We have seen previously that for continuous functions defined on closed intervals, the Fundamental Theorem of Calculus relates the process of finding antiderivatives to calculating certain areas. As it turns out the process used to transcribe Riemann sums that 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 $f(x)$ that is positive on a closed interval $[a,b]$ in its domain. Suppose that we are interested in finding the area bounded by the graph of $y=f(x)$ and the $x$-axis between $x=a$ and $x=b$. Below is an example of such an example.

We can write this area as the definite integral

$$A = \int _{x=a}^{x=b} f(x) dx ,$$

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 $y$ as an function of $x$, we divide the interval $[a,b]$ into $n$ pieces of uniform width $\Delta x$.

As a note, using rectangles of equal width 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. The top of the rectangle should coincide with the top curve, and the bottom of the rectangle should coincide with the lower curve above some common $x$-value on the base of the rectangle. For the sake of the picture, we use the lefthand endpoint to determine the height of each rectangle.

We can now find the area $\Delta A_k$ of the $k$-th rectangle.

$$\begin{align} \Delta A_k & = (height) \times (width) \nonumber \\ \Delta A_k &= f\left(x_k^*\right)\Delta x \end{align}$$ where $x_k^*$ here is the $x$-value of the right-hand endpoint of each rectangle. The height of the rectangle is thus $f(x_k^*)$.

Let $S_n$ denote the total area obtained by adding the areas of the $n$ rectangles together. Then, we can compute $S_n$ easily by adding up the areas of all of the rectangles. $$S_n = \Delta A_1 + \Delta A_2 + \cdots +\Delta A_n$$ or if you prefer using sigma notation,

$$\begin{equation} S_n =\sum_{k=1}^{n} \Delta A_k = \sum_{k=1}^n f(x_k^*) \Delta x. \end{equation}$$ Note that as we use more rectangles, the following occur simultaneously:

• The width $\Delta x$ 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 $A = \lim _{n \rightarrow \infty } \left [ \sum _{k=1}^n f(x_k^*) \Delta x \right ]$.

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 $y=f(x)$ is continuous on $[a,b]$, we may find the area by finding antiderivatives and evaluating the difference of an antiderivative of $f(x)$ at the endpoints.

$$A = \int _{x=a}^{x=b} f(x) dx = F(b)-F(a) ,$$

where $F(x)$ is an antiderivative of $f(x)$.

This can now be interpreted conceptually as follows:

• The notation “$\Delta x$” represents the finite but small width of a rectangle. The notation “$dx$” 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 $f(x) dx$ can be thought of as the area of an infinitesimal rectangle of height $f(x)$ and thickness $dx$. As such, we cannot think of a rectangle of width $dx$ 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 $y=-x$ and $y=0$ from $x=0$ and $x=1$ 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 $+\frac {1}{2}$.

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.

Integrating with respect to x

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

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

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 $x$ as we did before, we are slicing with respect to $y$. The area of one of these rectangles is $\Delta A = h \Delta y$. To find the exact area, we simultaneously need to shrink the widths of the rectangles and add all of them together. The same procedure as before produces a convenient result.

Now, let’s revisit the last example.

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 $x$ 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 $y$ to find the area.

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

We conclude the section by summarizing a few important facts.

• To find vertical distances, we always take $y_{top} - y_{bot}$. To find horizontal distances, we always take $x_{right}-x_{left}$.
• When we integrate with respect to $x$, we use vertical slices and when we use vertical slices, we integrate with respect to $x$. When we integrate with respect to $y$, we use horizontal slices and when we use horizontal slices, we integrate with respect to $y$.
• 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.