[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.76,0:00:02.40,Default,,0000,0000,0000,,- [Instructor] So,\Nwe've got a Riemann sum.
Dialogue: 0,0:00:02.40,0:00:04.92,Default,,0000,0000,0000,,We're gonna take the limit\Nas N approaches infinity
Dialogue: 0,0:00:04.92,0:00:06.16,Default,,0000,0000,0000,,and the goal of this video
Dialogue: 0,0:00:06.16,0:00:08.14,Default,,0000,0000,0000,,is to see if we can rewrite this
Dialogue: 0,0:00:08.14,0:00:09.76,Default,,0000,0000,0000,,as a definite integral.
Dialogue: 0,0:00:09.76,0:00:11.18,Default,,0000,0000,0000,,I encourage you to pause the video
Dialogue: 0,0:00:11.18,0:00:14.78,Default,,0000,0000,0000,,and see if you can work\Nthrough it on your own.
Dialogue: 0,0:00:14.78,0:00:16.05,Default,,0000,0000,0000,,So, let's remind ourselves
Dialogue: 0,0:00:16.05,0:00:20.43,Default,,0000,0000,0000,,how a definite integral can\Nrelate to a Riemann sum.
Dialogue: 0,0:00:20.43,0:00:24.34,Default,,0000,0000,0000,,So, if I have the definite\Nintegral from A to B
Dialogue: 0,0:00:27.29,0:00:29.12,Default,,0000,0000,0000,,of F of X, F of X, DX,
Dialogue: 0,0:00:34.05,0:00:36.39,Default,,0000,0000,0000,,we have seen in other videos
Dialogue: 0,0:00:36.39,0:00:38.90,Default,,0000,0000,0000,,this is going to be the limit
Dialogue: 0,0:00:38.90,0:00:43.07,Default,,0000,0000,0000,,as N approaches infinity\Nof the sum, capital sigma,
Dialogue: 0,0:00:44.74,0:00:47.08,Default,,0000,0000,0000,,going from I equals one to N
Dialogue: 0,0:00:47.92,0:00:49.57,Default,,0000,0000,0000,,and so, essentially\Nwe're gonna sum the areas
Dialogue: 0,0:00:49.57,0:00:51.72,Default,,0000,0000,0000,,of a bunch of rectangles
Dialogue: 0,0:00:51.72,0:00:55.09,Default,,0000,0000,0000,,where the width of each\Nof those rectangles
Dialogue: 0,0:00:55.09,0:00:57.26,Default,,0000,0000,0000,,we can write as a delta X,
Dialogue: 0,0:00:58.14,0:01:00.92,Default,,0000,0000,0000,,so your width is going to be delta X
Dialogue: 0,0:01:00.92,0:01:02.78,Default,,0000,0000,0000,,of each of those rectangles
Dialogue: 0,0:01:02.78,0:01:03.74,Default,,0000,0000,0000,,and then your height
Dialogue: 0,0:01:03.74,0:01:06.29,Default,,0000,0000,0000,,is going to be the value of the function
Dialogue: 0,0:01:06.29,0:01:08.43,Default,,0000,0000,0000,,evaluated some place in that delta X.
Dialogue: 0,0:01:08.43,0:01:10.13,Default,,0000,0000,0000,,If we're doing a right Riemann sum
Dialogue: 0,0:01:10.13,0:01:12.80,Default,,0000,0000,0000,,we would do the right\Nend of that rectangle
Dialogue: 0,0:01:12.80,0:01:14.41,Default,,0000,0000,0000,,or of that sub interval
Dialogue: 0,0:01:14.41,0:01:18.58,Default,,0000,0000,0000,,and so, we would start\Nat our lower bound A
Dialogue: 0,0:01:18.58,0:01:22.75,Default,,0000,0000,0000,,and we would add as many delta\NXs as our index specifies.
Dialogue: 0,0:01:23.78,0:01:25.20,Default,,0000,0000,0000,,So, if I is equal to one,
Dialogue: 0,0:01:25.20,0:01:26.94,Default,,0000,0000,0000,,we add one delta X,
Dialogue: 0,0:01:26.94,0:01:28.96,Default,,0000,0000,0000,,so we would be at the right\Nof the first rectangle.
Dialogue: 0,0:01:28.96,0:01:31.36,Default,,0000,0000,0000,,If I is equal two, we add two delta Xs.
Dialogue: 0,0:01:31.36,0:01:34.63,Default,,0000,0000,0000,,So, this is going to be delta X
Dialogue: 0,0:01:34.63,0:01:35.96,Default,,0000,0000,0000,,times our index.
