Applications of Integration - AP Calculus Unit 8 Review

hey everybody today we are going to be reviewing ap calculus unit 8 applications of integration this is going to be the last video on the series so i hope you enjoyed this series so far and let's just have a great ending so first thing i want to talk about is the average value of a function so we've talked about average rate of change uh actually i don't i don't think we specifically talked about that but the average rate of change is something from algebra or right where you take the slope by doing like um you know like y2 minus y1 over x2 minus x1 that's average slope i mean in algebra class that was just the slope of the function here in calculus if you do that that's just going to be the average rate of change of a function right versus instantaneous which is the derivative here we're talking about not the average rate of change but the average value of a function so basically if i have like a function that goes kind of like this if i average all of these values if i take like their heights their positions of all of them and then i put them like all on one kind of like line or something it's gonna look something like this something in the middle that's kind of halfway through if we take the average of all of these um and it's gonna look like that we're gonna get a numerical answer from that so this would make more sense if we talked about a finite amount right but this is calculus we're doing an infinite amount of stuff so we're taking all of the points on here adding them up finding the average and then put plotting it on one particular line so that's what basically we're going to be doing today um so how does that work well i'll give you the formula so this is going to be i'm going to call this the average value of f f with a bar on top um that's because it's very commonly denoted that way if in physics for example if we have average velocity it's going to look like that um if we have average acceleration it's gonna look like that and so on and so forth here in this class uh we don't really denote it like that but you can um it might be more helpful to call it f sub average like that uh if you want to do that you can you're fine with that um but i just want to kind of show you what's commonly used so this is going to be equal to we'll have 1 over b minus a times the integral from a to b of f of x dx and i'm directly giving you the formula because i'm about to explain what's going on here we can't really do anything to manipulate this formula right now but if we say that the integral of f of x dx is equal to capital f of x this can kind of help us rewrite some of this so the integral of f of x is just capital f so this is going to be capital f of x evaluated at b and minus a basically so this is going to be 1 over b minus a times the capital f of x but this guy is evaluated starting at a all the way to b right and so really what that means is we're going to take 1 over b minus a and multiplied by capital f of b minus capital f of a this is technically the fundamental theorem of calculus right and then i'm going to rewrite rewrite it this way just to make it look better so we got f of b minus f of a over b minus a and this this is capital f right so this doesn't totally help us because it's like okay we started with this lowercase and now we're ending with the capital left right but you can kind of see that if you if you think about if you think of capital f as the integral of lowercase f then you can kind of see well this looks like the average rate of change of our integral function right so it's kind of like we're doing the average so the rate of change is kind of like the derivative of the integral and it's not totally the derivative because the derivative is instantaneous this is just an average so it's kind of like we're you know it's kind of a weird thing going on but this does give us the average rate of change i'm sorry the average value of our function f of x so you can plug in any function let's say x squared and you can plug it into this formula and you have to do it on a particular endpoint uh you can't really you know like x squared looks like this and it keeps going off to infinity so if you try to plug in infinity it's just not going to work right but um this does this will work for any of the questions that ap will give you uh they'll give it to you on an on a specific interval where you just plug in there um and on that interval basically this is just that length of the interval so let's say your interval is i don't know from negative three to seven so then what you do is one over seven minus negative three so it's going to be ten because that interval is ten units long so you're going to do 1 over 10 times the integral from negative 3 to 7 of whatever function they give you so that's just a small little thing that i wanted to mention the average value of a function so the second section i want to discuss is the area between two curves this really isn't that hard to understand basically we know how to take the area under one curve so let's say we have this function here the area under this curve from like i don't know here to here we just find the area here we take the integral so let's say this is a to b we just take the integral and this is f of x from a to b of f of x bx this is going to give us the area right but let's say we had instead we had something like let's say we were asked to find the area between these two curves we have f and g here okay so the area between these two curves is this region here so we're going to just shade that in so now we just got to find the area well if you think about it so first obviously the end points so we're going to take an integral so the endpoints are going to be here and here so we can just draw these two lines there we got a and b so the integral from a to b of what exactly now well basically we're just going to find well you can think of us okay well we can find the integral of f of x dx that's going to be too much though because we're including this portion that we don't want well that portion that we don't want is exactly the integral of g of x from a to b so what we can do is we can just take f of x and then subtract off that g of x and then integrate that and that's it that's literally it um and that's our final answer now it's not always going to be f minus g it could be g minus f or whatever other functions but really you want to do the top function so the top minus the bottom function okay that way you're taking the air to the bigger area and then subtracting off whatever you don't want okay uh and it could also go under the x-axis it'll just become negative at that point but you can just treat it the same way and it's always good to draw a picture so you know exactly what what you're dealing with now there might be cases where you don't exactly know what's on the top and what's on the bottom so for example if you have a function like this and then another function and then like