How to do two (or more) integrals with just one

can you compress two integrals into one let's say you want to find what we'll call the second anti-derivative of this function that is you want to find the integral of the integral of this function so you'd have to integrate this function once and then integrate it again which is a bit of a hassle but is there a way to skip that intermediate integral and just calculate the entire second anti-derivative in just one step can you turn two integrals into one can one integral do the job of two as a matter of fact the answer is yes in fact any number of repeated integrals can be compressed into just a single integral expression how and what does that single integral look like [Music] if you've taken calculus before you probably spent a lot of time computing derivatives and integrals and after doing both you probably came to regard derivatives as the easier beast after all there are a plethora of shortcuts you can use to speed up taking derivatives like the power rule the product and quotient rules and above all the chain rule and together these shortcuts enable you to find the derivative of any elementary function that is a function that's a combination of familiar functions like roots exponentials logs and trig functions integrals on the other hand are much more ornery there are some shortcuts and tricks but they often require some tricky algebraic manipulation before they can be applied and ultimately they sometimes don't work at all as the integral of an elementary function doesn't have to be elementary even numerically derivatives are often easier as to approximate a derivative you can just take the slope of the secant line between two points that are very close to the tangent point whereas numerical integration pretty much always requires some kind of summation of a large number of small quantities as you can imagine this will only get worse with higher order integrals like taking two integrals one after the other or three or more even with computers calculating multiple nested integrals numerically can get slow and even inaccurate as roundoff errors can accumulate more the longer the calculation gets but if multiple repeated nested integrals can somehow be reframed as just a single integral this could help a lot but how exactly do you turn multiple integrals into one let's start with just two nested integrals how can you turn two integrals into one well there is admittedly one uselessly simple way to do it step one integrate the inner integral step two congratulations you now have a single integral but obviously this is no good since the whole point is to avoid doing any intermediate integration when turning the double integral into a single integral now if you've got some foresight you might already be getting the feeling that this task is impossible i mean if you turn a double integral into a single integral then however you did it surely the inside of our new single integral must simply be the result of integrating the inner integral so it seems like the task is basically to integrate the inner integral without actually integrating it which is probably impossible since that would effectively mean you've obsolated the core of integral calculus hmm well maybe i don't know but dang it i really don't want to go back to computing a billion integrals how about we see how far we can get if we try to do this anyway but how on earth should we approach it well how about we start by trying to come up with a good picture for what a repeated two-fold integral represents now picturing an indefinite integral that is an anti-derivative of a function f x is not so clear of course by definition it's whatever function has f of x as its derivative but that's a fairly indirect way to picture it a better idea might be to look at a definite integral instead where we put lower and upper limits on our integral the good old fundamental theorem of calculus says that if we make the upper integration limit a variable then the function it defines is an anti-derivative of f the upshot is we do have a really nice direct visual for what a definite integral represents it's the area under the graph of f between x equals a and x equals t and as t varies we can see how the area changes so we can represent a double antiderivative with two nested definite integrals where the inner integral integrates along x from x equals a to x equals y and the outer integral integrates the variable y from y equals a to y equals t where t is our final independent variable that defines the double anti-derivative those of you who have had multi-variable calculus might see where this is going a nested double integral represents the volume of a certain three-dimensional solid so let's see if we can visualize this solid now if you haven't seen double integrals before or would just like a refresher i have a video on my old channel about how to interpret and set up double integrals i'll leave a link to it in the description if you want to check it out [Music] to start let's focus on the inner integral alone for now here the lower limit a is just some constant turns out it doesn't really matter what a is as long as it's a constant so just to keep things simple for now let's take a to be 0. as for the upper limit y we should think of it as an independent variable we can control as you change y you change the area under the f x curve the integral is computing by the way take note that in this picture y is not meant to represent the vertical coordinate in this grid system rather it's meant to be an independent parameter controlling the area we're sweeping out under the f x curve you'll see in a minute why i chose to name this parameter y instead of something like s or t but for now just keep in mind that in this picture y doesn't have any connection to the vertical axis alright so for each value of y we plug in we get a different shaped area under the curve and the higher y value we plug in the further to the right this area extends but now let's place all these different integration areas together on their own separate y-axis where their position on the y-axis reflects the value of y used to define that particular area under the curve if i fill in the gaps with all the infinitely many other areas for all the other values of y we get a solid whose volume is the double integral we want and our original double integral computes the volume of the solid by first finding the area of an arbitrary slice of the solid along the x-axis and then integrating these slices along the y-axis starting from y equals zero and ending at y equals t to carve out the volume of the solid neat but there's more than one way to slice up a solid what if instead of slicing parallel to the x-axis and then integrating them up along the y-axis we try going the other way around what if we take slices first along the y-axis and then integrate them up along the x-axis well first of all what do slices along the y-axis look like hey it kind of looks like they're all just rectangles why do you suppose that is well think again about where the solid originally came from it came from taking various integration regions of the f x curve and stacking them all together but the key thing to note is that each integration slice came from integrating the same function f x that means if you focus on just one particular x value the point it corresponds to on the f x curve doesn't change as you increase y the integration region will extend further and further to the right but no part of the region that came before changes this constancy of the curve's height at a fixed x value translates into a flat contour of the solid when you walk on it in the y direction so a slice of the solid along the y direction will result in rectangles whose height is equal to f x where x is the position of the