We will always have STEM with us. Some things will drop out of the public eye and will go way, but there will always be science, engineering, and technology. And there will always, always be mathematics.
— Katherine Johnson
5.1 Motivation - Arbitrary Areas and Distance Traveled
Over the course of primary and secondary education, students learn how to find the areas of regular shapes like triangles, rectangles, and circles. Suppose I gave you a shape like the following:
Calculus will allow us to find the areas of irregular shapes like these, too. We will now need a new limiting behavior called integration. This will be the subject of this and the next few chapters.
Let’s take this idea a bit further and return to the car example from Chapter 3. Recall the following formula from basic physics
\[\text{Speed} = \frac{\text{Distance}}{\text{Time}} \hspace{5mm} s = \frac{d}{t}\]
Which can be rewritten as
\[\text{Distance} = \text{Speed}\times\text{Time} \hspace{5mm} d = s\times t \]
The diagram below is for your car if it is moving at a constant speed. We can use the formula above to compute the distance your car has traveled.
The total distance traveled is the area shaded in blue. It indicates the speed (\(4 \frac{m}{s}\)), and the time spent travelling (\(t\) seconds). In this case, the acceleration is \(a = 0 \frac{m}{s^2}\), since the speed is constant.
We can also very easily determine the total distance traveled when the acceleration is a non-zero constant. This case is illustrated below.
Now, we cannot blindly apply the formula \(\text{Distance} = \text{Speed}\times\text{Time}\). Instead, we can determine the distance by computing the area under the triangle. Recall that the area of a triangle is given by the formula \(A = \frac{1}{2}b h\). As indicated in the diagram, the base is given by \(b = t\), which is the time the car has spent travelling. The height of the triangle is given by \(h = v = 2t\), the height of the shaded triangle, which is the current speed of the car. Therefore, the distance traveled is \(d = A = \frac{1}{2}b h = \frac{1}{2}\cdot t\cdot (2t) = t^2\).
Now suppose the speed of the car looks like the following:
As before, we want to compute the distance \(d\) the car has traveled for each time \(t\). Once again, we can determine the area under the function to compute the distance that the car has traveled. In this case, we don’t have a formula handy to compute areas of complicated figures as we did for rectangles and triangles. Instead, we must use an integral to compute the area under the curve, which in turn gives us the distance the car has traveled.
5.2 The Basic Idea
We have nice, compact formulas for the areas of rectangles, triangles, and trapezoids. The fundamental idea is to use shapes like rectangles to approximate an arbitrary area, then allowing these rectangles to become smaller and smaller and allowing the number of such rectangles to become larger and larger. Below is an example of what this procedure looks like:
We can move the points \(a\) and \(b\) using the points on the \(x\)-axis. Furthermore, you can change the number of rectangles used to approximate the true area with the slider for \(n\). Notice that as \(n\) becomes larger and larger, the rectangles more and more closely approximate the area we’re interested in computing.
Let’s consider another function to illustrate the process. We want to determine the area under the curve \(f(x) = 1 + \sin x^2\).
Letting \(a = 0\) and \(b = 8\), we can see the sequence corresponding to the total area of all rectangles being used to approximate the area of interest:
This may remind the reader of the sequences we saw in Chapter 2. Indeed, integration is yet another limiting process, one in which the number of rectangles goes to infinity while the width of each rectangle goes to zero. Note that each element of the sequence \((s_n)\) corresponds to a different area, and that this area converges to \(s_{\infty}\) as \(n\) grows larger and larger. We call the value that \((s_n)\) converges to the integral of the given function over the interval of interest (in this case, \([a, b] = [0, 8]\)).
As we have stated repeatedly in this text, mathematicians are not fully satisfied with concepts. We need to express our thoughts in notation. Doing so will be the subject of the next section.
5.3 Limits of Riemann Sums
We can place the rectangles in different positions with respect to the function whose area we want to compute. The following illustrates this.
In the illustration above, we can define a Left Riemann Sum, a Right Riemann Sum, and a Midpoint Riemann Sum. In the plot above, Choice = 1 corresponds to a left Riemann Sum, Choice = 2 corresponds to a right Riemann Sum, and Choice = 3 corresponds to a midpoint Riemann Sum.
