 Welcome back everyone to lecture 47. We're talking about the Funnel of Theorem of Calculus part two. And as we saw previously, note that the Funnel of Theorem of Calculus part two tells us that if we take the definite integral from a to b of a function f of x dx, this will equal f of x evaluated at a and b where capital f of x is some anti-derivative, any anti-derivative of the function little f right here. So what we wanna do is do some calculations of these definite integrals using the Funnel of Theorem of Calculus. And so what this is gonna do is it's gonna require that we look for an anti-derivative of the various functions. Let's start with this one right here, e to the x. To find an anti-derivative capital f of x, we need to find a function whose derivative is e to the x. Well, since e to the x is its own derivative, it also serves as its own anti-derivative. Now the general anti-derivative would be e to the x plus c, but as we've talked about before, when it comes to definite integrals, you don't need the plus c because it just subtracts itself out anyways. So by the Funnel of Theorem of Calculus, the integral from one to three of e to the x dx, this is equal to e to the x evaluated at one and three, which this notation means we take e cubed minus e to the first, which really gets e cubed minus e. And as an exact answer, this is what we get, we get approximate if we want to, but exact answers will be sufficient for us. The area under this curve would be e cubed minus e. For the next example here, let us integrate from three to six, the function dx over x. Now sometimes, I mean, because this dx right here, this differential, is actually this, it's a factor that shows up in this definite integral. The dx is the limit as delta x. What that's like, when it comes to this dx, it's important to remember that the dx right here is the limit of the delta x as the number of rectangles is approaching infinity right here. And so this measures the width, the width of our rectangles. We don't want to forget about it. It's important in terms of our calculations. And again, especially as we go more and more into calculus two and applications of the integral, it's important that we keep track of that dx because it is a physical quantity that's measuring the width of these rectangles and things associated to rectangles. Again, we'll see all of this in the future. And so because of the length times width formula of rectangles at x multiplicatively, and so oftentimes we write dx over x would really mean one over x dx like so. And so to continue with this calculation, if we want to use the fundamental theorem of calculus, we want to find an anti-derivative of our function. And so this we could write as the integral right of one over x dx. And we've seen in the past that there is a function whose derivative is one over x and it's gonna be the natural log of the absolute value of x plus a constant. You want to remember to put absolute values on your natural log. That gives you the most correct anti-derivative. But again, we just need to have an anti-derivative. It doesn't matter exactly which one it is. So the plus c is okay to remove and that's what we're gonna do. So by the fundamental theorem of calculus part two, an anti-derivative of one over x, we can use the natural log of the absolute value of x as we go from three to six. We're gonna plug in six and three. So we get the natural log of six minus the natural log of three. You'll notice that both six and three are positive numbers. So if you had forgotten the absolute value inside of the natural log, you actually would have gotten away with it. I mean, you would have been right for the wrong reason and I guess I can live with that, right? So we get the natural log of six minus the natural log of three. By logarithmic properties, you can combine that together to get you the natural log of six over three. The difference of a log of logs is equal to a log of a quotient and that then simplifies to be the natural log of two as the area of the curve. And again, exact answers are okay right here. We don't need to worry about a decimal approximations. Probably anyone in their duck could plug that into a calculator. Looking at some other examples right here, let's look at the integral from one to two of four T cubed DT. Because we have to do these anti-derivatives, we don't usually like to segregate, oh, I did it, indefinite integral right here and then indefinite integral right here. When it comes to these calculations using the fundamental theorem, we usually just do them in line. So I first kind of think of this as I'm going to do a definite integral and I essentially sort of like nor, kind of ignore these bounds right here. If I was looking for the definite integral, right, we would be doing by the power rule, we get four over four T to the fourth. But since it's not actually a definite, it's not an indefinite integral, we do have to page into these bounds. We're going to get this one to two right here. Simplifying four divided by four of course is a one, so that goes away. And so we end up with two to the fourth minus one to the fourth. And this provides us the most difficult part of a lot of these anti-derivatives. It's called arithmetic, right? The scourge of many, many children, right? Learning how to do five digit multiplication. It can be a challenge right here. Feel free to use a calculator, these type of, these type of step, this step right here with the arithmetic. This one's not so bad here. Two to the fourth of course is 16. One to the fourth, well one to anything is going to be one. And 16 take away one is 15. So we've, and remember these calculations, we are calculating the area under the curve. So we're looking at the area under the curve, y equals four t cubed as we go from one to two. That area is gonna be equal to 15. And for this last example here, if we wanna integrate from two to five, the polynomial, six x squared minus three x plus five, by properties of integrals, we can actually break this up into three different integrals. We can take the integral from two to five of six x squared dx minus the, minus the integral from two to five of three x dx, and then plus the integral from two to five of five dx. We can do that. And each and every one of these coefficients, the six x we can bring out, the three x we can bring out, and the five x we can bring out. And so we can treat these as three separate integrals. But honestly, when people work through these things, that's not exactly how we consider this. What we kinda do is we just do our anti-derivative in line, is do these all together. So think it of the anti-derivative, all right? Anti-derivative of six x squared, we're gonna get six x cubed over three by the power rule, minus three x squared over two plus five x. We just kinda handle this all together and then write your limits right there, two and five. Simplify your expression if you can, three does go into six, two times, three halves, we're just kinda stuck with that one. And so now we're in a situation where we're gonna plug in, plug in the five into each of these places, plug in the two into each of these places. And so this one will show you a little bit more why the arithmetic will often be considered the most tedious part of this exercise right here. So we're gonna first do the case where we have five. So we're gonna get two times five cubed minus three over two times five squared plus five times five, five times five. That's the first expression. And then subtract from that, we're gonna get two times two cubed minus three over two times two squared plus five times two. Try to simplify these things the best we can. So notice what can we say about this? Well, one thing that can be helpful when you look at these type of calculations with these indefinite girls with the arithmetic part of it is you're gonna notice that because you're plugging in the same numbers in different locations, sorry that is you're gonna plug in different numbers in essentially the same location in the polynomial. There are some like coefficients you could use to try to simplify the arithmetic. I mean, you have a two five cubed minus a two two cubed. So you could write that as two times five cubed which is 125 minus two cubed, which is eight. Then the next part you're gonna have a negative three halves and there's gonna be a five squared which is 25 minus a two squared, which is a four. We already did eight there. And then lastly, you're gonna get plus five minus two. And this, if you break up the terms in this way, it can make things a little bit easier in terms of this arithmetic, right? So you're gonna get going on right here. You're gonna get 125 take away eight. That is gonna be 117. Next, you get 25 take away four, which is 21. And then lastly, five take away two, which is a three. And if we perform these, these multiplications, two times seven of 117 will give us 234, right? Two doesn't go into 21 evenly but we can take three times 21 to get 63 over two and then five times three is 15. We can add those together. And then we have to subtract 63 over two, right? I mean, that's really what's left to do. If we add the 15 and the 234 together, we get 249 minus 63 over two. And so in order to add these together, of course, you're gonna get two over two here. And again, I honestly wanna bore anyone with the arithmetic. It's the tedious part, but this is the part we probably know quite well. There's no calculus going on here. In the end, we'll end up with 435 over two as a fraction or you can write this as a decimal as well. Be patient with yourself when it comes to the arithmetic here. But again, this is, the power roll part was pretty straightforward with this because of the anti-dread was we've done before. Be very careful with the arithmetic. Feel free to use a calculator to help you at this stage because again, this is probably our most tedious part of the whole calculation. We'll do some more examples in the next video. Stay tuned. I'll see you then.