 So one of the reasons that anti-derivatives are important is that it's often easier to measure the derivative of a quantity than to measure the quantity itself. For example, suppose I have an object moving at a certain velocity, and we enter a tube with a length of 20 meters, and while inside the tube the object's velocity increases by some amount to t plus 1 meters per second squared. And we want to determine the object's velocity when it leaves the tube. And from a physical standpoint, one of the things that's important is that acceleration is easy to measure. Velocity is not as easy to measure. So let's take a look at that. v of t, our velocity, is going to be measured in meters per second. And a of t, our acceleration, is going to be measured in meters per second squared. And so that tells me that v of t is an anti-derivative of a of t. Well, I have a formula for a of t, so I can find v of t by anti-differentiation. How about this constant of anti-differentiation? Since we have an initial velocity of 5 meters per second, then I know v of 0 must be 5, and so this tells us what c is equal to. And that gives me my formula for the velocity of the object. Now we want to know the velocity when the object leaves the tube, so we need to know what t is at that instant. Well, we're given a length of 20 meters, so let's find the distance s of t that the object has traveled. Since s of t is measured in meters, and v of t is measured in meters per second, that means that s of t is the anti-derivative of velocity. So we'll find that. And since t is the amount of time since the object entered the tube, then s of 0 has to be 0. And that tells us that a constant also is equal to 0, and so I have my formula for s of t, the object's location within the tube. We want to find t when the object leaves the tube, so we need to solve s of t equals 20. And this is a cubic equation. And unless you know Cardano's formula, you won't be able to solve this equation algebraically. So we have a couple options, and let's go through them. First, we can use a computer algebra system. Second, we might do some calculus and use the intermediate value theorem. Or third, we might use even more calculus and use Newton's method. So we can use a computer algebra system like Wolfram Alpha. Again, I'm not a paid spokesperson for Wolfram Alpha, but if they want to send some money my way, I'm happy to accept. So we'll ask it to solve 20 equals 1 third t cubed plus 1 half t squared plus 5 t. And we'll hit Enter and let Wolfram Alpha compute for a minute. And our answer comes up as this rather frightening expression. Fortunately, we can find an approximate form of the expression, and that'll give us our approximate solution. Of course, if the zombie apocalypse hits and the internet shuts down, what are we going to do? So here you are in the middle of the zombie apocalypse, and you decide you need to solve this calculus problem. Well, because this is a continuous function, we can use the intermediate value theorem. So we'll try it a couple of t values. At t equals 2, s of t is. At t equals 3, s of t is. Since we want to solve s of t equals 20, we note that at 2 we're too small, and at 3 we're too big, so some place in the middle will be just right. And so we'll guess something in the middle. Now, if we were a computer, Our next guess might be exactly halfway between 2 and 3. But if we're a thinking human being, we might notice that 2 seems to be a little closer to the solution. So we may take a guess at something a little closer to, say at 2.4. And we find that at 2.4 we're still too small, so somewhere between 2.4 and 3 we're going to have just right. And again, we might suspect that our solution is closer to 2.4 than it is to 3. So maybe we'll guess 2.45, which is too big, but we're awfully close. So let's drop that down to 2.44, still too big. So maybe we'll try 2.435. So we know there's a solution between 2.435 and 2.44. And we'll stop at this point because it's obvious that we've run out of room. Remember the general question of how accurately do we want the answer is not something that can be answered easily by the mathematics. We have to make our best judgment call as to how accurate is good enough. And in this case, we'll say that T approximately 2.44 seems to be a good answer. And why not? Let's see what happens when we use Newton's method. So we're trying to solve our equation. So we need to make that an expression equal to zero. So we can phrase this as a problem of where the graph of a function crosses the axis. And so Newton's method uses a series of successive approximations. And we find the tangent line through T equals 2 has slope 11 and crosses the T axis at T about 2.48 mumble. If we go to that point, the tangent line through that point has slope 13 mumble and crosses the T axis at a certain location. And if we go to that point, we find a crossing of the T axis at 2. mumble. And if we try to go any further, we don't get any further improvements on our accuracy. So the best we can find is that T is around 2.4 mumble. So whether we use Wolfram Alpha, continuity in the intermediate value theorem, or Newton's method, we find the object reaches the end of the tube after about 2.44 seconds. And since we have V of T, we'll substitute that in to find the object's velocity when it leaves the tube.