Dialogue: 0,0:01:37.06,0:01:38.62,Default,,0000,0000,0000,,So, this is the general form
Dialogue: 0,0:01:38.62,0:01:40.65,Default,,0000,0000,0000,,that we have seen before
Dialogue: 0,0:01:40.65,0:01:42.38,Default,,0000,0000,0000,,and so, one possibility, you\Ncould even do a little bit
Dialogue: 0,0:01:42.38,0:01:44.37,Default,,0000,0000,0000,,of pattern matching right here,
Dialogue: 0,0:01:44.37,0:01:47.41,Default,,0000,0000,0000,,our function looks like\Nthe natural log function,
Dialogue: 0,0:01:47.41,0:01:49.13,Default,,0000,0000,0000,,so that looks like our func F of X,
Dialogue: 0,0:01:49.13,0:01:51.95,Default,,0000,0000,0000,,it's the natural log function,
Dialogue: 0,0:01:51.95,0:01:53.33,Default,,0000,0000,0000,,so I could write that,
Dialogue: 0,0:01:53.33,0:01:56.83,Default,,0000,0000,0000,,so F of X looks like the natural log of X.
Dialogue: 0,0:01:58.46,0:02:00.08,Default,,0000,0000,0000,,What else do we see?
Dialogue: 0,0:02:00.08,0:02:02.50,Default,,0000,0000,0000,,Well, A, that looks like two.
Dialogue: 0,0:02:03.58,0:02:05.58,Default,,0000,0000,0000,,A is equal to two.
Dialogue: 0,0:02:05.58,0:02:08.11,Default,,0000,0000,0000,,What would our delta X be?
Dialogue: 0,0:02:08.11,0:02:10.57,Default,,0000,0000,0000,,Well, you can see this right over here,
Dialogue: 0,0:02:10.57,0:02:12.37,Default,,0000,0000,0000,,this thing that we're multiplying
Dialogue: 0,0:02:12.37,0:02:14.57,Default,,0000,0000,0000,,that just is divided by N
Dialogue: 0,0:02:14.57,0:02:17.19,Default,,0000,0000,0000,,and it's not multiplying by an I,
Dialogue: 0,0:02:17.19,0:02:19.58,Default,,0000,0000,0000,,this looks like our delta X
Dialogue: 0,0:02:19.58,0:02:22.80,Default,,0000,0000,0000,,and this right over here\Nlooks like delta X times I.
Dialogue: 0,0:02:22.80,0:02:26.96,Default,,0000,0000,0000,,So, it looks like our delta\NX is equal to five over N.
Dialogue: 0,0:02:28.28,0:02:30.82,Default,,0000,0000,0000,,So, what can we tell so far?
Dialogue: 0,0:02:30.82,0:02:33.47,Default,,0000,0000,0000,,Well, we could say that,\Nokay, this thing up here,
Dialogue: 0,0:02:33.47,0:02:36.66,Default,,0000,0000,0000,,up the original thing\Nis going to be equal to
Dialogue: 0,0:02:36.66,0:02:38.11,Default,,0000,0000,0000,,the definite integral,
Dialogue: 0,0:02:38.11,0:02:41.09,Default,,0000,0000,0000,,we know our lower bound is going from two
Dialogue: 0,0:02:41.09,0:02:43.15,Default,,0000,0000,0000,,to we haven't figured\Nout our upper bound yet,
Dialogue: 0,0:02:43.15,0:02:45.08,Default,,0000,0000,0000,,we haven't figured out our B yet
Dialogue: 0,0:02:45.08,0:02:48.64,Default,,0000,0000,0000,,but our function is the natural log
Dialogue: 0,0:02:48.64,0:02:52.14,Default,,0000,0000,0000,,of X and then I will just write a DX here.
Dialogue: 0,0:02:53.58,0:02:55.20,Default,,0000,0000,0000,,So, in order to complete writing
Dialogue: 0,0:02:55.20,0:02:56.20,Default,,0000,0000,0000,,this definite integral
Dialogue: 0,0:02:56.20,0:02:58.89,Default,,0000,0000,0000,,I need to be able to write the upper bound
Dialogue: 0,0:02:58.89,0:03:00.55,Default,,0000,0000,0000,,and the way to figure out the upper bound
Dialogue: 0,0:03:00.55,0:03:03.19,Default,,0000,0000,0000,,is by looking at our delta X
Dialogue: 0,0:03:03.19,0:03:05.94,Default,,0000,0000,0000,,because the way that we\Nwould figure out a delta X
Dialogue: 0,0:03:05.94,0:03:08.48,Default,,0000,0000,0000,,for this Riemann sum here,
Dialogue: 0,0:03:08.48,0:03:12.18,Default,,0000,0000,0000,,we would say that delta X\Nis equal to the difference
Dialogue: 0,0:03:12.18,0:03:15.30,Default,,0000,0000,0000,,between our bounds divided\Nby how many sections
Dialogue: 0,0:03:15.30,0:03:17.61,Default,,0000,0000,0000,,we want to divide it in, divided by N.
