 Hello, and welcome to Screencast today about the fundamental theorem of calculus. And this screencast is going to be focusing on power functions. Okay, so the fundamental theorem of calculus, as I wrote out here, says that if f is a continuous function on the interval from a to b, and big f is any anti-derivative of f, okay? Because remember these come in families. Then the integral from a to b of little f of x dx is equal to big f of b, minus big f of a. Okay, so again we're going to be focusing here on the idea of the anti-derivative. So that is working our way backwards from the derivative, okay? And we can always check our answers because what you can do is just take the derivative of your answer and make sure you get back to where you started. All right, my first example here today says evaluate the integral from negative one to three of 3x squared minus 2x plus pi. And all of this is going to be done dx. So remember that just means then we are integrating with respect to x. So x is going to be our variable, okay? All right, we know f is a continuous function because it's a polynomial. So we don't have to worry about anything going funky there. So now we just need to find our anti-derivative, okay? So let's take it piece by piece because remember we can break this integral up. Or we can just do like we did with the derivative whenever we have sums and differences, you can just do the anti-derivative of each piece, okay? So the anti-derivative of 3x squared, so you've got to ask yourself, what function do I take the derivative of that will give me 3x squared, okay? Well, the power rule, remember, works in reverse. So that's going to be, we need to, excuse me, leave our three out front. We add one this time to our exponent. And instead of multiplying that, we're going to divide it, okay? And then hopefully you notice these trees will cancel. So the anti-derivative of 3x squared is x cubed, okay? We can check that because the derivative of x cubed is 3x squared. All right, next piece, so minus, the two again comes along because it's a constant with our constant rule. Then we'd have x, technically this is to the first power. So if I add one, I get two and I divide by that, I also get two. So again, those coefficients cancel. So the derivative, sorry, the anti-derivative of negative 2x is x squared. And that's because the derivative of x, negative x squared is negative 2x. All right, and the last one, the constant here plus pi. Well, so now I got to think to ourselves, all right, so what kind of a function must have a constant slope? Because remember, that was our derivative of pi. Well, that's got to be a linear function. So that's going to be the function pi x. And again, if you take the derivative of pi x, do you get back to pi? Absolutely. Okay, so let me go ahead and pretty this up a little bit. Let me throw my endpoints on here. So we've got negative one to three. So this anti-derivative then is x cubed minus x squared plus pi x. And again, I'm going from negative one to three. Now you notice the dx went away, the integral sign also went away. So these two pieces then are basically the integration idea. But those two pieces do not get carried along. They drop out whenever you do the anti-derivative. Okay, plugging in our numbers then. So we're going to have three cubed minus three squared plus pi times three. That's going to be our f of b, basically. So this is the anti-derivative evaluated at our top end point minus. Okay, now we got definitely got to watch our negatives on this one. So I'm going to do negative one cubed minus negative one squared. And then plus pi times negative one. So this big quantity here is my big f of a. So that's my anti-derivative evaluated at my bottom end point of negative one. Okay, all right, so anyway, crunch out some of these numbers. Let's see, so this gives us 27 minus nine plus three pi, okay. You want to leave that as exact, we can, or we can approximate it later. Then we have, let's see, negative one cubed will give us negative one. Minus negative one squared, though, gives us a positive one. And then that's going to be minus pi, okay. So that gives us, let's see, 18 plus three pi. And then minus negative two minus pi. Combining our like terms, that gives us a negative and a negative. So this negative here has to be distributed, which is why I keep putting these parentheses in here. So that's going to end up giving me 20 plus four pi. So that's my exact value of my integral. If you just want an approximate value, though, this gives us about 32.57, if I crunched everything correctly, okay, fantastic. All right, next example, again, evaluate the integral. This one looks a little bit uglier, though. So we've got the integral from two to five of one over the square root of p dp, okay. So again, this dp here tells me that I am doing my integral with respect to p, okay. Well, if we were doing the derivative of this function, I think it was smart to rewrite this first. So same thing happens when you do integrals. So we're going to have the integral from two to five. So those numbers don't change at all. And remembering the rules from algebra, we can rewrite this function as p to the negative one-half. And then again, that dp gets brought along, okay. So now this looks more like a power. So I'm going to add one to my power. So that gives me negative one-half plus one, so that's a positive one-half. And then I'm going to divide by that number, so I'm dividing by a half. And that's my anti-derivative. Again, I know this because if I do the derivative of this function, I'll get back to where I started. Okay, I'm evaluating this from two to five. And if we pretty that up a little bit, that's two p. Actually, let's go and call it two square root of p from two to five. So then that's going to give us two square root of five minus two square roots of two. Can't really do much with those. But if you want that to be an approximate value, that gives us about 1.64, okay. Now, you notice this one had an interval that was very well chosen, okay. Because if you think about this function here, one over the square root of p, that function is not continuous at zero because it's not defined. But you notice zero is not part of my interval on the integral, okay. So that was very well chosen and just always kind of make sure to double check those things as you're going. All right, thank you for watching.