Classes of Functions

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“Indefinite Integrals are a class of functions” - What?

We’ve seen how indefinite integrals are linked with antiderivatives, but there is a little more to it than that. In particular, we mentioned that there is a “ $+C$ ” that shows up for some reason. We discuss why that “ $+C$ ” is, and what it means for the subtle (but important) difference between an indefinite integral, and an antiderivative.

Video Lecture

Text with Additional Details

Consider the functions $f(x) = x^2 + 3x - 7$ and $g(x) = x^2 + 3x + 19$ . On the one hand, we know these two functions are obviously very different - just trying a couple $x$ -values we get different results. Nonetheless, if we take the derivative of both functions we get: $f'(x) = 2x + 3$ and $g'(x) = 2x + 3$ - the same function. So what gives? To answer this, we return to the originating motivation for calculus, Newtonian Mechanics.

1 : What is it about $f(x)$ and $g(x)$ above, that made their derivatives result in the same function, even though they were different functions to start with?

They are both polynomials, so they would result in the same derivative. They are both quadratics, so they would result in the same derivative. If you subtract one from the other you get a constant - since derivatives of constants are zero, the difference between them will be zero when you take a derivative. Aren’t you suppose to tell me that?

One of the key equations taught in first year physics is the equation for the (vertical) position of an object that is thrown with some initial (vertical) velocity and from some initial (vertical) position. The equation is $s(t) = -4.9t^2 + v_0t + h_0$ , where $v_0$ is the initial velocity, and $h_0$ is the initial height. (No, you don’t need to memorize this or know it for later, we’re just going to use it as an educational example here.)

Now, a reasonable question might be, what does this have to do with antiderivatives and the indefinite integral... and even moreso, why are we discussing this in a math class rather than a physics class? That’s fair, but bear with me for just a bit, this context will ultimately shed light on what is going on with our antiderivative issue and this weird “ $+C$ ” craziness that cropped up.

Let’s take our position function $s(t)$ and use it to get the other two equations for projectile motion that are often taught in a first semester physics class. As we discussed earlier, we can accomplish this by taking derivatives. $s'(t)$ (the change in position with respect to time, a.k.a. the velocity) is often denoted as $v(t)$ and with our equation we get $s'(t) = -9.8t + v_0$ (where, again, $v_0$ is the initial velocity). Finally $s''(t)$ , or $a(t)$ is the acceleration, and for our equation we have $s''(t) = -9.8$ , which is the acceleration (in meters per second squared) of an object due to gravity (negative here meaning it falls toward the Earth).

Now we get to the part that’s helpful. Let’s go back a step from $s''(t)$ , but rather than writing out the $s'(t)$ we know, let’s do it via integration. Thus $s'(t) = \int s''(t) \dt$ , that is, taking an integral of the second derivative should get us back to the first derivative... at least intuitively. After all, taking an integral is “reversing” the process of a derivative.

But if we calculate this integral, we get $s'(t) = \int s''(t)\dt = \int -9.8\dt = 9.8t + C$ . Remember that the antiderivative for a constant is the line, and we also have this $+C$ that always shows up. But what is up with that $+C$ ?

Remember that the $C$ here is suppose to represent some kind of constant value - like a $7$ or $\pi$ or $-13.8$ . Keeping this in mind then, we can compare our integral result “ $\int s''(t)\dt = -9.8t + C$ ” to the original version of $s'(t)$ that we had: $s'(t) = -9.8t + v_0$ . We can see that $C$ is taking the place of “ $v_0$ ”, the starting velocity of the object!

In fact, this is exactly what this $+C$ in an indefinite integral represents, at least in principle. The $+C$ represents the fact that there was some kind of constant value lost, but more importantly, that constant value represents some kind of initial condition of the equation.

Initial Condition An initial condition is a value for an arbitrary constant at the beginning of a mathematical model - like the speed, acceleration, or position of an object at time zero for a Newtonian Mechanics model/equation.

In our original $s(t)$ equation we needed two pieces of information, the initial velocity, and the initial height of the object. We can see here that, if we integrate the acceleration function, we can get the velocity function, except that going backwards like this means the initial velocity is missing. Since the integration process results in an unknown initial velocity, the only thing we can do is leave it as a generic term, the $+C$ .

2 : The reason the $C$ is an arbitrary constant instead of a variable or an actual number is because...