Dialogue: 0,0:03:17.61,0:03:20.03,Default,,0000,0000,0000,,So, it's equals to B minus A,
Dialogue: 0,0:03:21.89,0:03:23.89,Default,,0000,0000,0000,,B minus A over N, over N
Dialogue: 0,0:03:29.40,0:03:31.10,Default,,0000,0000,0000,,and so, you can pattern match here.
Dialogue: 0,0:03:31.10,0:03:34.39,Default,,0000,0000,0000,,If this is delta X is\Nequal to B minus A over N.
Dialogue: 0,0:03:34.39,0:03:35.52,Default,,0000,0000,0000,,Let me write this down.
Dialogue: 0,0:03:35.52,0:03:38.44,Default,,0000,0000,0000,,So, this is going to be equal to B,
Dialogue: 0,0:03:39.36,0:03:41.61,Default,,0000,0000,0000,,B minus our A which is two,
Dialogue: 0,0:03:43.14,0:03:44.72,Default,,0000,0000,0000,,all of that over N,
Dialogue: 0,0:03:45.84,0:03:47.01,Default,,0000,0000,0000,,so B minus two
Dialogue: 0,0:03:50.51,0:03:52.52,Default,,0000,0000,0000,,is equal to five
Dialogue: 0,0:03:52.52,0:03:55.75,Default,,0000,0000,0000,,which would make B equal to seven.
Dialogue: 0,0:03:55.75,0:03:57.86,Default,,0000,0000,0000,,B is equal to seven.
Dialogue: 0,0:03:57.86,0:03:58.70,Default,,0000,0000,0000,,So, there you have it.
Dialogue: 0,0:03:58.70,0:04:02.45,Default,,0000,0000,0000,,We have our original\Nlimit, our Riemann limit
Dialogue: 0,0:04:03.81,0:04:05.55,Default,,0000,0000,0000,,or our limit of our Riemann sum
Dialogue: 0,0:04:05.55,0:04:08.65,Default,,0000,0000,0000,,being rewritten as a definite integral.
Dialogue: 0,0:04:08.65,0:04:09.62,Default,,0000,0000,0000,,And once again, I want to emphasize
Dialogue: 0,0:04:09.62,0:04:11.20,Default,,0000,0000,0000,,why this makes sense.
Dialogue: 0,0:04:11.20,0:04:13.18,Default,,0000,0000,0000,,If we wanted to draw this
Dialogue: 0,0:04:13.18,0:04:14.58,Default,,0000,0000,0000,,it would look something like this,
Dialogue: 0,0:04:14.58,0:04:19.36,Default,,0000,0000,0000,,I'm gonna try to hand draw\Nthe natural log function,
Dialogue: 0,0:04:19.36,0:04:21.69,Default,,0000,0000,0000,,it looks something like this
Dialogue: 0,0:04:27.39,0:04:30.26,Default,,0000,0000,0000,,and this right over here would be one
Dialogue: 0,0:04:30.26,0:04:33.21,Default,,0000,0000,0000,,and so, let's say this is two
Dialogue: 0,0:04:33.21,0:04:35.66,Default,,0000,0000,0000,,and so going from two to seven,
Dialogue: 0,0:04:35.66,0:04:37.66,Default,,0000,0000,0000,,this isn't exactly right
Dialogue: 0,0:04:38.58,0:04:42.93,Default,,0000,0000,0000,,and so, our definite integral\Nis concerned with the area
Dialogue: 0,0:04:42.93,0:04:46.57,Default,,0000,0000,0000,,under the curve from two until seven
Dialogue: 0,0:04:46.57,0:04:48.15,Default,,0000,0000,0000,,and so, this Riemann sum you can view
Dialogue: 0,0:04:48.15,0:04:51.50,Default,,0000,0000,0000,,as an approximation when N\Nisn't approaching infinity
Dialogue: 0,0:04:51.50,0:04:52.54,Default,,0000,0000,0000,,but what you're saying is look,
Dialogue: 0,0:04:52.54,0:04:55.08,Default,,0000,0000,0000,,when I is equal to one,
Dialogue: 0,0:04:55.08,0:04:59.05,Default,,0000,0000,0000,,your first one is going to\Nbe of width five over N,
Dialogue: 0,0:04:59.05,0:05:01.57,Default,,0000,0000,0000,,so this is essentially\Nsaying our difference
Dialogue: 0,0:05:01.57,0:05:02.65,Default,,0000,0000,0000,,between two and seven,
Dialogue: 0,0:05:02.65,0:05:04.27,Default,,0000,0000,0000,,we're taking that distance five,
Dialogue: 0,0:05:04.27,0:05:06.32,Default,,0000,0000,0000,,dividing it into N rectangles,
Dialogue: 0,0:05:06.32,0:05:11.10,Default,,0000,0000,0000,,and so, this first one is\Ngoing to have a width of five
Dialogue: 0,0:05:11.10,0:05:14.17,Default,,0000,0000,0000,,over N and then what's\Nthe height gonna be?
