We introduce the procedure of “Slice, Approximate, Integrate” and use it study the area of a region between two curves using the definite integral.

*approximate*such areas to definite integrals that give an

*exact*answer is a fundamental procedure that applies to numerous examples in both mathematics and other STEM fields. In the upcoming sections, we will apply this process to many different cases of interest. As an opening remark, we deal only with piecewise continuously differentiable functions unless otherwise noted.

### The Fundamental Theorem of Calculus and Areas

We begin the section with an example that illustrates the concepts that are fundamental in setting up definite integrals. Consider a continuous function that is positive on a closed interval in its domain. Suppose that we are interested in finding the area bounded by the graph of and the -axis between and . Below is an example of such an example.

We can write this area as the definite integral

and we may use the Fundamental Theorem of Calculus to evaluate it. However, recalling how this result was obtained in the first place is instructive. Understanding the logic behind it is essential in order to apply a similar method to set up integrals to model other types of situations. We thus give a detailed conceptual outline of the argument here.

Step 1: Slice Since we have expressed as an function of , we divide the interval into pieces of uniform width .

As a note, using rectangles of equal width is not is not a requirement. In a more theoretical treatment, this can be treated, but it suffices for us to use uniform widths in order to present more conceptually tractable examples. As such, we adopt this convention here as well as in the coming sections.

Step 2: Approximate We cannot determine the exact area of the slice, but we can approximate each slice by a rectangle whose heights are determined by the value of the function at some -value on the base of the rectangle. Here, we use the righthand endpoint to determine the height of each rectangle.

We can now find the area of the -th rectangle.

where here is the -value of the righthand endpoint of each rectangle. The height of the rectangle is thus .

Let denote the total area obtained by adding the areas of the rectangles together. Then, we can compute easily by adding up the areas of all of the rectangles: or if you prefer using sigma notation:

Note that as we use more rectangles, the following occur*simultaneously*:

- The width of each rectangle decreases.
- The total number of rectangles increases.
- The sum of the areas of the rectangles becomes closer to the actual area.

The actual area is then .

Step 3: Integrate While this infinite limit can be quite cumbersome to work out in even the simplest cases, the Fundamental Theorem of Calculus comes to the rescue. It guarantees that since is continuous on , we may find the area by finding antiderivatives and evaluating the difference of an antiderivative of at the endpoints.

where is an antiderivative of .

This can now be interpreted conceptually as follows:

- The notation “” represents the
*finite*but small width of a rectangle. The notation “” represents the*infinitesimal*width of a rectangle. - The procedure of definite integration can be thought of conceptually
as follows. We
*simultaneously*shrink the widths of the rectangles while adding all of the areas together. - The integrand can be thought of as the area of an
*infinitesimal*rectangle of height and thickness . As such, we cannot think of a rectangle of width as having width zero since we must add infinitely many such rectangles together.

This same procedure can be used to model many other situations, which will be the subject of the coming sections. It is therefore important to understand the logic behind it.

### The Area Between Two Curves

The above procedure also can be used to find areas between two curves as well. Henceforth, by “area”, we will mean “total area”; the area bounded by the curves should be taken to be positive. For example, the area bounded by and from and is shown below.

By noticing that this is the area of a triangle whose base and height have length 1, the total area should be .

As it turns out, the previous procedure can be used to find the total area between two curves. We explore this in the context of another motivating example.

So how should we do this? Let’s apply the procedure of “Slice, Approximate, Integrate”.

Step 1: Slice We divide up into pieces of equal width .

Step 2: Approximate We cannot determine the exact area of the slice, but just as before, we can approximate that each slice by a rectangle whose heights are determined by the value of the function at some -value on its base. Here, we choose to use lefthand endpoints.

The area of one the th rectangles is given by

where is the -value of the lefthand endpoint of the -th rectangle.

To find the height of the darkly shaded rectangle, notice that this height is just the vertical distance between the curves. Vertical distances can always be found by taking the top -value minus the bottom -value. Since these -values lie on the graphs of the given functions, at a given -value, we have and .

Thus, the height of the rectangle is .

The approximate total area obtained by adding the areas of the rectangles between
and together. Note that as we use more rectangles, the following occur
*simultaneously*.

- The width of each rectangle decreases.
- The total number of rectangles increases.
- The sum of the areas of the rectangles becomes closer to the actual area.

The approximate area is given by the finite sum .

Step 3: Integrate The actual area is found by taking the limit of the above sum, that is

and we can write this as the definite integral

Once again the Fundamental Theorem of Calculus allows us to find the area using antiderivatives.

