Definite integral evaluation
Area between a curve and the x-axis: negative area
What I've got here is the graph of y is equal to cosine of x. What I want to do is figure out the area under the curve y is equal to f of x, and above the x-axis. I'm going to do it over various intervals. So first, let's think about the area under the curve, between x is equal to 0 and x is equal to pi over 2. So we're talking about this area right over here. Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And remember, all this is, is kind of, it's reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. And that's kind of what this notation is trying to depict. But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is. Or what an antiderivative of cosine of x is. Evaluate it at pi over 2, and from that subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative? Well, we know if we take the derivative-- let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant, and still get cosine of x. Because the derivative of a constant is 0. This could be pi, this could be 5, this could be a million, this could be a googol, this could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we just have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case the constant is 0. So let's evaluate. So one way that we can denote this, the antiderivative cosine, or an antiderivative of cosine of x is sine of x. And we're going to evaluate at pi over 2. And from that, subtract and evaluate it at 0. So this is going to be equal to-- let me do it right over here. This is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting. Let's think about the area. Let's think about the area under the curve between, let's say pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative-- or an antiderivative of cosine of x is sine of x. Evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 pi over 2? If we visualize the unit circle really fast, 3 pi over 2 is going all the way 3/4 around the unit circle. So it's right over there. So sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We got a negative area. Now how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now this is just denoting this entire area. Going from 0 all the way 3 pi over 2. And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0. Which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under, all this orange area that I just outlined, is clearly not negative, or any area isn't a negative number. And the area isn't even 1. But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1, or another way to say it, your net area is negative 1. So it's taking the 1 region above the x-axis and subtracting the 2 below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get a negative number, that means that the net area is act-- that actually most of the area is below the x-axis. If we get 0, then that means it all nets out. And if you want to see a case that's 0, take the integral from 0 all the way all the way 2 pi. And this will evaluate to 0 because you have an area of 1 and another area of 1, but it nets out with this area of negative 2. Let's try that out. So if we go from 0 to 2 pi of cosine of x dx, this is going to be equal to sine of 2 pi minus sine of 0, which is equal to 0 minus 0, which is indeed equal to 0. Now clearly there was area here, but all of the area that was above the x-axis netted out with all the area that was below the x-axis.