another function like that like that so you don't really know what's on the top of the bottom because this is kind of under this guy but they're the same function so it's like what well what you can do is turn these into functions of y okay so they might give you these in terms of functions of y you just have to make sure you know what variable you're talking about but then what you're going to do is the right function minus the left function so the area here would be from a to b and in this case a is here and b is here these are the y values okay so you're going to do the right function with respect to y minus the left function with respect to y and then d y you're integrating with respect to y this means that you cannot have any x's in here so you're going to turn if you had something like i don't know if f x was equal to x squared then so this this is kind of like y equals x squared so then you're going to square root both sides so then you have the square root of y is equal to x and you really can think of this as the square root of y equals like g of y so g of y equals square root of y okay so this is going to be your your new function it's not going to be x squared it's not going to be square root of whatever it's going to be square root of y right so you just have to make sure that if you're dealing with stuff like this it's always going to be in terms of y only so either x only or y only so it's either top minus bottom or right minus left okay uh and yes graphing it's it's typically better to try to convert these into functions of x to graph them and then convert them back into functions y because they might give it to you they might say y equals this or x equals this okay if they give you y equals this and you realize you have to do right minus left graph it as a y equals function because you know how to graph functions of x and then convert the form the formula into something in terms of y okay that way then you can plug it into here and then integrate as well as the endpoints too make sure your endpoints are also y coordinates okay so just be careful with that i'm not going to go through any examples because that's there are tons of other videos on youtube and examples on the internet about that kind of stuff but i just want to give you guys the foundation of what's going on all right for this next section we're going to talk about disks and washers specifically they are methods of finding volume so yeah we're going to we're going to find volume using integrals now you might be like whoa that's that's a large step but it's okay it's just kind of like um pretty fairly basic volumes uh relative to other things um you'll go more into depth with this in calculus 2 and way more in depth in calculus 3. all right so for now this is just kind of introductory stuff um so yeah let's just get into it so if we have a function actually let me let me draw the whole axis here so let's say we have something like this and we are taking some region i'm going to say we're gonna take like this this region here okay and we're going to revolve it around the x-axis so it's going to basically go kind of like that we're going to revolve it and think of it going like outside of this two-dimensional world because this is volume so it's going to be three dimensional so think of it as being rotated all around like this kind of like line here so it's going to go kind of like this and it's going to go all the way around 360 degrees so uh this region if i can kind of just try to draw it so it's going to kind of hold on i'm really bad at drawing 3d things but uh hopefully i can try to it's gonna go like that and it's gonna have like um i don't know it's gonna look kind of like a football you can kind of think of it that way or a lemon maybe um but yeah it's gonna it's gonna be this kind of football looking shape that i think that's the best way to describe it right so the way we're gonna do this is what we call the disc method okay because it's like we're rotating rotating a disc because if you if you took a cross section uh like like a cross section like this this way if you just sliced it here and you looked at it this way you would see just circles because they're being rotated it's the whole thing is being rotated around the axis so you're going to see kind of the circular cross section so we know the area of a circle there is a circle so i'm going to say area of a circle is equal to pi r squared okay but the radius changes the radius is smaller here and larger like over here because you can see this radius is different than that radius this is definitely bigger than that so the radius changes it is actually a function of x because depending on where you are depending on where you are in in that particular interval the radius is going to be different so the radius is a function of x and the radius is exactly whatever f of x is so if this is f of x we can just change this radius for f of x so the area of the circle for one of our cross sections is pi times f of x squared for one particular cross section but we want to know what it is for all of the cross sections added together that's specifically what an integral is so if we want to find the volume this is going to be the integral of pi times f of x squared dx and of course we're going to go from a to b so here in this case a looks like 0 and then b looks like i don't know 5 or whatever but yeah it's going to be pi times f of x squared basically this is the area of a circle and we're integrating all of these and so it's we're integrating a bunch of disks basically um and that's precisely our formula now again it could also be rotating around the y-axis in that case you're just going to take the area from a to b where a and b are both like y coordinates of pi times f of y squared it's going to be functions of y squared and then d y so the same thing is just going to be different so make sure if if you if it tells you it's rotated about the x-axis or the y-axis okay make sure of that because if it's the x-axis it's just this it's the y-axis you change everything for y's but you don't again you just don't you don't just substitute a y for an x you actually have to convert your your function of x into a function of y okay then you can just do everything that way and make sure if you have these x coordinates that you get the y coordinates as well okay i cannot stress that enough it's really important that you make sure everything's in terms of y and you actually convert it properly okay so this is the disc method so what what is the washer method well if you know what a washer is it's kind of like this like ring type thing so it's like a circle it's a whole disk but then there's like a middle portion cut out so the way that would kind of work here is let me just kind of draw this again basically if we had our axes here and our figure was like this something like that and we were rotating this about the x-axis so it would look you know kind of like this and then uh it's being like i can't