 So as the song goes, what can you do with an antiderivative? Well, it turns out that you can do quite a bit with it. I won't afflict you with any more of my singing. So let's take a look at the antiderivative. So again, remember that what an antiderivative is, is it's a function whose derivative is whatever the antiderivative we're taking. And it's helpful, again, to think about what those units are going to be. So remember that differential, that dx indicates that when we are differentiating capital F of x, that derivative is going to be taking place with respect to x. So this relationship that the antiderivative is a function whose derivative is F of x says that if I differentiate with respect to x, I get back my original function. And what that translates into is the following. The units of F of x are going to be the units of capital F over the units of x. And equivalently, if I think about cross-multiplying the units, again the units are treatable just like algebraic quantities, that says the units of capital F will be the same as the units of F of x times the units of x. And again, here's where having that notation is really nice, because I can actually read that right off of our notation. The antiderivative is this thing, units of F times units of x. And this dimensional analysis is a really helpful starting point in almost any problem. So let's suppose I have the velocity of an object, and that's going to be measured in meters per second, t seconds after the object begins moving, and we can find the derivative, the units of the antiderivative, and the units of the derivative. So let's see, the units of the antiderivative, well those are going to be the units of v of t, meters per second, multiplied by the units of t, seconds. So those units will be meters per second times seconds, and the seconds are just like algebraic quantities, and they're going to cancel. The antiderivative is going to have units of meters. Whatever the antiderivative represents, it's going to be something that is measured in units of meters. How about the derivative itself? Well the units of the derivative are going to be the units of the function divided by the units of the independent variable. So v of t has units meters per second, t has units seconds, so the derivative is going to have units meters per second over seconds, which we can simplify as meters per second squared, or meters per second per second. And so whatever the derivative measures is going to be something that's measured in meters per second squared. And this bit of an additional analysis is going to help us in any sort of problem that involves, well, anything. So here's an example, the velocity of a thrown object is given by some function, meters per second, t is measured in seconds after the object begins moving, and we have some information about the height of the object, and well here's the problem, we want to find the height function that gives the object's height t seconds after we launch the object. So first thing to note is that we want a function that gives the object's height. And that is something that should be measured in meters, how do you know? Well, part of it is the only height that you're given is actually measured in meters. So whatever the height is, it's going to be something that's measured in meters. Now this is calculus, so we don't have too many choices of what we can look at. We can look at the derivative, and we found that the derivative has units meters per second squared, which is not a height. We measure height in meters, not in meters per second squared. So whatever the derivative tells us, it's not the height. On the other hand, we can look at the units of the antiderivative, and the units of the antiderivative are going to be meters, and that could potentially be the height. And what that suggests is that the height has something to do with the antiderivative of V of t. So let's go ahead and find that first. So my antiderivative of V of t, after all the dust settles, is going to look something like this. And here's why that added constant is important, because there's an infinite number of antiderivatives. But there's only one height function for this object. Also a very important thing to note, finding the antiderivative is step one out of however many steps this is going to be. If you don't get the antiderivative correct, then everything that follows from that point is wrong. So you want as much reassurance as you can get that you've done the antiderivative correctly. Well, how can you do that? Well, remember, the claim is that if I find an antiderivative, the derivative of this function should be what I started with. Well, we have just spent most of a semester learning how to differentiate, so it should only take a couple of seconds to see that if I differentiate this, I do actually get 150 minus 32t. So I have some assurance that the first step in the problem is done correctly. Well, again, all this really tells me is that the height function looks something like this. But until I identify what that constant is, I don't have a function. So to determine what that constant is, I can use the fact that I have some information about when the object is launched. So the object is launched from a height of 100 meters. I know something about the height when the object is launched. Well, I actually need to know a t value. And thinking about that, when the object is launched, that is the same as t equals 0, because t is the amount of time after the object begins moving. So I know the height at t equals 0, and I know that that height is going to be 100 meters. So that tells me that h of 0 is 100, and I can find c from this information. So I'll substitute that in, h of t looks like that. If I let t equals 0, h of 0 should be 100. And after all the dust settles, I find that c is equal to 100. And so that tells me that h of t, 150t minus 16t squared. And I know that c for the specific problem that we have is going to be equal to 100. Now, the problem is not actually complete until we identify the units of h of t. Unfortunately, we already know what those units are. H of t is going to give the object's height, and we're measuring height in meters. So we do want to indicate that in our final answer.