Note further that those rectangles below the \(x\)-axis are colored red, while those rectangles above the \(x\)-axis are colored green. Those rectangles colored green add positively to the final sum, while the red rectangles add negatively to the final sum. This is because the function \(f(x)\) that we are integrating is positive when the rectangles are upright and negative when the rectangles are upside down.
For all three values of Choice, we want to compute the area for an arbitrary number of rectangles. We usually assume that all rectangles are of the same length to simplify calculations.
5.3.1 Leftpoint Sum
In the illustration above, when Choice = 1, we are drawing the arrows such that their upper-left-hand corners make contact with the function. In that case, the height of each rectangle is equal to \(f(x_i)\)
We want to compute the area covered by the rectangles. We force all rectangles to have the same width, say \(s\). To compute the width of these rectangles, we simply divide the total length of the interval \([a, b]\) by the number of rectangles \(n\) used to make the approximation:
\[x_{i+1} - x_{i} = s = \frac{b - a}{n}\]
The height of each of the rectangles is given by \(f(x_i)\), which is the function evaluated at the left endpoint of each rectangle. Therefore, the area of the \(i^{th}\) rectangle is given by
where we let \(L_n\) denote the left Riemann Sum for \(n\) rectangles.
Rightpoint Sum
The formula for the right Riemann Sum is very similar. Consider the diagram below.
Once again, we want to determine the area covered by the rectangles above. As before, the width of each rectangle is given by \(x_{i + 1} - x_i = s\). Because there are \(n\) rectangles over an interval of length \(b - a\), the width of each rectangle is given by
\[s = x_{i + 1} - x_i = \frac{b - a}{n}\]
Now, it is the upper right hand corner of each rectangle making contact with the function. This is \(f(x_{i + 1})\), as illustrated above. Therefore, with the heights and the bases of each rectangle, we compute the right Riemann sum as
This is almost identical to the left Riemann sum \(L_n\), but we replace \(f(x_i)\) with \(f(x_{i + 1})\) in the formula for the right Riemann sum. Note that as the number of rectangles becomes larger (that is, as \(n \rightarrow \infty\)), the rectangles become narrower and narrower. If the rectangles become narrower, then \(x_{i+1} - x_{i} \rightarrow 0\).
Midpoint Sum
Taking Limits of Sums
We aren’t particularly interested in the Riemann sums themselves; instead, we are interested in what happens when the number of rectangles becomes arbitrarily large; furthermore, if we are stuffing more and more rectangles into the same interval \([a, b]\), then the widths of those rectangles must go to zero. Therefore, we need a limit. We have the following:
The \(\int\) symbol was introduced by one of the fathers of Calculus, Gottfried Wilhelm Leibniz. It is an elongated S and stands for “sum”. In words, the last set of symbols means “integrate the function \(f(x)\) from \(x = a\) to \(x = b\)”.
The \(dx\) might look a little odd. Remember that \(s = x_{i + 1} - x_i\) are the widths of the intervals over which we’re summing. \(dx\) corresponds to the lengths of the intervals as we let the number of rectangles go to infinity, or \(x_{i + 1} - x_i \rightarrow 0\). You might hear this referred to as an infinitesmal, or a quantity of arbitrarily small length.
It is not true that all functions have Riemann sums which converge to an integral in the limit. We have glossed over many details involving \(\delta, \epsilon\) proofs which provide the conditions under which the integral of a function exists. (Very) roughly speaking, a function is Riemann integrable if
and if the points at which the function is discontinuous is not too large (this is a very subtle point; if you wish to learn more, take a course in real analysis).
5.4 The Fundamental Theorem of Calculus, Part 1
We now understand what an integral is and where it comes from. This is nice, but we can’t actually do anything useful with what we’ve learned thus far; we’ve only learned the concepts. The following theorem, the Fundamental Theorem of Calculus, is the bridge between differentiation and integration. It is the theorem that actually allows us to use Calculus to solve interesting problems. The proof of this theorem is somewhat difficult, but the reader is strongly encouraged to read the proof and do their best to understand it. It is impossible to understand what Calculus is all about without it.
