 Hi, and welcome to this screencast where we're going to use anti-derivatives to find the distance traveled by a moving object. Let's recap what we know so far about finding distance traveled by a moving object. Mainly, what we know is that to do this we need to have a graph or table present that gives us the velocity of the object. Then we estimate the distance traveled by the object over a time interval by subdividing the time interval appropriately and then forming a rectangle sum, and then the sum of the areas present estimates the distance traveled. Now we said earlier and you read in the section that what we're doing here is going backwards from our usual method of derivative taking. Earlier in the course we learned that if we are given position and take a derivative we end up with velocity. Now we are given velocity and we're trying to find out something about position. So we're going backwards through the derivative process and we're going to call this backwards process anti-differentiation. Here's an important definition. If we're given a function g then the function f is said to be an anti-derivative of g if f prime equals g. So f is a function whose derivative is the given function g. And this is our situation here. We're given velocity which is the derivative of position and we need to work backwards to find position. We'd say in this language that position is an anti-derivative for velocity because taking the derivative of position gives velocity. We can use anti-derivatives to find distance traveled as follows. If we're given an object's velocity function as a formula then we first try to find an anti-derivative for velocity which will be another formula. Let's call that anti-derivative s of t and this s of t in some way represents the position of the object relative to a starting point because s prime is velocity. And so s of t has the right fingerprints to be the object's position. Then we'll simply compute s evaluated at the ending time minus s evaluated at the starting time. This gives us a difference in position which is another way of saying the distance traveled. So let's do a concrete example. Here's the function for velocity that we used in screencast 412. Now unbeknownst to you that function had a formula behind it namely v of t equals negative 1.5t plus 10 plus sine 2t. Now I claim that the following function is an anti-derivative for that velocity. Negative 0.75t squared plus 10t minus 1 half cosine of 2t. That is if I take the derivative of the second function I'll get my velocity function. Now on the next slide I'm going to verify this is indeed the case that if I take the derivative of this anti-derivative I end up with my velocity that I started with. And then I'm going to use the fact to find the distance traveled by the moving object on the interval from 0 to 5. So we were given a velocity function v of t equals negative 1.5t squared plus 10 plus sine 2t. I'm going to claim that the function s of t equals negative 0.75t squared plus 10t minus 1 half cosine 2t is an anti-derivative for that velocity. So all this requires is simply that I take the derivative of the thing that I claim is the anti-derivative and see if I end up with the velocity function. That would make the new function s of t an anti-derivative for velocity. So here I have set up the derivative of my anti-derivative and hopefully they will sort of cancel each other out and I'll end up with v of t. So let's chug through the derivative process here. So I'm going to take the derivative first of all of this first term and that would give me negative, there's a 2 times 0.75 times t. The derivative of 10t is 10. I have a minus 1 half and now the derivative of cosine 2t is going to be negative sine 2t. And then I have to use the chain rule and take the derivative of the 2t on the inside and that gives me a 2. Now let's just see what this shakes out to be. I have several cancellations I can do. The 1 half and the 2 cancel. I have a minus and a minus and makes a plus. And let's just keep working here. Negative 2 times 0.75 is negative 1.5. That's times a t. I was adding on a 10 and now all that's left over here is the sine of 2t. And that is indeed equal to what I said it should be equal to velocity. So this function s of t is actually an anti-derivative for velocity because its derivative is velocity. Now the point of finding this anti-derivative for velocity is that this is in some ways something like a position function for my moving object. And so I can use it to find the distance traveled. We're going to look at the distance traveled by this object over the interval from 0 to 5. So I'm going to take my position and figure out where it was at 5 and subtract off where it was at 0. And that difference there should give me the difference between the starting position and ending position. That's the total distance traveled. So now I have a formula for s, which I had on the other side. Let's carry that over to here. I'm going to put 5 in and put 0 in and then subtract. So s of 5 would be negative 0.75 times 5 squared plus 10 times 5 minus 1 half cosine of 2 times 5, which is 10. And now I'm going to subtract off and in this bracket is going to go s of 0. That would be negative 0.75 times 0 squared. That's just a 0 plus 10 times 0. That's also just a 0. Minus 1 half cosine of 2 times 0 and that's also a 0. A lot of things in that second set of brackets will go away for us. The first set of brackets will be negative 0.75 times 5 squared is about equal to, should make that approximately equal to negative 18.75. 10 times 5, of course, is 50. Minus 1 half cosine of 10 is not anything sort of natural. It just comes out to be negative 0.419536. That's accurate to six decimal places. And inside here I have 0, 0, and cosine of 0 is equal to 1. And so this becomes a negative 1 half inside the square brackets there. Now I'll skip all the arithmetic here and just say this comes out to be about 32.1695. That's in meters because this estimate here is of a distance travel. And if you remember back to screencast 4.1.2, we did some pretty high level estimates of that distance travel. For example, using 25 rectangles. And it was coming out to be pretty close to 32. So this is a very good estimate here. In fact, it's highly accurate. We could actually make this an exact quote unquote estimate by instead of writing this decimal expansion here, just keeping it at cosine of 10. So with antiderivatives, we have a way of actually finding the exact value of the distance traveled by this moving object. So with all that, why should we ever use estimates at all when finding distance traveled? When we can get a lot more accuracy using antiderivatives. Well, there are two main reasons. First of all, we have to have a formula for velocity in order to do any of this. And that just doesn't always happen. In fact, most functions in real life are given as graphs or tables of data or verbal descriptions. And formulas are very hard to come by sometimes. So this method of antiderivatives is not a method that applies in many cases. Second, even if we do have a formula for velocity, finding an antiderivative for that formula could be hard or even impossible. You might be wondering, rightly so, where did I even come up with the antiderivative for velocity that we saw in the previous slides? And that would be a good question. Guess and check would not be a very good approach. But systematic ways of finding derivatives can be very complicated. And much of second semester calculus is spent on techniques for finding these. And even so, again, some functions don't even have antiderivatives. So we can find distance traveled with great accuracy using antiderivatives in this process. But don't throw away the estimation techniques just yet.