draw it it's just it's hard to draw but like you're rotating this you're gonna let me just get rid of this actually because it just makes it looks worse um but you're rotating this guy about about the x-axis so there's gonna be this hole here because you're not this this section is like blank so like kind of this like region here like around that area it's it's we're not gonna count that area but this area we're going to count so in this case we're going to be subtracting off this area and maybe it's not constant like i like i'd right here maybe it's it's kind of weird maybe it's like this right so these are kind of two curves here so let me just kind of make them look better like that so these are two curves so i can call this f and this one g and we are rotating them about the the x-axis here so uh the way we would do this it's very similar to what we had earlier but we're going to subtract off this g area so it's kind of a combination of that area between two curves and the disc method which we just did so it's going to be so the area i'm sorry the volume so the volume is going to be equal to the integral from a to b again a is kind of here and then b is kind of here basically the integral from a to b of we're going to take the top minus the bottom actually first we have to multiply by pi right since we're rotating everything and then we're going to take f of x minus g of x square that and then dx so this is our radius because our radius is some like a line like that so that's basically f minus g because if we if we took this line all the way to the bottom that's the whole height of f and then i'm subtracting off the height of g and then i have that radius basically so it's going to be f of x minus g of x that's our radius square it multiply by pi that's the area of one of our cross sections and then we mult and then we we add up all of them with our integral and that's going to give us the full thing and again same thing works with the you know if you're rotating about the y axis so that's kind of the washer method there and there's one more method to finding volume and that's going to be using cross sections so let's say now we have um i don't know some let me just draw this again let's say we have some region here right and it's and we have these cross sections that are squares that are perpendicular to the x-axis so basically we're going to have like a square they're going to be poking out basically so it's going to be just imagine a square with this like width and this height just kind of poking out outside like towards you um and so we're going to have like an infinite number of cross sections so let me kind of just draw it like that you know since we uh since we're looking at it kind of this way it's going to be kind of like different again i'm really bad at drawing these kinds of 3d figures but uh it's something like this it looks like um you can just imagine like an infinite number of them okay so how would we figure out what the volume of that figure is well uh basically we just take one of the cross sections and then integrate that so we know the area of a square is going to be uh the side and i'll just call it uh i don't want to call it s uh i'll just call it uh l it's gonna be l squared the length squared basically okay so basically to figure out what i mean all we have to do is really just um find you know since our length will depend on the function here so again we're going to have like an f and a g here that's going to be our length so our length is going to be specifically it's going to be f of x minus g of x right and then we're going to square that and that's the area of one of these squares it's going to be f of x minus g of x squared because it's kind of f minus g and then that's kind of our height there and then we're going to square it and then to find all of them we're going to integrate it and then dx this is no longer the area this is now the volume of the whole thing so it's going to be the integral of f of x minus g of x squared dx right and it's typically it's just going to be top minus bottom it's not always going to be f minus g it could be just top minus bottom whatever your actual functions are called and again if they're perpendicular to the y-axis then you would do the integral with respect to y and everything else is in terms of y okay well what if your cross sections were not squares they could be other figures like the equilateral triangle so if you had that the volumes this is the volume of the square this is volume of a triangle this is going to be equal to the first what is the area of an equilateral triangle well it's an area of equilateral triangle it's going to be a equals of four thirds pi no four thirds uh times s squared i believe let me just let me just double check that okay i was wrong is root three over 4 times s squared okay where s is just our side and so our sine in this case is just going to be f minus g so we can do root 3 over 4 times f of x minus g of x squared dx okay this way we're taking one of these equilateral triangles and then integrating them to find all of them combined together and yeah so you should know this area formula i will make a video soon on where to get that area formula from uh where it comes from basically but you can kind of try to play around and see if you can figure it out okay well uh there's also you can find the volume with respect to a semi-circle all right so pretty much um that's going to be well the area of the semi-circle that's so the area of the circle is pi r squared but the area the semi-circle is one-half so i'm going to say pi over 2 times the so it's going to be the integral and again these are from a to b always make sure a to b all right so a to b of pi over 2 times um it's not just going to be f of x minus g of x because that's the diameter we're going to take the square we're going to have to take it put cut it in half to find the radius basically so it's going to be f of x minus g of x over 2 squared dx right so you could think of this as 2 squared is 4 and then just move it over here and say it's pi over 8 times f of x minus g of x squared you could also think of it that way but these are the three main things that you would find cross sections for i just want to make sure you guys are familiar with the formulas and you might have to use a combination of any things we talked about today so your problems can get pretty annoying and pretty difficult but some of them use graphing calculators which is good so you can kind of use that to your advantage but yeah so that's the end of this video and that is the end of the ap calculus review series i hope you enjoyed all the videos and i really enjoyed making them so thank you guys for watching and i'll see you later bye guys

Channel: Math Café

Views: 8,786

Keywords: math, calculus, ap, ap calculus, integral, integration, applications, average value, volume, cross section, disk, washer

Id: 4SOGj1WS3sw

Length: 24min 54sec (1494 seconds)

Published: Sat Mar 26 2022