Theorem 5.1 (The Fundamental Theorem of Calculus, Part 1) Let \(f\) be a differentiable function. Then
Notice that the parts in red are the same. Since we have one in the numerator and another in the denominator, these quantities cancel. Therefore, we have:
Recall that the limit of a constant is equal to that constant; that is, \(\lim\limits_{x_{i + 1}\rightarrow x_i} c = c\). In the example above, \(f(x_n) - f(x_1)\) contains no\(x_i\)or\(x_{i + 1}\). Therefore, the quantity \(f(x_n) - f(x_1)\) is constant relative to the innermost limit. Hence, it can be “dropped”. Therefore, we have
Recall that the limit of a difference is equal to the difference of limits. Furthermore, \(f(x_1)\) is constant relative to the limit \(\lim\limits_{n\rightarrow \infty}\). Therefore, we have
Now, recall that as the number of rectangles becomes larger and larger, the rectangles themselves become narrower and narrower. With this in mind, look at the image in the section on left Riemann sums, 5.3.1. It is clear that if the rightmost rectangle becomes more and more narrow, \(x_n \rightarrow b\). Meanwhile, \(x_1 = a\). Therefore, we finally find
Addition and subtraction are inverse operations. If I add three then take three away, I have not changed the quantity I am considering. Multiplication and division are also inverse operations. We have just shown that integration and differentiation are inverse operations. Integration adds things up, while differentiation takes them apart. Therefore, if I apply an integral to a derivative, it returns the function (evaluated at the endpoints of the integral).
5.5 Learning to Use Integrals
Now that we have the fundamental theorem, how do we use it? For instance, suppose I ask you to find the area under the function \(f(x) = x^5 + 3x^2\). The next example described how to use the Fundamental Theorem to achieve this.
Example 5.1 (Integration Example 1) We want to integrate the function \(f(x) = x^5 + 3x^2\) from \(a = 1\) to \(b = 5\), as illustrated above.
The integral we wish to compute is given by
\[\int_{1}^{5} x^5 + 3x^2\, dx\]
What we want to find is a function \(F(x)\), called the antiderivative, so that
\[F'(x) = x^5 + 3x^2\]
If we find such a function \(F\), then we can use the Fundamental Theorem of Calculus and compute \(F(5) - F(1)\), which is equal to the value of the integral. So the question we need to answer is: “Which function has derivative \(x^5 + 3x^2\)?”
First, we find a derivative which is equal to \(x^5\). Recall from Chapter Three, 3.2.3, that
\[\frac{d}{dx} x^6 = 6x^5\]
That’s close! We just need to remove the “6”. To do so, we simply divide both sides of the equation by 6, to obtain
\[\frac{d}{dx}\frac{1}{6}x^6 = x^5\]
Wonderful! Now we must find a function whose derivative is \(3x^2\). Once again, we use 3.2.3 to obtain
\[\frac{d}{dx} x^3 = 3x^2 \]
Great! We’re almost there. We found that \(\frac{d}{dx}\left(\frac{1}{6}x^6 + x^3 \right) = x^5 + 3x^2\). Therefore, by the Fundamental Theorem of Calculus, we have
If you let \(n\rightarrow\infty\) in the diagram above, you should find it converges to \(2728\), as we would expect.
We can currently integrate any function whose derivative we found in the previous chapter. The following example illustrates how we can find the integral of the cosine function, \(f(x) = \cos x\).
Example 5.2 (Integration Example 2) Suppose we want to integrate the function \(f(x) = \cos x\) from \(a = 0\) to \(b = \pi\). That is, we wish to evaluate
\[\int_{0}^{\pi} \cos x \, dx\]
Recall from last chapter that \(\frac{d}{dx} \sin x = \cos x\). Therefore, by the Fundamental Theorem of Calculus, we have
\[\int_{0}^{\pi} \cos x \, dx = \sin x \big|_{a = 0}^{b = \pi} = \sin(\pi) - \sin(0) = 0\]
The area is zero, since we are adding as much positive stuff as we are negative stuff. Indeed, for each \(n\), we are adding a green and a red rectangle which nearly cancel. If the reader lets \(n\rightarrow\infty\) in the figure above, the area will be seen to go to (or be nearly equal to) zero.
