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Express the limit as a definite integral on the given interval.

$ \displaystyle \lim_{n \to \infty} \sum_{i = 1}^n \frac{e^{x_i}}{1 + x_i} \, \Delta x $, [0, 1]

$$\int_{0}^{1} \frac{e^{x}}{1+x} d x$$

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okay, we have this limit and I'll explain what all these two different little symbols and pieces mean, but we want to take this limit and express it as a definite integral. Um so this is a re months um and we're taking the limit of these Riemann sums. Uh so we take the limit as N approaches infinity. Um but n is a fixed number and could be seven and could be 80. Eventually. You want to keep finding these sums as an approaches infinity. Uh but for each particular end, whether you know, and there's a million and there's a million one and it's a million to you're taking the sum uh from I equals one up to some number. End what end is uh N is the number of partitions that you break up the interval 01 into. For example, if end was just four, you would be breaking up the interval 01 from 0 to 1 into four parts Yes, So let's pretend just for a moment and is four. So you would be taking to some uh from I equals 124 of e uh to the excel 1/1 plus X L one times delta X. Uh and then plus E x 0 2/1 plus X +02 delta X. Let me show you what this means. Let's pretend N is for If N is for I'm going to divide the partition from 0 to 1 into four parts four different sub intervals. So this is the first interval corresponding to I equals one, second interval, I equals two, third interval, I would be three and then the fourth interval I would be four, which is what end would be because we broke it up into four pieces. So E except by what is the exit by ex of I is an X. Value in the I. F Interval. So this is the first interval, so that's going to be X one. This is the second interval, pick any point in this interval, Call it X two, pick any point in the third interval, call it X three, pick any point in the fourth interval. Call it X four now four X one. Uh If you look at this, this little piece of the sum basically corresponds to your function. So what function can we see in here? E to the X over one plus X. So that's our function that we're working with. E to the X over one plus X. So this blue curve is the graph of E T D X over one plus X. Now for X one we have the corresponding point on the graph. And this point on the graph is uh the function value. Well if the X value is X one, then the function value is E to the X. Sub 1/1 plus X. Sub one. E to the X sub one. As the value of this function at this point over one plus X. Sub one. And so what we have is we have a little rectangle. Now the width of this rectangle is delta X. And the height of this rectangle is the E to the X sub 1/1 plus X. Someone. So the area of this rectangle is going to be the height. This E TX tough times to with delta X. So this E T D X stuff E T D X. So by over one plus X. L. By is the area is the height of a rectangle. Delta X. Is to within a rectangle. So what happens is when X is one, this becomes the area of the first rectangle. When I is to remember we're going up to four because we broke to partition up into four when I is too Okay, here's X. Up to, here's the function corresponding to X. Up to so when I is to E to the X. Up to over one plus X. Up to is the height of this rectangle, delta X. Is the width of this rectangle. And so height times with gives you the area of the second rectangle. So we're going to do this for a total of n equals four times. And then here's XO four. Here's the point on the craft. So in the end this some from I equals one to end. We just did the example where and was a total of 1234 partitions. This sum is really just a some of the four rectangles and to some of the areas of these rectangles, uh is an approximation for the area underneath the blue curve. Now, as N approaches infinity, all that simply means is you're breaking up this interval from 0 to 1 into smaller partitions. More partitions, uh that will be all smaller in size and so you're rectangles will fit better underneath the curve and when you take the some of the areas of all the rectangles, you get closer to the area underneath the curve. Now, uh the area underneath the curve, the limit of these Riemann sums as end goes towards infinity is given by the definite integral. The bounds the boundaries for this integral are the X values from 0 to 1. The function that we would be integrating would be f of X E to the X over one plus X and delta X just basically gets replaced with dX. So if we calculate this definite integral, the integral from 0 to 1 of E T D X over one plus X with respect to X. This definite integral will be the limit of these Riemann sums as an approaches infinity. There will also be the area, the exact area underneath that blue functions curve

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