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Determine a Riemann sum that would determine the area of the region bounded by $f(x)=\ln x$ and the $x$ -axis, between $x=1$ and 2 (do not evaluate this limit).

$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n} \ln \left(1+\frac{i}{n}\right) \frac{1}{n}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 5

Sigma Notation and Areas

Integrals

Missouri State University

Harvey Mudd College

Baylor University

University of Nottingham

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

07:00

Use the limit of a Riemann…

06:49

Evaluate the integral by c…

04:59

06:55

were given that into girl from 0 to 2 for the function x squared and and were asked to value this integral using the limit of Riemann Sums. So we know that to do this are formula is gonna be the limit as an approaches infinity as and as the number of rectangles were making. And then Sigma equals one. Since we're doing right hand to dreaming sums and then end is on top, and then inside the Sigma, we're gonna have Delta eggs, times f of a plus Delta X high. So to get stuff to plug into this formula, As you can see, we're definitely going to need a Delta X. So our first step will be to find out two x So Delta X is always B minus a over where B is the upper limit. Ever integral is the lower limit of our integral and is the amount of rectangles were making. However, we're not gonna worry about and in this case, so we're just gonna have tu minus zero over and which is to over. And so now that we know are Delta X. We could just plug this into our formula over here, so we have to write through rewrite the limit and the signal. And once we do that, we have to ever. And instead of Delta X and then effort a plus Delta X High, we know that our original function is X squared. So it's genetic. So we're just gonna have a plus tell tax I all of that squared. So we know that is zero. Since it's the lower limit of our integral and then plus Delta X, I is simply going to be too high over and and then all of that squared, since that was our original function X squared. So now that we have this, we can simplify this a bit to make it easier for us. So well, once earlier at the signal, we have to end to her and sorry times two out over end squared and that's simply four squared over and squared. Then we can simplify this even further by multiplying both of these together. So once we do that, we get e I squared over n cubed. So now that we have the simplified version of our limit of the Sigma weekend, get rid of the Sigma about isolating are in keeping I inside the Sigma and taking everything else out of it. So to do this, we're still gonna have our limit on the outside. But then we can take a over an cubed outside of the Sigma just like that. And then we're gonna have our Sigma, and then we're just gonna have I squared on the inside of her sigma. So now that we have this, we know that the Sigma of I squared is always gonna be end times and plus one turns to in plus 1/6 So we can rewrite this and get rid of the Sigma and we get right it, uh, and times and plus one times Chu and this one over six. So now let's multiply the end times and plus one times two n plus one out. So we're still Did I have our limit over here actually, before we can? Before we do that, we can simplify this down a little bit and make this floor, and this could be three. So now we have four over and cubed times. We know that end times M plus one. Let's just write that on the side is and squared plus and and then that times to m plus one equals to end squared, plus and squared, plus or two in Q plus and squared plus two and squared plus on which simplifies out to two and squared or two in cubed plus three and squared plus and then all of the over three. So now we can just bring both of these together our multiply both of these together and once we do that, we have or a n cubed plus 12 ends word plus for on over three on cubed So now we can divide all of these and some fly them How much ever we can So let's do that. And for the first part an cubed over three and cubed we're just left with a over three plus 12 and squared over three and cubed which equals for over and and then lastly, foran over three in cubed is 4/3 and squared So now we could just solve out our limits. So if we plug infinity into this we know that any number over infinity is always gonna be equal to zero. So four over infinity is going to turn into Syria and for over three and squared or 4/3. Infinity squared, I suppose, is gonna also be equal to zero. So the answer that we get once we plug in our limit is 1/3 plus zero plus zero, which is simply a third's. So this is our answer.

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