 All right then, let's look at some applications of the anti-derivative process. We're gonna look at something which is commonly referred to as the initial value problem. Initial value problem. Cause what we're asked to do is we have to first find an anti-derivative, but what we've been doing before is we found the general anti-derivative. What we wanna do now is find a particular anti-derivative because what we know, we wanna find our function F. What we know about it is we have its derivative. We also know it goes through a specific point and that's what we mean by this initial value. So as sort of like an analogy here, we know how fast something's going and we know where it's located so therefore we can compute its distance function. Let's say I was trying to drive from Cedar City, Utah where I live to visit some relatives in Boise, Idaho which I do on occasion. And so my relatives give me a call and of course I wouldn't answer it cause I'm driving but maybe my wife answers the phone and they ask us or my wife will put it on speakerphone so we can all talk or something hands-free, totally safe there. But they ask us, oh, where are you? And we're like, oh yeah, we're driving 80 miles per hour. We'll be there soon. And it's like, okay, that's great, but where are you? We don't, knowing your speed is not enough to know when we're gonna get there. It's like, oh, we're in Salt Lake right now. It's like, okay, you're driving that speed in Salt Lake and I know Salt Lake speed limits not 80 miles per hour but it's Utah. If you're going 80 miles per hour, you're going slow. But anyways, if you know the speed and location then you could determine, oh, okay, you're gonna be here in like five hours. Okay, okay, that's good to know. So we could determine where you'll be if we know your speed and your location. So for this initial value problem to find f of x we first are gonna calculate the antiderivative of the derivative f prime of x. Well, we know what f prime of x is. It's this six x squared plus four, dx. And using the antiderivative power rule and other properties we've seen, the antiderivative will look like six x cubed, excuse me, the power goes up, divide by three plus four x plus a constant. Don't forget that constant there. If we simplify the fraction, three goes into six two times we get two x cubed plus four x plus c. This is equal to f of x. But this is the issue that my family member was having. If I don't know where you're located then I don't know how to finish this problem. And this is where the initial value comes into play. Because this initial value one one is on the graph that means that when x equals one the y-corner will be one. We have this relationship that f of one equals one. If we plug that into this equation right here we then see that f of one equals one but we can also plug in one for x. So we get two times one cubed plus four times one plus a constant. I'm gonna slide this up a little bit. If we simplify, well the left hand side's still one, the right hand side we get two plus four plus c. This will equal six plus c. And so let's subtract six from both sides here. We'll end up with that c equals negative five. And so that's then our function. f of x equals two x cubed plus four x minus five. And so we can determine the exact value of that plus c. And if we had forgotten the plus c we'd have the wrong function right now. Let's look at another example. So suppose a particle moves in a straight line has acceleration given by the acceleration function a of t equals six t plus four. We also know it's initial velocity. The velocity at time zero is negative six centimeters per second. We also know it's initial position. It's located at nine centimeters at the beginning of the problem. Let's find s of t. Well the position function, how is it related to velocity and acceleration? Well velocity is the derivative of position which means positions the anti-derivative velocity. And acceleration is the derivative of velocity which means velocity is the anti-derivative acceleration. And as acceleration is the second derivative of position that means position is the second anti-derivative of acceleration. So we're gonna start off by finding the velocity function. Let's do that first. So velocity is gonna equal the anti-derivative of acceleration which by the formula we're given the acceleration is six t plus four centimeters per second squared. We're gonna integrate that thing for which we're gonna get six t squared over two plus four t plus a constant. We see that the velocity is gonna equal because two goes into six three times three t squared plus four t plus c. We have that c again. What are we gonna do with c? Well like we did on the last example because we know the initial velocity we can plug that in for the velocity right here and we can solve for this constant. v of zero is equal to negative six. Well then we plug in zero for all the t values. So we get three times zero squared plus four times zero. It's always generous when they give you zero there because the right hand side will simplify just to be c because of all the zeros there. So c is negative six and therefore our velocity function v of t will equal three t squared plus four t minus six. And now that we have the velocity function we can then look for position because position s of t will equal the integral of velocity. It's the anti-derivative of velocity. And so taking the anti-derivative of this three t squared plus four t minus six d t by the power rule we're gonna get three t cubed over three plus four t squared over two minus six t plus another constant a plus c there. I'm going to just use the same plus c as before even though we already had a plus c but again it's just a constant we have to figure out right now. And if you remember from what we saw above the initial position was s of zero equals nine centimeters. And so we set this equal to I guess I should simplify this thing a little bit first. The threes there cancel and then two goes into four two times. So we end up with a zero cubed plus two times zero squared minus six times zero plus c. And again because all those zeros the right hand side again becomes c so c is nine. And so to finish this thing we end up with our position function s of t it will equal t cubed plus two times t squared minus six times t plus nine. And this gives us our position function that if we know how long the particles been moving in t seconds then we'll know it's been moving s centimeters like here. And so we can use anti-derivatives to solve many physics problems, economics problems other examples we're not seeing right here but it is important that we know that plus c. The plus c is by forgetting it we're assuming the initial value is zero and when the initial value is not zero that can be a problem. So thanks for watching this video here today. Lecture 41 in our series about anti-derivatives. If you have been liking our videos and liking these calculus lectures so far please like them. Subscribe if you wanna see some more updates in the future if you wanna see more videos please post any comments. If you have any questions about any of the things you see here just post them in the comments below and I'll be glad to answer them in the future. I will see you next time, bye.