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Main content
Current time:0:00Total duration:6:35
AP.CALC:
LIM‑5 (EU)
,
LIM‑5.B (LO)
,
LIM‑5.B.1 (EK)
,
LIM‑5.B.2 (EK)
,
LIM‑5.C (LO)
,
LIM‑5.C.1 (EK)
,
LIM‑5.C.2 (EK)

Video transcript

so we've got a Riemann sum we're going to take the limit as n approaches infinity and the goal of this video is to see if we can rewrite this as a definite integral I encourage you to pause the video and see if you can work through it on your own so let's remind ourselves how a definite integral can relate to a Riemann sum so if I have the definite integral from A to B of f of X f of X DX we have seen in other videos this is going to be the limit as n approaches infinity of the sum capital Sigma going from I equals 1 to N and so we're essentially going to sum the areas of a bunch of rectangles where the width of each of those rectangles we can write as a Delta X so your width is going to be Delta X of each of those rectangles and then your height is going to be the value of the function evaluated someplace in that Delta X if we're doing a right Riemann sum we would do the right end of that rectangle or of that sub interval and so we would start at our lower bound a and we would add as many Delta X's as our index specifies so if I is equal to 1 we add 1 Delta X so we'd be at the right of the first rectangle if I is equal to we add 2 Delta X's so this is going to be Delta x times our index so this is the general form that we have seen before until one possibility you could even do a little bit of pattern matching right over here our function looks like the natural log function so that looks like our func f of X it's the natural log function so I could write that so f of X looks like the natural log of X what else do we see well a that looks like two a is equal to two what would our Delta X be well you can see this right over here this thing that we're multiplying that just is divided goodbye n and it's not multiplying by an eye this looks like our Delta X and this right over here looks like Delta X times I so it looks like our Delta X is equal to five over N so what can we tell so far well we could say that okay this thing up here up the original thing is going to be equal to the definite integral we know our lower bound is going from two two we haven't figured out our upper bound yet we haven't figured out our B yet but our function is the natural log of X and then I will just write a DX here so in order to complete writing this definite integral I need to be able to write the upper bound and the way to figure out the upper bound is by looking at our Delta X because the way that we would figure out a delta X for this Riemann sum here we would say that Delta X is equal to the difference between our bounds divided by how many sections we want to divide it in divided by n so it's equal to B minus a B minus a over N over N and so you can pattern match your if this is Delta X is equal to B minus a over N let me write this down so this is going to be equal to B B minus R a which is 2 all of that over n so B minus 2 is equal to 5 which would make B equal to 7 B is equal to 7 so there you have it we have our original our original limit our riemann limit or a limit of our Riemann sum being rewritten as a definite integral and once again I want to emphasize why this makes sense if we wanted to draw this it would look something like this I'm going to try to hand draw the natural log function it looks something like this this right over here would be one and so let's say this is two and so we're going from two to seven this isn't exactly right and so our definite integral is concerned with the area under the curve from two until seven and so this Riemann sum you could view it as an approximation when n isn't approaching infinity but what you're saying is look when I is equal to one your first one is going to be of with five over N so this is essentially saying our difference between two and seven we're taking that distance five dividing it into n rectangles and so this first one is going to have a width of five over N and then what's the height going to be well it's a right Riemann sum so we're using the value of the function right over there right at two plus five over n so this value right over here this is the natural log the natural log of two plus five over N and since this is the first rectangle times one times one and we could go keep going this one right over here the width is the same five over N but what's the height well the height here this height right over here is going to be the natural log of two plus five over n times two times two this is for I is equal to two this is i is equal to one and so hopefully you are seeing that this makes sense the area of this first rectangle is going to be natural log of two plus five over n times one times five over n the second one over here natural log of two plus five over n times two times five over N and so what this is calculating the sum of the areas of these rectangles but then it's taking the limit as n approaches infinity so we get better and better approximations going all the way to the exact area
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