 So why do we care about the anti-derivative? We'll take a look at some of its uses. Before we do that, it's useful to ask yourself, self, what are the units of the anti-derivative? So suppose capital F of X is an anti-derivative of F of X. Since the derivative of capital F of X is F of X, then F of X will be a rate of change of capital F of X. And to determine what's changing, we can then look at the units of capital F of X. For example, suppose V of T is the velocity in meters per second of an object. Let's find the units of the anti-derivative of V of T and provide a suitable interpretation. So one of the nice things about math is I might not know what something is, but I can at least name it. So let Y be the anti-derivative of V of T. Then the derivative of Y with respect to T will be V of T. And since V of T has units of meters per second, then the units of the derivative will be the units of Y over the units of T. That'll be meters over seconds. And this means the units of Y will be meters. And V of T will be the rate of change of something measured in meters with respect to time. So there's many things that we could measure in meters. For example, the length, width, and height of the object. The position of the object. The color of the object. No. The weight. No. The age. No. Now the other thing to keep in mind is that since V of T is a rate of change, we can focus on the properties that might change over time. And since V of T is the velocity of an object, then the thing that could change with respect to time is the position of the object. And so we might conclude that the units of the anti-derivative will be meters and this anti-derivative itself might be interpreted as the position of the object at a given time T. So let's consider a problem. We have an object thrown from a tower of some height and with upward velocity of some amount. And we know something about the upward velocity change. We want to find H of T the height of the object and determine when the object hits the ground. Now given all of this information, it's helpful to consider the units of H of T and since we've learned how to differentiate, we might consider the derivatives of H of T. H of T is the height of the object. And in this problem it seems that we're measuring height in meters. So H of T is going to have units of meters. H prime of T will have units of meters per second. H double prime of T will have units of meters per second squared. Now the given information has units of meters meters per second or meters per second squared. And what that tells us is that we have information about H of T, H prime of T, and H double prime of T. In particular, the tower 120 meters in height, well that's an H of T value because the units are the same as the units of H of T. So we have to figure out what value T is. And since T indicates the number of seconds after the object is thrown and the object itself is thrown from a tower 120 meters in height, that must mean that when we throw the object at T equals 0, the height must be 120 meters. And so that gives us our height at time 0 is 120 or in function notation H of 0 equals 120. Likewise, the object is thrown with an upward velocity of 14 meters per second. So again, at T equals 0 when the object is thrown its velocity is 14 meters per second. And so that tells us H prime of 0 is 40. Finally, we're given this information that the upward velocity decreases by 5 meters per second. Now this information is not associated with any specific time. It's not when the object was thrown. It's not the initial upward velocity. And so this looks like a second derivative at any value of T is equal to 5. Well, not quite. Note the use of the term decreases by. And this suggests that our velocity should be getting lower and lower. And what that translates into is that the second derivative should be negative. So H double prime of T should be minus 5. So let's see what we can find. We want to find H of T and so our new toy is anti-differentiation. So we know that H of T is the anti-derivative of H prime of T. Unfortunately, I don't know what H prime of T is. I know H prime of 0, but I don't know H prime of T. And until I know what functioned anti-differentiation, I can't use this. So can we find H prime of T? And the answer to that is sure, why not? So I know H prime of T will be the anti-derivative of H double prime of T, and I do know what H double prime of T is. It's negative 5. And so I'll find the anti-derivative of negative 5, which will be minus 5 T plus. Don't forget that all important constant. And here's where it's important to know that H prime of 0 is equal to 40. And so I can solve for that constant. And I get the value C is equal to 40. And H prime of T is minus 5 T plus 40 meters per second. And now I can find H of T by anti-differentiation. H of T will be the anti-derivative of H prime of T. I know what H prime of T is, so I'll find the anti-derivative. And again, don't forget that plus C, which we can find because we know that H of 0 is equal to 120. And so I have my formula for H of T minus 5 halves T squared plus 40 T plus 120 meters. And I want to know when the object hits the ground. So that's going to be when the height is 0. And so this gives us the equation and get two solutions. But since the negative value doesn't make any sense, only the positive solution is relevant. And so the object will hit the ground about 18.583 seconds after it's thrown.