 Alright, let's talk a little bit more about the Fundamental Theorem of Calculus, and we'll start off in a strange way. Here's how to get a wrong answer when applying the Fundamental Theorem of Calculus. So here's a problem. Find the area between y equals sine x and the x-axis over the interval 0 less than or equal to x, less than or equal to 2 pi. And it's easier to take a look at the problem and think, ah, well, here I have top, I have bottom, I have left, I have right. I've completely described the integral. Let's go ahead and set down something and get the wrong answer. And so we can do that by setting down the definite integral. So again, I have my definite integral, my function sine x, x-axis, little portion of that from 0 lower limit, 2, 2 pi, upper limit. And there's my definite integral, and now I know how to apply the Fundamental Theorem of Calculus. I'll find an anti-derivative of sine of x. So how about negative cosine of x is an anti-derivative. And I can substitute that in and evaluate. So the definite integral is 0 to 2 pi sine x. Here is an anti-derivative of sine x. I'm going to evaluate it with 0, evaluate that at 2 pi. And so that's going to be cosine 2 pi minus negative cosine of 0. And I go through all the algebra and the trigonometry. And after all the dust settles, I get a final answer of 0. And I circle it and say, I'm done with the problem. Well, actually, that's kind of a strange answer. It doesn't make a lot of sense because you know there is an area between y equals sine of x and the x-axis. You've seen the function y equals sine of x. It seems a little strange that our computation gives us an area of 0. And the question you should ask at this point, what happened? Well, let's try and get the correct answer. So I'm looking for the area between y equals sine of x and the x-axis over some interval. Let's graph that region. So we know what the graph of y equals sine of x looks like. And the interval 0 to 2 pi is one complete cycle of the sine function. So the area of interest is going to look like this. So I'll graph the region. I do want to sketch some representative rectangles. And so here's a representative rectangle in here. Here's another one a little bit later on. And the thing to notice is that the representative rectangles undergo a very important change. In this first interval here, from 0 up to pi, the height of the rectangles is, well, height is always top minus bottom. So here, in this first region, the top is our sine curve. The bottom is our x-axis. So the height of the rectangle will be sine of x minus 0. So the height of our rectangles in that first region, sine x minus 0, otherwise known as just sine x. On the other hand, if I look at the second part, this interval from pi up to 2 pi, height is still top minus bottom. But in this case, the top is the x-axis. And now the bottom is our sine curve. And so the height of the representative rectangles is going to be 0 top minus sine of x. That's our bottom. And the height is going to be negative sine of x. In general, every time a boundary curve changes, you have to write a new integral. So in this first region, our boundary curves were top sine of x bottom x-axis. But in the second region, our top curve changed. It became the x-axis. Our bottom curve changed. It became the y equals sine of x graph. And our boundary curves changed, which means that we have to consider writing a new integral. So let's go ahead and write those integrals down. So let's see. I have two regions. In this region, the boundary curve is always top sine of x bottom x-axis. So let's focus on this first region. So it's this interval. And the area, again, is going to be found by looking at those representative rectangles. Height is top sine x minus bottom 0. So the height is sine x. The width is a tiny portion of the x-axis. We're going to sum those rectangles up from 0 up to x equal to pi. And so there's my first integral. Quick check. Our differential dx. Our x values 0, x equals pi. That is the correct region of integration. So that's the area of the first region. But I actually want the area of the entire interval. So let's go ahead and take a look at that second interval. And so here for this part, my height top is x-axis. Bottom is sine of x. The height is going to be 0 minus sine of x, or just negative sine of x. Our width is going to be a tiny portion of the x-axis. That's going to be dx. So the area of the rectangle minus sine of x height times dx width. And so there's our areas. We're going to sum those areas up from differential controls x equal to pi up to x equals 2 pi. And so those are our limits. And now I need the antiderivative of sine of x and the antiderivative of minus sine of x. Well, we've already figured out antiderivative of sine x is minus cosine x, antiderivative of minus sine is cosine. And we're evaluating these antiderivatives from 0 to pi, or from pi to 2 pi. And so I'll evaluate those. Again, sometimes this is the highest part of the problem, because there's a lot of arithmetic and trigonometry and algebra involved. But it's generally not too difficult. So after all the dust settles, I get an area equal to 4. And that is actually what I would expect, that the area of the region is going to be some non-zero, non-negative number.