In the previous example, we used notation that looks like this
\[\sin x \big|_{a = 0}^{b = \pi}\]
This notation just indicates that we have found that the antiderivative is equal to \(F(x) = \sin x\), and the endpoints at which we’re evaluating this antiderivative are \(a = 0\) and \(b = \pi\).
One last bit about vocabulary. The following diagram and the accompanying text illustrate all of the vocabulary we will use when evaluating integrals for the rest of this book.
\(a\) and \(b\) are the lower and upper limits of integration, respectively.
\(\int\) is the integral.
\(f(x)\) is the integrand or the kernel of the integral.
\(dx\) indicates the variable of integration. In this case, we are integrating with respect to \(x\).
5.6 Basic Integral Formulas
We do not want to have to determine which function has a derivative equal to the integrand for every integral we encounter. Over the course of the next two chapters, we will build a little toolbox of integral rules that can be applied to problems you may encounter in all sorts of math and science problems. Generally, the author does not want you to memorize the material in this book; rather, you should try to develop a feeling in your bones for the subject and deeply understand it. That said, it is probably worthwhile to have these rules memorized so that you can solve problems quickly.
5.6.1 Sum and Difference Rule
The following illustrates the sum rule for integrals.
We are imagining that we have three functions: \(g(x)\), \(h(x)\), and \(f(x) = g(x) + h(x)\). We want to determine the integral of \(f(x)\). The illustration above shows us that the integral of \(f(x)\) is unsurprisingly the integral of \(g(x)\) plus the integral of \(h(x)\). We know this because the green rectangles for the integral of the function \(g(x)\) extend from the orange rectangles for the integral of the function \(h(x)\). When the orange rectangles plus the green rectangles are added together, they fill in the area under the blue curve, which is \(f(x)\). Therefore, we have the following theorem:
Theorem 5.2 (Integral Sum Formula) Let \(g(x)\) and \(h(x)\) be two integrable functions. Then
Meanwhile, we know that the derivative of \(f(x) = \sin x\) is \(\cos x\). Because the integral and the derivative are inverse operations, the integral (2) is given by
This theorem states that if we have a constant in our integral, we can pull it out.
5.6.3 Breaking the Interval Rule
Suppose we have an interval \([a, c]\) and another number \(b\) such that \(a < b < c\). We can split the interval into multiple pieces an integrate the function of interest on each interval separately. This intuitive idea is expressed as a theorem below.
Theorem 5.4 (Breaking the Interval) Let \(f(x)\) be an integrable function on the interval \([a, c]\), and let \(a < b < c\). Then
Let \(a\) be the lower limit of integration and \(b\) the upper limit of integration. If \(a = b\), then we are integrating the function only at a single point. The result necessarily has no area, so the integral is equal to zero. In the illustration below, move the slider for \(a\) such that it corresponds to the slider \(b\). You will find that, upon doing so, the corresponding sum is equal to zero. Said another way, no rectangles will fit to approximate the desired area as \(a\) and \(b\) become closer and closer, so the area approximated by those rectangles for any \(n\) must be zero.
This idea is captured in the following theorem.
Theorem 5.5 (Integral over Interval of Length Zero) Let \(f(x)\) be an integrable function. Then
Often we aren’t concerned with integrating a function over an interval; instead, we are simply interested in computing the antiderivative of some function of interest. Consider the following example.
Example 5.4 (Indefinite Integrals) Find the antiderivative of the function \(f(x) = \frac{1}{7} x^7 - x\).