We know that the value is a constant of some kind because it represents some kind of initial condition or starting value, but we don’t know what that value actually is so we can’t put in a number. We know it doesn’t change because then we’d have to integrate it, but we don’t know the number, so it’s got to be an arbitrary constant. It could be a variable, but it just isn’t for Newtonian Mechanics. Because you said so.

Similarly, if we integrate our original velocity equation $s'(t) = -9.8t + v_0$ , we get the following: $s(t) = \int -9.8t + v_0\dt = -4.9t^2 + v_0t + C$

Again, we can see that the $+C$ appears, and when we compare this integrated version of the $s(t)$ function: $-4.9t^2 + v_0t + C$ against our original $s(t)$ function: $-4.9t^2 + v_0t + h_0$ we can again see that the $+C$ has taken the place of a starting value - in this case the $h_0$ term, the initial height of the object. Since the initial height is not part of the $s'(t)$ function, when we integrate we don’t have the initial height information. Thus the initial height is the “initial condition” that is missing in the integrand, which is consequently represented by the $+C$ term in the end.

So, hopefully by now, we can see how the $+C$ represents some kind of initial condition that we don’t know (at least, it isn’t part of the function that we are integrating), which means we need to get it somewhere else. Usually we accomplish this by knowing a point on the graph of the function, either explicitly or implicitly, which applies to the function after integrating. Consider the following example:

Example: Find the function whose derivative is $f'(x) = 3x^2 + 1$ and whose graph passes through the point $(1,5)$ .

To solve this, we begin by integrating the function: $\int f'(x)dx = \int 3x^2 + 1dx = \frac {3x^{2+1}}{3} + \frac {1x^{0+1}}{1} + C = x^3 + x + C$ So we know that our function looks something like $f(x) = x^3 + x + C$ for some specific value $C$ . But we also know that $(1,5)$ is on the graph of $f(x)$ , which means that $f(1) = 5$ . So we can use this to solve for $C$ . $5 = f(1) = (1)^3 + (1) + C = 2 + C$ So, $C = 3$ , which means: $f(x) = x^3 + x + 3$ .

Keep in mind though, that without that initial condition (that the point $(1,5)$ was on the graph of $f(x)$ ) we couldn’t have narrowed it down any better - we would have been stuck with the equation $f(x) = x^3 + x + C$ . But, what does this equation actually represent? If we don’t know what the $+C$ value is (i.e. we don’t have the initial condition) what do we have? The answer is that we have a “class of functions” with the form $x^3 + x + C$ ... helpful right?

So, what in all that’s holy is a “class of functions”? Remember that the $+C$ is really a stand-in for any possible constant - importantly it is not a variable, we just don’t know which constant it should be. The weird part then, is that the $+C$ notation is actually a kind of shorthand representation for all possible antiderivatives of whatever we are integrating. This means that, when we write the “function” $x^3 + x + C$ we actually aren’t writing a function at all! In fact, what we are writing here is every possible function of the form $x^3 + x +$ (some constant).

For example, say we wanted to write out every possible antiderivative of the function $2x + 1$ . Then we could start with $x^2 + x$ (after all, the derivative of $x^2 + x$ is indeed $2x+1$ ). But we could also include $x^2 + x + 1$ , and $x^2 + x + 2$ , and $x^2 + x + 7$ and $x^2 + x - 31$ ... and so on. Forever. Obviously it’s impossible to write down every possible antiderivative, and this just goes to reinforce the idea that there is no single antiderivative, but rather infinitely many valid antiderivatives, i.e. the antiderivative is not unique. But we may also notice that the only difference between all these antiderivatives is the constant being added to the function. So, when we write $x^2 + x + C$ what we are really trying to capture is that there are infinitely many antiderivatives, and in particular the collection of all antiderivatives (often referred to as the “class of antiderivative” is:

$\{ x^2 + x + C : C \in \mathbb {R} \}$ In human speak, this set is: “all things of the form $x^2 + x +$ (some real number).

So, really, the answer to an indefinite integral isn’t a function, but rather an infinite collection of functions. In practice, this distinction is rarely important (at least until graduate level mathematics) and so we tend to gloss over this detail (and no, you won’t need to explain the details of classes of functions for this class - although you do still need to remember the $+C$ ), but it is important to understand that the indefinite integral can’t be a specific function because it’s always missing necessary information (that initial condition information we discussed earlier) to pinpoint which of the antiderivatives is the desired function. So the best we can do is narrow it down to a function we know, plus some unknown constant value that must necessarily come from somewhere else - somewhere other than the integration process.