Note that can be interpreted as follows:

- The integrand is the area of a rectangle, whose height is determined as
the difference between the top and bottom -values of the bounding curves,
and whose thickness is
*infinitesimal*. - Since we integrate with respect to , the limits of integration tell us the range of
the -values of the rectangles to be added.
- The lower limit gives the -position of the first rectangle.
- The upper limit gives the -position of the last rectangle.

A similar procedure can be used in many other examples for the rest of the chapter.
The major point here is that once we find the approximate area for a *single* rectangle,
we can immediately write down the integral that gives the *exact* area of the
region.

*every*quantity (limits of integration, functions in the integrand)

*must*be written in terms of that variable!

### Integrating with respect to *x*

We can summarize the above procedure with a formula that respects the geometrical reasoning described previously.

where is the height of each rectangle as a function of , gives the -value of the first slice, and gives the -value of the last slice.

By using vertical rectangles, we have where is found from the top curve, expressed as a function of , and is found form the bottom curve.

Let’s look at a few more examples that demonstrate how to apply the formula.

Note that there is a “natural” righthand boundary here given by the -value where the curves intersect.

From the picture, note that the other intersection point is not relevant.

We must now express in terms of the variable of integration. Since is a vertical distance, . The function used to determine the upper -value is . The function used to determine the lower -value is

The height of the rectangle is thus , and the area is given by:

Evaluating the integral, we find that the area is .

As usual, we begin by drawing a picture and indicating the type of rectangle that will be used to build the area:

As you can see, something interesting happens here; the curve used to determine the height of the top rectangle changes. In order to express this area by integrating with respect to , we have to split it into two pieces.

The top curve will change at the -value where the two curves intersect. To find this -value, we first must express each curve as a function of . The function is already a function of . For the line , we can solve .

We can now find the -value of the intersection point. Write with me:

Hence, or .

Note that by squaring both sides to eliminate the square root, we may have introduced an extraneous root. We can check this easily enough:

By substituting into the equation , we obtain , which is a true statement. However, doing the same for gives , which is not true (though it should be clear why is a solution to the equation that results from squaring both sides!)

Thus, we use .

For the second region, we find the rightmost -value is (set ).

We can now think of the original region in two separate parts and write down an integral that gives the area of each.

Using the picture above:

Putting this together, we can write down a sum of integrals that gives the area.

### Integrating with respect to *y*

We needed to use two integrals to find the area because we used vertical slices to build up the area of the region and the top curve changed. Mercifully, there is an easier way to find this area. Instead of using vertical slices to build the area, we could instead use horizontal ones.

So what are we doing? Instead of making slices with respect to as we did before, we are slicing with respect to . The area of one of these rectangles is . To find the exact area, we simultaneous need to shrink the widths of the rectangles and add all of them together. The same procedure as before produces a convenient result.

where is the height of each rectangle as a function of , gives the -position of the lowest rectangle, and gives the -position of the highest rectangle.

By using horizontal rectangles, we have where is found from the right curve, described as a function of , and is found form the left curve.

Now, let’s revisit the last example.

Since we are going to integrate with respect to , we must describe the curves as functions of :

For , we can solve for to obtain .

For , we can solve for to obtain .

We can note that the lowest slice occurs at and the upper slice occurs at the -value where the curves intersect. Setting these new expressions equal to each other gives .

Now that we have our limits of integration, we must express in terms of the variable of integration! Since is a horizontal distance,

The function used to determine the rightmost -value is and the function used to determine the leftmost -value is .

The length of the rectangle is thus and the area is given by the integral Evaluating this (which you should work out yourself and confirm) gives

### Choosing a variable of integration

As we have seen, choosing a particular type of slice may be more advantageous than another. To make this more explicit, draw a vertical rectangle or horizontal rectangle in the region.

- If the top or bottom curve of the region depends on where you draw the slice, you’ll need more than one integral with respect to to find the area.
- If the left or right curve of the region depends on where you draw the slice, you’ll need more than one integral with respect to to find the area.

How many integrals with respect to are needed to compute this area?

How many integrals with respect to are needed to compute this area?

Sometimes, you will need multiple integrals to find the area of a region no matter which type of slice you use.

How many integrals with respect to are needed to compute this area?

How many integrals with respect to are needed to compute this area?

We conclude the section by summarizing a few important facts.

- To find vertical distances, we always take . To find horizontal distances, we always take .
- When we integrate with respect to , we use vertical slices and when we use vertical slices, we integrate with respect to . When we integrate with respect to , we use horizontal slices and when we use horizontal slices, we integrate with respect to .
- Once we choose a variable of integration, everything in the integrand must be expressed in terms of that variable! This includes both the limits of integration and any functions that arise in the integrand.

“Mathematics is not about numbers, equations, computations, or algorithms; it is about understanding.” - William Paul Thurston