\frac{1}{2}(0)^2+\frac{3}{2}(0)+2 \) | 2 |
\( f(1) \) | \( -\frac{1}{2}(1)^2+\frac{3}{2}(1)+2 \) | 3 |
\( f(2) \) | \( -\frac{1}{2}(2)^2+\frac{3}{2}(2)+2 \) | 3 |
\( f(3) \) | \( -\frac{1}{2}(3)^2+\frac{3}{2}(3)+2 \) | 2 |
You can now substitute these values into the formula for the left-endpoint approximation, that is
$$\begin{align} L_4 &= (1)(2)+(1)(3)+(1)(3)+(1)(2) \\ &= 10. \end{align}$$
You found an approximation using just 4 subintervals. The more, the better!
Fig. 10. Left-endpoint approximation for the area below the function using 8 subintervals.
If you use 8 subintervals instead of 4, the left-endpoint approximation would give you an area of \(L_8=9.75.\)
Right-Endpoint Approximation
In the right-endpoint approximation, you use the rightmost value of \( x\) for each calculation. That is, you use
$$x_1, x_2, \dots, x_N$$
to calculate the height of each rectangle, which would be
$$f(x_1), f(x_2), \dots, f(x_{N}).$$
Fig. 11. Right-endpoints with its corresponding rectangles.
The right-endpoint approximation uses each right-endpoint to calculate the height of the rectangles. The total area is named \( R_N \) instead.
The right-endpoint approximation \( R_N \) of an area divided into \( N \) equally-spaced subintervals is given by:
$$R_N=\sum_{i=1}^{N} \Delta x f(x_i).$$
Let's find the area of our previous example using the right-endpoint approximation instead.
Approximate the area below the function
$$f(x)=-\frac{1}{2}x^2+\frac{3}{2}x+2$$
in the interval \( [0,4] \) by dividing it into four equally-spaced subintervals. Use a right-endpoint approximation.
Answer:
Begin by graphing the function and drawing the rectangles.
Fig. 12. Right-endpoint approximation for the area below the function using 4 subintervals.
Next, find the width of each rectangle for the approximation, so
$$\begin{align}\Delta x &= \frac{b-a}{N} \\[0.5em] &= \frac{4-0}{4} \\[0.5em] &= 1.\end{align}$$
To find the height of each start by evaluating the function at \(x_1=1\)
$$\begin{align}f(1) &=-\frac{1}{2}(1)^2+\frac{3}{2}(1)+2 \\[0.5em] &= 3\end{align}$$
Find the height of the rest of the rectangles similarly.
| \( f(x) \) | \( -\frac{1}{2}x^2+\frac{3}{2}x+2 \) | \( h \) |
| \( f(1) \) | \( -\frac{1}{2}(1)^2+\frac{3}{2}(1)+2 \) | 3 |
| \( f(2) \) | \( -\frac{1}{2}(2)^2+\frac{3}{2}(2)+2 \) | 3 |
| \( f(3) \) | \( -\frac{1}{2}(3)^2+\frac{3}{2}(3)+2 \) | 2 |
| \( f(4) \) | \( -\frac{1}{2}(4)^2+\frac{3}{2}(4)+2 \) | 0 |
Finally, substitute these values into the formula for the right-endpoint approximation, that is
$$\begin{align}R_4 &= (1)(3)+(1)(3)+(1)(2)+(1)(0) \\[0.5em] &= 8.\end{align}$$
This time the approximation gave you an area of 8 rather than the area of 10 you got using the left-endpoint approximation. For contrast, take a look at the right-endpoint approximation with 8 subintervals.
Fig. 13. Right-endpoint approximation for the area below the function using 8 subintervals.
The right-endpoint approximation using 8 subintervals would give you an area of 8.75. In general, the right-endpoint approximation and the left-endpoint approximation will differ!
Which approximation is better?
Unfortunately, you cannot tell which one of the approximations is better because the exact value of the area below the curve is not known. You need further information of the function in order to make a conclusion. You can, however, make better approximations by making partitions with more points.
Finding the midpoint in approximating areas
There is yet another approximation called the midpoint approximation. Rather than using endpoints, the midpoint approximation uses the midpoint of each pair of consecutive values of the partition. That is, you use
$$\frac{x_0+x_1}{2}, \frac{x_1+x_2}{2}, ..., \frac{x_{N-1}+x_N}{2}$$
to calculate the height of each rectangle.
Check out the details of this approximation by taking a look into our Forming Riemann Sums article!
Approximating Areas - Key takeaways
- Some figures do not have formulas for finding their area, so an approximation is required.
- One common way of approximating areas is to use rectangles.
- The first step to approximating areas is to make a partition of the given interval.
- A partition made of \( N+1 \) values divides the interval into \( N \) subintervals.
- The subintervals do not necessarily have to be of the same length, but it is the most common practice.
- If the subintervals have the same length it can be found using \( \Delta x = \frac{b-a}{N}.\)
- The first value is also called left-endpoint, while the last is also called right-endpoint.
- When approximating areas using rectangles, two approximations are more commonly used:
- The left-endpoint approximation, which uses the leftmost value of each rectangle to calculate its height.
- The right-endpoint approximation, which uses the rightmost value of each rectangle to calculate is height.
- The area of all the rectangles is added up to give an approximation of the area below the curve. The more rectangles, the better approximation you will get.
- You cannot tell which approximation is better beforehand. You need further information of the function in order to make a conclusion.
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