 Welcome back everyone to lecture 47 or lecture one, whether you're a calc one student or a calc two student at Southern Utah University. I wanted to talk about more about the fundamental theorem of calculus and how we can use it to calculate areas under a curve. So imagine we have this function f of x equals x squared minus four, and we want to calculate the area from zero to two. You can see the illustration displayed on the slide right here. And although this illustration is not necessary to calculate the area of the region, it can be very helpful to draw the picture so that we have a clear idea of what we're looking for. Cause especially in this context, when they actually find the area of a region, one thing to remember is that area is always considered a positive quantity. But yet this region you see right here falls below the x-axis. And so it'll actually be considered a negative area when we calculate, we'll get a negative area when we calculate with the definite integral. The definite integral is going to give us the net area, area above the x-axis is considered positive but area below the x-axis is considered negative. For many applications, this is a very preferable method like in science and chemistry engineering and the like. But if we're trying to find geometric area, we do want to make this be positive. And so we're going to have to make sure we take absolute values of our final result here. So in the meanwhile though, we know that the area is going to equal the integral from zero to two. These are the bounds that we were given of our function x squared minus four dx. And again, we're doing to have to take some absolute values of this because we're anticipating a negative area, but the area is going to equal this integral right here. And so using the fundamental number calculus, we can calculate this area by finding an anti-derivative x squared minus four, which will look like x cubed over three minus four x as we range from zero to two. Plugging in that zero and two, we're going to get the following. We will get two cubed over three minus four times, four times two, excuse me. That's the first part when we plug in the two. When we plug in the zero, we're going to end up with zero cubed over three minus four times zero. And you're going to notice this a lot that when we plug in zero, we often will just make the whole thing disappear. So we're going to love it when we integrate at the bound zero. That's sort of our favorite number to go from. Now continue on with the arithmetic. We'll get two cubed, which is eight, eight over three, four times two is also an eight. If you factor out the eight, you're left with one third minus one, which the one you might prefer to write it as three over three. And so taking that difference, you end up with a eight times negative two thirds. That is to say we get negative 16 over three. And remember that we said that we were taking absolute values, the whole time I wasn't writing the absolute value, you notice the whole time. And you can kind of get away without doing that. The idea here is if you're calculating these areas, you end up with something negative at the end, just make sure to take the absolute value. And so that we get in the end, that the area under the curve is a positive 16 thirds squared units of area. Looking at another example, this time let's find the area under the curve. That is we want to find the area between the x-axis and the function given x squared minus three x. This is what we mean by under the curve when you take the area between a function and the x-axis. This is kind of a weird way of saying it because when you look at the graph, the curve actually is below the x-axis. So it looks at the areas above. And that's actually why we're getting negative area here again, make sure to take absolute values when we're done. So to find this area, we're going to integrate from one to three our function x squared minus three x dx. And again, by the funnel of theorem calculus, we can use an anti-derivative to help us out here. Anti-derivative of x cubed minus three x would look like x cubed. What did I say x cubed earlier? Maybe I'm putting the carton from the horse there. We want to take the anti-derivative of x squared minus three x. So we're going to get x cubed over three minus three x squared over two divided by one and three. Ratio-phobics, maybe pause and fast forward to the next part of this video because there will be a lot of fractions coming forward here. We're going to get three cubed over three minus three times three squared over two when we plug in the three. And then we subtract from that, we're going to get one cubed over three minus three times one squared over two. In which case, we're going to end up with a 27 over three minus 27 over two. One thing to watch out for when you do these definite integrals is that everything in the second part of the function when you plug in the one is going to be subtracted one third minus three halves. Like so make sure you subtract all of these. Now we could simplify 27 over three but since we already have a one third right here, I'm just going to leave it as it is and we're going to combine these common denominators right there. So you're going to end up with 27 thirds minus a third which is 26 thirds. And then you're going to have a negative 27 plus, a negative 27 halves plus three halves. And so that's going to give us a negative 24 halves right there, all right? And so we can simplify the 24 halves there that of course just become negative 12. Right there, but we do have to add these together. So finding common denominator, we'll get three so we can make this 36 over three. And so finishing up this calculation, we end up with a negative 10 over three. That of course gives us the integral. We're trying to find the area. So if we want to find the area here, the area is going to equal the absolute value of all that. So we end up with 10 thirds right there. Always make your area be a positive quantity. If they ask you just to calculate the integral, then leave the integral negative if it turns out to be negative, but then ask you to find the area of the region, do take it to be a positive, take absolute values of that. And so let's look at one more example of this. This time let's use a trigonometric function cosine, find the area under the cosine curve from zero to B, right? Well, what if we leave B as some unknown, right? If you wanted to do that, we're looking for the integral from zero to B of cosine of x dx. In this situation by the fundamental theorem of calculus, we're going to look for an anti-derivative. This we have to be very careful about. While the derivative of cosine is negative sign, the anti-derivative of cosine is a positive sign because the derivative of sine is cosine. Be very careful about the signs right there. We're going to get zero to B. And so when you plug that in there, you're going to get sine of B minus sine of zero, right? And then, well, sine of zero is just zero. So that actually is just going to disappear. And so the area under the cosine function is just going to be cosine of that number B. So if we wanted to do the specific value, like you see in the illustration here, if we wanted to go from zero to pi halves right here, cosine of x dx, by our calculation, we see that this is going to be, well, shoot, when did I switch to cosine right there? That should be sine of B, sorry. The area under the sine function, under the cosine function from zero to any value is just going to be sine of that value. And so this would give us sine of pi halves if we wanted to go all the way at the pi halves, which is going to equal one. So this region you see highlighted right here, it has an area of just one. So you've seen like the unit square before, which is just one square unit. That's what this is going on right here. This region underneath the cosine right here is just one square unit of area. It's kind of interesting there. And so this time the area was naturally positive, so we didn't have to really worry about absolute value. But when you work with area, make sure you're always taking absolute values of everything to make sure you always end up with a positive value. See you next time, everyone.