 Welcome back to the second part here of lecture 45 for our series Math 1210. In the previous part, we had just recently looked at computing a definite integral by its definition. I want to do another example of that. So let's evaluate the integral from 0 to 1 of the function x squared dx. Now let's think about this geometrically for a second. What does this look like? The function f of x equals x squared, it's going to look like a standard parabola, which you can see something like this right here. It's just your standard parabola goes to the origin, has the standard curvature and such. And if we want to go from the point 0 to 1, we're looking at the area under the curve from 0 to 1. We're going to get a region that looks something like this. So we can anticipate probably a positive area under this thing. So we're trying to calculate the area of this region and how do we do that? Well, the integral here is the limit of the Riemann sum. And so there's a little bit of the ingredients we have to look at first. Let's first consider delta x, delta x by its formula looks like b minus a over n, which remember this b and a right here. This is going to be the limits of the integral, right? So we're going from 0 to 1 here because the lower limit of the integral is a 0 and the upper limit is a 1. And so as we go from 0 to 1, the numerator simplifies just to be 1 and we give 1 over n for our delta x there. Next we need to consider the xi. This tells us how high the rectangles are going to be in the Riemann sum. Well, the general formula for xi is a plus this i delta x. And so plugging in the specific values, we get a is 0, we get that i is just a variable to be dealt with later and delta x from before is 1 over n. So we get that. And so simplifying that xi will just become i over n. It probably goes without saying, but we should mention what the function is. By f of x in this situation, we are describing that quadratic function x squared. And so if we plug in the xi into the squaring function, we end up with i over n quantity squared, which if we square the numerator, we get an i squared. If we square the denominator, we get an n squared. And so that's our f of xi right there. So now we're ready to start computing the integral by this definition. The integral is a limit as n goes to infinity of the Riemann sum sigma as i goes from 1 to n. We're going to get f of xi right there times delta x. But f of xi from what we saw before can actually be replaced with the i squared over n squared. And then we have the times that by the delta x, which is 1 over n as we saw before. And so multiplying those together, we're still taking the limit as n goes to infinity of sigma where i goes from 1 to n. And then we're going to get i squared over n cubed right there. Now this n cubed is constant with respect to the variable i. So the sum, the sigma, it changes the i, it doesn't change the n. And so this n cubed can be factored out of the sigma. And when we do that, we get the limit as n goes to infinity of 1 over n cubed times sigma. Again, sigma goes from i equals 1 to n. And then we get an i squared right there. Let me kind of partition those from each other right there. Now to proceed from here, we have to remember the sigma power rule with respect to i squared. So if we take the sum as i goes from 1 to n of i squared, recall this is equal to n times n plus 1 times 2n plus 1 all over 6. And so we can make that substitution in for the sum of i squared above. And if we do that, we're going to get that this is equal to the limit as n goes to infinity of 1 over n cubed times n times n plus 1 times 2n plus 1 all over 6. All right. And so now we want to take the limit as n goes to infinity of this rational expression. I'm going to make a slight tweak right here. Let's just put the n cubed in the denominator. And yeah, that does kind of leave sort of a big gap between the limit and the function there. I think we can survive with that though. I mean, that's all right. We're now considering a balanced rational function, a balanced rational function which we have an n and 2n on top. So that's going to look like a 2n cube. Don't forget the 2 there. And then on the bottom, there's only one term, which of course is the leading term. We get this 6n cubed. And so as n goes towards infinity, this will just look like 2 over 6. That is to say 1 third, which is the area under this parabola from 0 to 1. And so this one wasn't as bad as the last example we did, but there is still a lot that goes into these Riemann sum type calculations. We have to first identify what is the delta x, what is the xi, what is the f of xi. Then we multiply f of xi and delta x together. Pull out the things from the sum that we can using properties of summation, the linearity there. Eventually you're going to get a sum of powers of i. Use those power rules we had before to evaluate those. This if you've done it correctly should give you a balanced rational function as n goes to infinity. And so then evaluate the limit as n goes to infinity. Find that horizontal asymptote. In this case it was 1 third. And that gives you the area under the curve. We'll do some more examples of this in the next video. Stay tuned. See you then.