Dialogue: 0,0:05:14.17,0:05:16.27,Default,,0000,0000,0000,,Well, it's a right Riemann sum,
Dialogue: 0,0:05:16.27,0:05:19.97,Default,,0000,0000,0000,,so we're using the value of\Nthe function right over there,
Dialogue: 0,0:05:19.97,0:05:22.30,Default,,0000,0000,0000,,write it two plus five over N.
Dialogue: 0,0:05:22.30,0:05:24.64,Default,,0000,0000,0000,,So, this value right over here.
Dialogue: 0,0:05:24.64,0:05:26.79,Default,,0000,0000,0000,,This is the natural log,
Dialogue: 0,0:05:26.79,0:05:30.12,Default,,0000,0000,0000,,the natural log of two plus five over N,
Dialogue: 0,0:05:32.16,0:05:33.100,Default,,0000,0000,0000,,and since this is the first rectangle
Dialogue: 0,0:05:33.100,0:05:35.75,Default,,0000,0000,0000,,times one, times one.
Dialogue: 0,0:05:36.69,0:05:38.56,Default,,0000,0000,0000,,Now we could keep going.
Dialogue: 0,0:05:38.56,0:05:40.31,Default,,0000,0000,0000,,This one right over here
Dialogue: 0,0:05:40.31,0:05:43.32,Default,,0000,0000,0000,,the width is the same, five over N
Dialogue: 0,0:05:43.32,0:05:45.22,Default,,0000,0000,0000,,but what's the height?
Dialogue: 0,0:05:45.22,0:05:47.99,Default,,0000,0000,0000,,Well, the height here,\Nthis height right over here
Dialogue: 0,0:05:47.99,0:05:49.56,Default,,0000,0000,0000,,is going to be the natural log
Dialogue: 0,0:05:49.56,0:05:53.31,Default,,0000,0000,0000,,of two plus five over\NN times two, times two.
Dialogue: 0,0:05:55.15,0:05:57.65,Default,,0000,0000,0000,,This is for I is equal to two.
Dialogue: 0,0:05:58.48,0:06:00.80,Default,,0000,0000,0000,,This is I is equal to one.
Dialogue: 0,0:06:00.80,0:06:02.72,Default,,0000,0000,0000,,And so, hopefully you are\Nseeing that this makes sense.
Dialogue: 0,0:06:02.72,0:06:04.88,Default,,0000,0000,0000,,The area of this first rectangle
Dialogue: 0,0:06:04.88,0:06:06.87,Default,,0000,0000,0000,,is going to be natural log of two
Dialogue: 0,0:06:06.87,0:06:09.04,Default,,0000,0000,0000,,plus five over N times one
Dialogue: 0,0:06:09.04,0:06:10.46,Default,,0000,0000,0000,,times five over N
Dialogue: 0,0:06:12.20,0:06:13.56,Default,,0000,0000,0000,,and the second one over here,
Dialogue: 0,0:06:13.56,0:06:17.30,Default,,0000,0000,0000,,natural log of two plus\Nfive over N times two
Dialogue: 0,0:06:18.90,0:06:20.32,Default,,0000,0000,0000,,times five over N
Dialogue: 0,0:06:21.50,0:06:23.41,Default,,0000,0000,0000,,and so, this is calculating the sum
Dialogue: 0,0:06:23.41,0:06:25.38,Default,,0000,0000,0000,,of the areas of these rectangles
Dialogue: 0,0:06:25.38,0:06:28.26,Default,,0000,0000,0000,,but then it's taking the\Nlimit as N approaches infinity
Dialogue: 0,0:06:28.26,0:06:30.05,Default,,0000,0000,0000,,so we get better and better approximations
Dialogue: 0,0:06:30.05,0:06:33.13,Default,,0000,0000,0000,,going all the way to the exact area.