slice on the x-axis but what about the rectangle's width well based on how we constructed our solid the far side of the solid represents the integration slice for picking y to be equal to t so the far end of our rectangular slice along the y axis must be located at y equals t as well but what about the near end of our rectangle well first observe this remember how y represents how far to the right our integration region sweeps out [Music] but then each of these integration regions or slices was placed along the y axis at exactly the same y value to form the solid that is the position of an integration slice on the y axis is exactly equal to how far to the right that slice extends that means when viewed as a slice of the solid the far right endpoint of each integral slice occurs where the x and y coordinates are equal meaning the lower boundary of our solid's base is the line y equals x so when looking at any given rectangular slice along the y direction its near end must be located at y equals x where x is the position of the rectangle along the x-axis since the far end of the rectangle is always located at y equals t it means the width of the rectangle is t minus x and since the height of the rectangle is f x the area of the slice must be t minus x times f of x [Music] so that's the area of the xth slice of the solid in the y direction and so the volume of the solid can be computed by integrating this area from 0 to t along the x-axis so there you have it we managed to represent a repeated double integral as a single integral and we got away with it without having to do an intermediate integration because it turned out when you slice the solid in the y direction you just get rectangles and you don't need integration to find the area of a rectangle let's test it out let's use this formula to find the second anti-derivative of six t we know the answer should be t cubed first rewrite the double anti-derivative as two nested definite integrals then according to the formula this should equal the integral from zero to t of t minus x times six x dx distributing the six x through we get six times t x minus six x squared now since we're integrating here with respect to x the variable t is treated as a constant so we get three t x squared minus two x cubed evaluated from x equals zero to t plugging both bounds in for x and subtracting we get three t cubed minus two t cubed minus zero minus zero so the end result is just t cubed exactly what we expected or well almost if we're computing a general double anti-derivative here we actually have to tack on an arbitrary linear term to the final answer to account for the first integration constant itself getting integrated but the salient part of the result is the t cubed part alright so it seems to work but something about this may still seem kind of fishy think about what this formula is doing we started with a repeated double integral and somehow we wrote it as a single integral we compressed two integrals into one so like we suspected earlier does that mean that the integrand of our new single integral expression somehow represents the result of integrating the original function one time is this expression the first anti-derivative of f in disguise well not really the reason why is a little bit subtle the single integral expression on the right is actually quite a bit different from a regular integral expression for one thing there are two variables in the integrand instead of just one and one of those variables also appears in one of the integration valves what this means is we have the strange situation where the thing being integrated is itself changing depending on the upper integration bound that means the usual interpretation of a definite integral with a variable upper bound as just being a simple anti-derivative of the integrand no longer applies because the integrand itself changes with the upper bound so the single integral expression really is an unusual beast it fundamentally doesn't represent just a single anti-derivative of something as a matter of fact it represents two so that's how you can compress two repeated integrals into one but as i said at the beginning it's actually possible to compress any number of repeated integrals into just a single integral expression first rewrite the n fold integral as a nesting of indefinite integrals where the lower bound of each integral is a constant and the upper bound is the integration variable of the next integral that contains it up until t that's a bit of a mouthful so for simplicity i'll abbreviate it like this as it turns out n repeated definite integrals of f x can be rewritten as the following single integral expression it's pretty similar to the two integral case except the t minus x factor is raised to the n minus one power and we also divide the whole thing by n minus one factorial this more general formula is known as koshi's formula for repeated integration though not to be confused with koshi's integral formula which is a totally different beast now i don't have a nice visual derivation for this more general version of the formula but from an algebraic point of view it basically just comes down to recursively applying integration by parts to generate compressions of higher order repeated integrals to give the basic idea i'll flash the derivation of the n equal 3 case on screen for a moment so those of you who want can pause and read it also i should be up front here that although this formula allows you to turn multiple repeated integrals into just one it's probably not that useful for calculating them symbolically most of the time this is because other than a few special cases computing the integral of t minus x to the n minus 1 times a function will most likely mean doing integration by parts n minus 1 times and so you'll end up doing n integrations anyway however i believe this formula can be useful for computing repeated integrals numerically because instead of having to compute n many nested integrals numerically which is computationally intensive and potentially error prone you can just get away with integrating a single modified integral i think that's pretty cool furthermore despite their reputation for being tougher to work with than derivatives integrals seem to be rather unique in being compressible like this derivatives don't have a similar compression formula at least as far as i know but before we go there is one other application of koshi's repeated integral formula worth mentioning you see there's this function called the gamma function written like this and one of its properties is that evaluating it at a positive integer n results in n minus 1 factorial but what's neat is you can actually plug in non-integer numbers as input into the gamma function like one-half five-thirds or even pi effectively giving you a way to take the factorial of non-integer numbers for example gamma of one-half turns out to be the square root of pi which is kind of like saying negative one-half factorial equals the square root of pi pretty weird now i won't go into how the gamma function works in this video but all you need to know for now is that such a function exists and we can compute it and that it relates to the classic factorial in that evaluating gamma at a positive integer n is equivalent to n minus one factorial okay but if you remember our cauchy formula had an n minus 1 factorial in it so if we want we can replace it with gamma of n but now that we have gamma in there what do you think this formula will compute if i plug in say n equals one-half hmm [Music] [Music] you

Channel: Morphocular

Views: 366,681

Keywords: math, calculus, integration, integrals, repeated integrals, cauchy's formula for repeated integration, integral compression

Id: jNpKKDekS6k

Length: 18min 3sec (1083 seconds)

Published: Thu Jun 16 2022