We are interested in finding the following:
\[\int \frac{1}{7} x^7 - x \, dx\]
We can use the sum and difference rule, 5.6.1, to change the integral above into
\[\underbrace{\int\frac{1}{7} x^7 \, dx}_{(1)} - \underbrace{\int x \, dx}_{(2)}\]
To find (1), we must find a function \(F_1(x)\) whose derivative is \(\frac{1}{7} x^7\). Recall from 3.2.3 that
\[\frac{d}{dx} x^8 = 8 x^7\]
This is close to what we want. If we multiply both sides of the equation above by \(\frac{1}{56}\) (\(56 \div 8 = 7\)), we have
Notice that, in the previous example, we weren’t interested in calculating an area. Instead, we found the antiderivative of a function. That is, if I compute the derivative of the function \(F(x) = \frac{1}{56} x^8 - \frac{1}{2} x^2 + C\), I get
\[\frac{dF}{dx} = \frac{1}{7} x^7 - x,\]
once again illustrating that derivatives and integrals are inverse operations.
The Constant of Integration
One more point to note. The inclusion of the \(C\) in the previous example may seem a little odd. This \(C\) is a constant and is appropriately named the constant of integration. We must include this \(C\) when we evaluate indefinite integrals. Recall that when evaluating integrals, we are ultimately looking for a function whose derivative is the integrand (the function inside the integral). That way, we can apply the Fundamental Theorem of Calculus (5.1) and evaluate the integral. But there are many functions whose derivative is the integrand. Consider the illustration below.
All of these functions have exactly the same derivative. This is because the derivative of a constant is zero (3.2.1). Therefore, to be clear that we do not know which constant is the right one after performing an indefinite integral, we include the variable \(C\). Another way of expressing this is provided below:
we don’t know which constant (\(6\) or \(-\frac{1}{2}\)) is the “right” one. Therefore, to indicate this lack of knowledge we include a constant of integration, \(C\), when evaluating an indefinite integral:
Below we list a few integrals that arise frequently in practice. We simply use the Fundamental Theorem of Calculus to identify these integrals. We will add to this list next chapter.
Theorem 5.7 (Indefinite Integrals) Let \(a\) be a constant. Then
We already showed through the first part of the Fundamental Theorem of Calculus that if you apply an integral to a derivative, the integral cancels the derivative. We also have the reverse: the derivative cancels the integral. The second part of the Fundamental Theorem is provided below. We then provide intuition for why the second part of the fundamental theorem should hold, followed by a proof.
Theorem 5.8 (The Fundamental Theorem of Calculus, Part 2) Let \(f(x)\) be an integrable function. Then
There is something peculiar here that we have not yet encountered: namely, the integral is itself serving as the function. In particular, the input \(x\) is the upper limit of integration of the integral.
Before proving this result, we will need an idea called the Mean Value Theorem for Integrals. We will not be proving this result, as it is relatively intuitive. Consider the diagram below:
The idea is this: let’s suppose we are integrating the function \(f(x)\) from \(a\) to \(b\). We want to find some other \(x\) value, \(c^{*}\), such that the area of the rectangle (\(\color{red}{f(c^{*})(b - a)}\)) is equal to the integral \(\color{blue}{\int_{a}^{b} f(x) \, dx}\). It turns out this is always possible as long as \(f(x)\) is continuous. We write this as a theorem below:
Theorem 5.9 (The Mean Value Theorem for Integrals) Let \(f(x)\) be a continuous function. Then we can find some \(c^{*}\) in the interval \((a, b)\) such that
\[\int_{a}^{b} f(x) \, dx = f(c^*) (b - a) \]
This theorem will be indispensable in our proof of the second part of the Fundamental Theorem of Calculus:
We wish to evaluate the derivative
\[\frac{d}{dx} \int_{a}^{x} f(t) \, dt\]
We can simplify things using the definition of the derivative, 3.2. We have the following:
We’re almost there! We now must apply the Mean Value Theorem for Integrals, 5.9. By the result of that theorem, we can find some \(c^*\) in the interval \((x, x+h)\) such that
We know that \(c^*\) is between \(x\) and \(x + h\). Therefore, as \(h\) comes closer and closer to zero, \(c^*\) is forced to be closer and closer to \(x\). Moreover, because \(f(x)\) was assumed to be continuous, it must be that \(\lim\limits_{h\to 0} f(c^*) = f(x)\). Hence, we have:
