in the previous video we were looking at the the problem of finding the area underneath a curve and what we saw was that finding the area under the curve amounted to finding a limit so our objective or our approach to finding the area under a curve was to approximate the area with a collection of rectangles and we noticed that as the number of rectangles increased that our approximation was improving and then what we saw we saw that picture at the very end of the last video that as the number of rectangles was increasing that all three of the sums that we were finding the left some the right some and the midpoint some all were converging to the same destination so what we were observing here is a limit so the object here to find our area under a curve is to find a limit of one of these Riemann sums so again we define a Riemann sum this way so when we're talking about a Riemann sum we have a partition of some interval so the Riemann sum is the sum of this form so as to sum I equals 1 to n we say f of X I start times Delta X I don't let the X I star confuse you the X I star just is a notation for an arbitrary point in the ice-sub interval and then we have Delta X I being the width of the ice-sub interval so together f of X I star gives me a height Delta X I gives me a width multiplying those two together gives me an area and then we sum up those areas we looked at three particular ways of choosing this excise star we looked at the left right in the midpoint sums so if the partition is uniform any Riemann sum is going to converge to that same value as the number of the points in the partition or the number of rectangles goes out to infinity all right so that's the object here if we have a uniform partition send our rectangles to infinity and we should converge to the area that we are looking for so I want to talk brief we hear about the general case what if we don't have a uniform partition so if we take any partition let's define the norm of the partition this notation here is the norm of the partition right what we define the norm to be is we define the norm to be the maximum width of each subinterval so I take all those possible widths that I might have the norm is going to be the maximum of all of those widths so if I want the area for a arbitrary partition the characteristic that I want is the norm of the partition goes to zero so in other words the maximum width sub interval has to be going to zero in order for this limit to give me the area all right so again we talked about this absolute value of the partition minus 1 being the number of rectangles now in a uniform partition Delta X I is the same so we say dis Delta X and we define Delta X to be B minus a over in if we have a uniform partition n going to infinity in the norm going to zero are equivalent alright the converse is not going to be true however it's not good enough for the number of rectangles to go to infinity right what has to happen is the norm of the partition has to go to zero in other words the maximum width sub interval has to be getting dragged down to zero all right but again we're not going to look at partitions that behave this way we're only going to look at uniform partitions all right so in a uniform partition our area is going to be expressed as this limit the limit as n goes to infinity of our Riemann sum right provider that limit exists an X I star is any point inside the ice-sub interval all right so in general it's not good enough for the number of rectangles to go to infinity but if I do have a uniform partition which is the only type of partition that we're going to work with right the number of rectangles going to infinity is going to be sufficient in order to find our area under the curve so we're let's make this statement here suppose F is a non-negative continuous function on a closed interval so the area which will denote as a between the graph of F and the x-axis on that interval is found by calculating this limit so the limit as n goes to infinity we have the sum I equals 1 to N of f of X I star times Delta X Delta X for a uniform partition is going to be B minus a divided by n now algebraically it's easiest to work with the right-hand endpoint in my opinion so again we saw that picture it doesn't matter if I use right left or mid to converge to the same destination so we're going to use X I star or select X I star to be the right-hand endpoint all right so we will calculate this limit and this limit should give us the area exactly under the curve so before we do a couple of examples with this limit let's make sure that the limit at least matches our expectation so I don't want this to give me surprising results right if I have a triangle in this case I want to see that this limit matches what I expect for the area of a triangle so here's how we'll set the triangle up we'll set the triangle up as 8 a line segment all right so maybe this is our just a segment of the line that we're working with to give us a triangle so this is gonna have some height to it all right so up here on the y-axis this will be H for height down here on the x-axis this will be B for base now without any calculus I know the area of a triangle is 1/2 times the base times the height so what would this be let's write this as a as a function so what this is just a line segment I'm passing through the origin and so how do I get to the next point I have a rise over a run so this should just be the height over the Ace times X okay so we're saying that the area is going to be able to be found through this limit all right using Delta X to find this way in our right hand endpoint so Delta X all right now notice our this is our interval or an oval is going to be 0 to be so the width of that interval is going to be B minus 0 divided by n so we're gonna use a uniform partition so that's just gonna be B over in all right and then our right-hand endpoint X I star is gonna be the a which in this case is 0 plus I times Delta X well Delta X is B over in so I Delta X will be B I over in all right so we're claiming that the area should be the limit as n goes to infinity we've got the sum I equals 1 to n here we've got f of X I star times Delta X now I want to see that this limit is going to be 1/2 base times height all right so let's work through this so we have the limit as n goes to infinity so our function f of X is equal to H over B times X so I have H over B now we're gonna plug in X I star into the function and that's going to be B I over in and then Delta X is gonna be B over in all right so let's take care of some things these bees are going to cancel and I have the limit as n goes to infinity of the sum I equals 1 to N now what's left over I have h I be over in squares so this is probably feeling like a bunch of gobbledygook here but let's see what we can do now notice that this is a limit as n goes to infinity so H is acting as a constant B is acting as a constant and so is the end the end is something that's fixed as far as the Sun goes the only thing changing in regards to the sum is this I term so remember one of those properties of sums we can pull out constants so I can pull out HB over N squared and I'm just gonna be left with the sum I equals 1 to N of I right how do we finish this up so let's go back to the last video we had some of these special sums so one of those special sums was this sum right here so the sum I equals 1 to N of I can be replaced with this formula N squared plus n over 2 so let's do that so now what I have is the limit as n goes to infinity so I have HB over N squared and now we're going to replace this sum with in squared plus n over 2 the formula that we have all right so one last step here let's multiply these two together so on the top I'm gonna have H be in squared plus h be in divided by 2 N squared so I just distributed this HB to these two terms distributed or multiply two times N squared now what do we notice here I have a limit of n going to infinity this is effectively behaving like a rational function now I have a degree two over degree two so the limit is just going to be the ratio of these coefficients right so that's gonna be HB / - right which you should notice is the same as just one-half base times height so we do see that this limit this limit of the Riemann sum that we've formed right given our knowledge of this right here as a triangle we know the result should be 1/2 base height we know that from basic geometry and now we have showed that that matches through our limit here of the Riemann sum that we've set up and worked through okay so let's finish up this section with a couple of examples working through this limit of a Riemann sum to find the area under a curve exactly so this is the function that we were working with yesterday in our estimates so f of x equals x squared plus 1 on 0 2 or and I told you in the last video that this area is going to be 14 thirds so I want to show that exactly using our limit of the Riemann sum that we have now alright so we want to find this area trapped between the curve and the x-axis on the interval 0 to 2 so let's set up our Delta X our Delta X for a uniform partition is going to be 2 minus 0 over N which is going to be 2 over in and then our right-hand endpoint it's gonna be a plus I Delta X so that's going to be 0 plus 2i over in so again we'll always always use the right-hand endpoint because it will typically give us the easiest algebra left-hand and midpoint will also work but sometimes the expressions that result are a little less advantageous to work with so we have our width and we have the expression for our right-hand endpoint so we know the area is going to be the limit of the Riemann sum so we have f of X I star times Delta X and that sum runs from 1 to n so I want to work through this and see that we get to 14 thirds to find our area under under the curve so let's begin to work through this now this again this looks complicated but what do we have here this is an expression for a height this is the expression for the width together that gives me an area we're gonna sum those areas and we're gonna send the number of rectangles to infinity all right so again this may look quite complicated but break it down into pieces to understand what's going on here height width area some of the areas take the number of rectangles to infinity so our function is x squared plus 1 so we're gonna have to I over N squared plus 1 times 2i over excuse-me times 2 over in so let's just work through the algebra here we'll have the limit as n goes to infinity that's gonna be 4i squared over n squared plus 1 times 2 over in summing I equals 1 to n now let's distribute that 2 over in so that's gonna be the sum I equals 1 to n we're gonna have 8i squared over in cubed plus 2 over alright one of those properties of sums we can break a sum apart and we can pull out all terms that do not hold the index I so I'm gonna do that in two in one step here so 8 over N cubed comes out and we're just left with the sum I equals 1 to N of I squared the 2 over n comes out and we're just left with the sum I equals 1 to N of just a loan 1 