Learn about Definite integrals - part 4

Khan Academy Presents: Examples of using definite integrals to find the area under a curve

Welcome back, I'm not going to use definite integrals to figure out the areas under a bunch of curves and if we have time maybe even between some curves. So let me write down the fundamental theorem of calculus, I know I covered it really fast in the last presentation. Just to make sure you understand this formula the last couple of presentations were really to give you intuition for this exact formula. Let's say that big F prime of X is equal to f of x, right? That’s also like saying that’s equivalent to saying that f of x, big F of x is equal to the antiderivative of f of x, right? And the reason why well let's just say it's one of the possible antiderivatives of f of x, right because there's always a constant term here and you're not sure whether it is and this is why people tend to use this standard because we know that f of x is the derivative of big F prime of x. Big F of x is just one of the antiderivative of f and x so this is a little bit not true but I think you get the idea. But the fundamental theorem of calculus tells us if this top line is true then the definite integral from A to B of f of x, dx is equal to its antiderivative evaluated at B minus its antiderivative evaluated at A and I know I said here that F, big F isn’t the only antiderivative, right because you could have any constant to this and that would also be the antiderivative. But when you subtract here the constants will cancel out, so it really doesn’t matter which of the constants you pick the constant actually doesn’t matter. So that’s why I actually said the antiderivative. But let's apply this, because you might be confused right now. So let's draw a graph. There you go and look how kind of straight that is. Draw the x axis, not perfect but it will do. It will do and let's say that my f of x is equal to x squared plus one. So f of x looks like this, this is one, this is one so it will start with one and it’ll just be a parabola and look let's see if I could draw this. I've done worst, okay. So that’s f of x. It is parabola y intercepted one. And let's say I want to figure out the area under the curve between the curve and really the x axis. Let's say I want to figure out the area between the curve and the x axis from x equals negative one to I don’t know x equals three. So this is the area I want to figure out, I'm going to shade it in, so this is the area. All of this stuff, I want to figure out this area and you could imagine before you knew calculus figuring out an area of something with the curve is kind of top boundary. It would have been very difficult but we will now use a fundamental theorem of calculus and hopefully you have intuition of why this works and how the integral is really just the sum of a bunch of little, little small squares with infinitely small basis but if you watch the last videos hopefully that that hit the point home. But we’ll now we’ll just mechanically compute because actually understanding it is it would be hard than just doing it. But this is mechanically computed. So we are essentially just going to figure out the integral from minus one to three of f of x which is x squared plus one, dx. Well what's the antiderivative of the x squared plus one? Well this is equals the antiderivative so it's just x to the third you could say one third x to the third or actually third over three plus x, right the derivative X is one. And then we don’t have to worry about plus C because were going to subtract out the C’s. You’ll see and you could get the point. It doesn’t matter you could picked an arbitrary C right here and it will just cancel out. We’re going to evaluate that at three and negative one and we’re going to subtract out big f of negative one from F of three, this is just the notation that you use. You figure out the antiderivative and you say where you're going to evaluate it and then this is the equal two—so if I evaluate three, three to the third power is well, that’s 27, 27 divided by three is nine and then nine plus three is 12, right? This is just big f of three right, kind of figure out the end but this is big F. This is big F of x, you can kind of notice this is big F of x but not to be confuse with small cursive f of x. This is big F of X, so this is big F of three and then from that we’ll subtract big F of negative one minus big F of negative one. And if we put minus one here let's see minus one to the third power is minus one. So it's minus one third and then plus minus one, right. So minus one third plus minus one I think that equals minus four thirds, correct. I think so, maybe I'm making a mistake with negative signs. Minus one third, minus four third I'm going to subtract that right. So if I'm subtracting minus four thirds it's the same thing as adding minus four thirds and then we have our answer to it. We actually don’t even have that. I mean it's 12 and four thirds, whatever units, square units 12 and four thirds square units we can write this as an x number as well. Let's do, let's do another one and I’ll do the slight, slight variation. Okay, so let me draw again some coordinates. I don’t have enough time to do in this video but I’ll try. I always try. And let's say I have f of x is equal to the square root of x right, so it looks something like this. That’s actually pretty nice looking kind of side ways brave I think. This is f of x. And let's say I have another function g of x which equals x squared. So g of x is actually going to look something like this. I was doing well and then something happened and of course it will continue on this side as well, right because it is defined for negative numbers. But anyway, my question to you or my question to myself really is what is the area between the curves where they intersect right here. So what is this area? Well, the first thing you have to figure out is just what are the boundary points? What is this point and what is this point? Well this point I think it's pretty clear, it's a zero-zero right. They both intersect in zero-zero and even at this point you could probably do it from intuition. But if you don’t I guess want to do it through intuition you could just set the two equations equal to each other as you could say x squared is equal to the square root of x, right? And then you could do a bunch of things you could square both sides or well actually this is the same thing in doing by it intuition. But I think it's pretty obvious that the only places where s squared is equal to the square root of s are the points x equal zero which you already known and x equals one. So this is the point one comma one which is due for both of them. It is more algebra so I don’t go into that too much detail I've kind of running out of time. So we want to figure out the area between these two curves. So what we can do is maybe you want to pause it and think about it yourself. We can figure out the area under this, under the great curve, we could figure out this area so if we want to figure out this is the boundary between zero and one, we could figure out this area and then we could figure out the entire area under the green curve separately and then we could subtract the difference which is exactly how we’re going to do it in the next video because I have run out of time.