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- [Instructor] So,[br]we've got a Riemann sum.

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We're gonna take the limit[br]as N approaches infinity

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and the goal of this video

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is to see if we can rewrite this

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as a definite integral.

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I encourage you to pause the video

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and see if you can work[br]through it on your own.

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So, let's remind ourselves

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how a definite integral can[br]relate to a Riemann sum.

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So, if I have the definite[br]integral from A to B

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of F of X, F of X, DX,

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we have seen in other videos

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this is going to be the limit

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as N approaches infinity[br]of the sum, capital sigma,

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going from I equals one to N

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and so, essentially[br]we're gonna sum the areas

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of a bunch of rectangles

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where the width of each[br]of those rectangles

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we can write as a delta X,

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so your width is going to be delta X

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of each of those rectangles

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and then your height

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is going to be the value of the function

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evaluated some place in that delta X.

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If we're doing a right Riemann sum

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we would do the right[br]end of that rectangle

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or of that sub interval

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and so, we would start[br]at our lower bound A

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and we would add as many delta[br]Xs as our index specifies.

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So, if I is equal to one,

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we add one delta X,

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so we would be at the right[br]of the first rectangle.

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If I is equal two, we add two delta Xs.

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So, this is going to be delta X

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times our index.

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So, this is the general form

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that we have seen before

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and so, one possibility, you[br]could even do a little bit

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of pattern matching right here,

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our function looks like[br]the natural log function,

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so that looks like our func F of X,

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it's the natural log function,

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so I could write that,

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so F of X looks like the natural log of X.

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What else do we see?

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Well, A, that looks like two.

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A is equal to two.

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What would our delta X be?

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Well, you can see this right over here,

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this thing that we're multiplying

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that just is divided by N

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and it's not multiplying by an I,

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this looks like our delta X

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and this right over here[br]looks like delta X times I.

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So, it looks like our delta[br]X is equal to five over N.

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So, what can we tell so far?

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Well, we could say that,[br]okay, this thing up here,

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up the original thing[br]is going to be equal to

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the definite integral,

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we know our lower bound is going from two

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to we haven't figured[br]out our upper bound yet,

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we haven't figured out our B yet

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but our function is the natural log

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of X and then I will just write a DX here.

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So, in order to complete writing

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this definite integral

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I need to be able to write the upper bound

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and the way to figure out the upper bound

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is by looking at our delta X

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because the way that we[br]would figure out a delta X

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for this Riemann sum here,

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we would say that delta X[br]is equal to the difference

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between our bounds divided[br]by how many sections

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we want to divide it in, divided by N.

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So, it's equals to B minus A,

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B minus A over N, over N

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and so, you can pattern match here.

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If this is delta X is[br]equal to B minus A over N.

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Let me write this down.

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So, this is going to be equal to B,

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B minus our A which is two,

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all of that over N,

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so B minus two

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is equal to five

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which would make B equal to seven.

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B is equal to seven.

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So, there you have it.

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We have our original[br]limit, our Riemann limit

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or our limit of our Riemann sum

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being rewritten as a definite integral.

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And once again, I want to emphasize

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why this makes sense.

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If we wanted to draw this

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it would look something like this,

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I'm gonna try to hand draw[br]the natural log function,

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it looks something like this

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and this right over here would be one

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and so, let's say this is two

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and so going from two to seven,

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this isn't exactly right

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and so, our definite integral[br]is concerned with the area

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under the curve from two until seven

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and so, this Riemann sum you can view

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as an approximation when N[br]isn't approaching infinity

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but what you're saying is look,

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when I is equal to one,

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your first one is going to[br]be of width five over N,

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so this is essentially[br]saying our difference

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between two and seven,

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we're taking that distance five,

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dividing it into N rectangles,

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and so, this first one is[br]going to have a width of five

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over N and then what's[br]the height gonna be?

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Well, it's a right Riemann sum,

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so we're using the value of[br]the function right over there,

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write it two plus five over N.

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So, this value right over here.

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This is the natural log,

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the natural log of two plus five over N,

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and since this is the first rectangle

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times one, times one.

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Now we could keep going.

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This one right over here

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the width is the same, five over N

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but what's the height?

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Well, the height here,[br]this height right over here

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is going to be the natural log

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of two plus five over[br]N times two, times two.

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This is for I is equal to two.

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This is I is equal to one.

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And so, hopefully you are[br]seeing that this makes sense.

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The area of this first rectangle

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is going to be natural log of two

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plus five over N times one

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times five over N

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and the second one over here,

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natural log of two plus[br]five over N times two

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times five over N

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and so, this is calculating the sum

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of the areas of these rectangles

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but then it's taking the[br]limit as N approaches infinity

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so we get better and better approximations

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going all the way to the exact area.