now here's where we're going to use our formulas so here the sum I equals 1 to N of I squared has a special formula and that's going to be the formula here so I equals 1 to N of I square it has this formula to in cubed plus 3n squared plus n over 6 so let's make that replacement we have the limit as n goes to infinity we have 8 in 8 over N cubed right now we have 2 in cubed plus 3 in squared plus n over 6 and now for the second sum well if I add up a 1 in times if I have 1 plus 1 plus 1 in times that's gonna come out to in now notice these ends are going to cancel alright to finish this up let's go ahead and multiply across so we're gonna have 8 times 2n cubed so we'll have 16 in cubed plus 24 N squared plus 8 in over 16 n cubed and then a plus 2 now again what do I notice power of 3 power of 3 the limit as the n goes to infinity is going to be the ratio of those leading coefficients so we're gonna have 16 over 6 plus 2 alright 16 over 6 plus 2 that's gonna be the same as 16 plus 6 excuse me 16 over 6 plus 12 over 6 that's gonna be 28 over six which simplifies to fourteen thirds that I promised so we have now found the area under the curve for f of X equals x squared plus one on the interval zero to two that area is now determined to be exactly fourteen thirds squared units so let's work through one more of these let's find the area under the graph of f of x equals x cubed plus 2x on the interval 0 to 3 halves so here's that segment of the curve we want to know the area of this shaded region here so we're going to take the same steps so our Delta X is going to be B minus a divided by the number of rectangles in so we're gonna have 3 over 2 in for our width expression now for our right-hand endpoint again we're gonna have a plus I Delta X so we're gonna have 3i over 2 in so we'll set our limit up so the area is going to be the limit as n goes to infinity summing I equals 1 to n we're gonna have f of X I star so f of 3 I over 2 in times the width 3 over 2 in so again let's step through this this is gonna be you know we're thinking of putting rectangles under the curve here f of 3 I over 2 in is going to be our height at the right-hand endpoint and 3 over 2 in because we're using a uniform partition is going to be our width height times width is area I'm gonna sum those areas and take the number of rectangles send it out to infinity so now this is really just a big algebra problem it's a matter of simplifying this expression down into something that is much easier to work with alright so our function says X cubed plus 2x so we're gonna have 3i over 2 in cubed plus 2 times 3i over 2 n times our width of 3 over 2 in okay so let's work through this I have the limit as n goes to infinity so we're going to cube this expression that's going to be 27 I cubed over 8 n cubed now notice these twos will cancel so I'm just gonna have a 3i over in and then we're gonna multiply by 3 over 2 in so let's distribute this now so it's 27 times 3 so we'll distribute this term 27 times 3 it's gonna be 81 I cubed and we'll have 16 into the 4th and now we'll have 9i over 2n squared all right now again let's do the same thing let's pull out all non I terms so we'll have the limit as n goes to infinity we're gonna have 81 over 16 into the 4th and then the sum I equals 1 to n I cubed here we're gonna have n over 2 n square excuse me 9 over 2 N squared and then the sum of I now again we've got formulas here so for the sum of the cube terms we have this formula n to the fourth plus 2n cubed plus N squared and then for the sum of the I terms we have that formula in N squared plus n over 2 so we have the limit as n goes to infinity we have 81 over 16 into the fourth now the formula again we're going to replace the cube and the just I to the first power with their formulas so for the it I cubed we have into the fourth plus two in cubed plus N squared divided by four and for the I we have n over 2 N squared times N squared plus n over two alright so we're almost ready to finish this let's just go ahead and multiply those terms out so that we see our result a little bit more clearer we're going to have 81 into the fourth plus 162 in cubed plus 81 N squared divided by 64 into the fourth power and then plus 9 N squared plus 9 in over 4 in squared all right so as n goes to infinity notice I have degree 4 over degree for that limit is going to be 81 over 64 and then we have degree 2 over degree to that limit is going to be 9 over 4 and adding these together in simplifying we're going to have the fraction 225 over 64 square units for our area so at the moment this is our approach to finding the area under the curve we're going to set up the a limit of a Riemann sum using a uniform partition and the expression for the right-hand endpoint and then we slog our way through that limit to find our area right pretty soon we're going to find a way to do this a lot quicker we're gonna make some connections with this idea of area under the curve back to that topic where we sort of started this transition the topic of the antiderivative so ultimately we will make this a much simpler process but for now we don't have that connection so for now when we find the area under a curve we're thinking of looking at the limit of a Riemann sum and these examples demonstrate how we will do that using a uniform partition and our expression for the right-hand endpoints all right and going through the entire limit were able to come up with an expression for a final numerical result for the area that we were intending on finding at the beginning