welcome to Mac TV with Professor V this video is a comprehensive review for calculus 2. and it'll have to be done in multiple parts so here's part one now I'm going to begin with area between curves and volumes because that's the first topic that we cover at the campus that I teach but I know a lot of schools start with integration techniques and this topic I'm about to go over is in calculus one so if that's the case for you then just skip ahead till you get to all the integrals they're coming up don't worry but if you're one of my students then you need to know this stuff for the final so first off finding areas between curves so we'll get we'll warm up with a nice basic problem find the area enclosed by the given curves we have y equals 2x minus x squared and y equals 2x minus 4. now nowhere in the directions does it say graph but you absolutely must graph every single time okay don't even think about trying to get away with doing the problem without graphing especially if you go to calc 3 forget it so before we start putting a graph together I need to figure out the intersection for these two curves so I'm going to set 2X minus x squared equal to 2x minus 4. I can see 2x is going to cancel out on both sides negative x squared equals negative 4 that means x squared is positive 4. so X is plus or minus 2. and then let's figure out what the Y coordinates are for these points of intersection so I'll just substitute them in here if x is positive 2 then Y is going to be 0 and if x is negative 2 Y is going to be negative 8. tell y equals 2x minus x squared that's going to be a parabola opening downward y equals 2x minus 4 that's just a line and I know where this Parabola and line are going to intersect fabulous here's x axis y axis negative 1 2 and then negative eight one two three four five six seven eight okay they intersect right here and then they also intersect at positive 2 0. so right here okay well I already know how to graph the line I'll make this a little bigger yeah because the y-intercept is negative four boom and then the parabola 2x minus x squared just to get a nice graph going I can factor out an X and then you have 2 minus X so I can see here the x-intercepts all right x equals zero and at x equals two so here's an intercept here's an intercept that means the vertex of the parabola that's going to be halfway between at one the y coordinate Will Be One so it looks something like this and it comes all the way down over here maybe I should match it on the other side let's make it come down just as far beautiful okay so we have to find the area enclosed by these two curves that's going to be this area and based on the way the functions are written it looks like we're going to want to integrate with respect to X which means we have to set up our area so that the base of each little rectangle is Delta X and the height would be top curve minus bottom curve and yeah that works out so the height of each of these rectangles to give us the area is going to be top curve minus bottom curve which curve is on the top the parabola 2x minus x squared minus the bottom curve is 2x minus 4. so this is going to simplify to negative x squared plus 4. and then the base of each of these rectangles is Delta X which becomes DX when we integrate okay now we're ready to set up our integral so the area is going to go from since we're integrating with respect to X they need to be limits in terms of X that bound the region and this region here begins where x equals negative 2 and it ends where x equals positive 2. so those are my limits of integration so area equals definite integral from negative two to two of Base times height so negative x squared plus 4 and Delta X becomes DX okay good now a lot of the time if if we have a region that is symmetric with respect to the y-axis you can double it and go from zero to two we cannot do that here because this region is not symmetric with respect to the y-axis so don't do anything illegal um let's just go ahead and anti-differentiate term by term anti-derivative of negative x squared is negative one-third X cubed plus this is 4X from negative 2 to 2. okay now let's substitute in our limits of integration we have negative 1 3 times 2 cubed that's eight plus four times two is eight minus negative one-third times negative 2 cubed is negative eight plus four times negative 2 that's minus eight so this gives me let's see here I have negative eight thirds plus eight minus another eight thirds plus another eight so that's negative sixteen thirds plus sixteen so that's going to be 32 thirds okay now depending on the region sometimes you're going to have to integrate with respect to Y and then you would do right minus left um for the sake of time so that this video isn't five hours long I'm not going to cover every possible scenario for every single section so if you want more details more examples then you can go back through the video lectures I'll link the playlist in the description box along with the titles for some of the key video lectures okay moving on next topic volumes and these are volumes from taking a region a bounded area and revolving it about a certain axis when you're in calc 3 you'll compute volumes that arise from different methods okay not spinning stuff so first method we're going to go over involves using discs or washers to compute the volume and just remember for discs and washers we slice perpendicular to the axis of rotation okay so if you have a disc then the volume is given by definite integral from A to B of pi r squared this could be DX or d y depending which way you're slicing but if you're spinning around the x-axis then perpendicular slices mean you integrate with respect to X if you have a washer because you have a hole in the middle of the region then the definite integral for the volume would be Pi capital r squared minus Pi lowercase r squared capital r is the outer radius outer radius and then lowercase R is the inner radius okay all right good so let's look at an example here find the volume of the solid generated by revolving the region bounded by the given lines and curves about the x-axis and then I add it in you must include a clearly labeled sketch of the region of course you must okay so y equals negative 7x plus 14 and Y equals 7x those are the two curves and we're also bounding the region with x equals zero that's the y-axis first things first we need to figure out the intersection for these curves so let's do it so I'm going to set negative 7x plus 14 equal to 7X that gives me 14 equals 14x so they intersect when X is one okay if x is one y is going to be 7. all right so let's see we're gonna primarily stay in the first quadrant right yes because we're bounding these two curves that intersect in the first quadrant with the y-axis and then let's see so one of them is negative seven X Plus fourteen that means the y-intercept is at 14. and then the other one is just y equals 7x now this is my graph I'm going to scale it how I wish in a way that makes sense for me so you know I'm scaling by sevens in the y direction and by one on the X Direction and that's totally appropriate they intersect at one seven this would be y equals 7x right here and then negative seven X Plus fourteen slope here boom boom okay now also you have to identify the region that you're spinning correctly x equals zero that's the y-axis so the y-axis this curve and this curve bound the region this is the region that we're spinning okay don't pick the wrong triangle don't be a goober and think it's this no one said y equals zero is bounding the region okay this is getting spun about the x-axis so we're going this way so just imagine if you have spun it that original region would get revolved or reflected below the x-axis and if we're gonna slice perpendicular to the axis of revolution then notice we have a washer situation okay if you're spinning around the x-axis perpendicular slices move in the X Direction so we're going to integrate with respect to X which is easier notice the functions are given to me in terms of x so I want to use washers in this case if I if I were trying to do shelves that's coming next you would peel parallel to the axis of Revolution so you would integrate with respect to Y and I I don't want to have to rearrange these equations so they're in terms of Y and do the problem that way okay plus you'd have to set up two integrals so anyways let's see if we can identify outer radius in our radius okay always look at the original region before you spun it to figure out what the radius is so outer radius goes from there to the center okay you could always think top minus bottom to figure out the top curve here is yes negative 7x plus 14 what's on the bottom we're just spinning around y x axis excuse me y equals zero so it's negative 7x plus 14 minus zero which You Don't See okay what about the inner radius inside radius goes from here down there so what's bounding it on the top y equals 7x what's bounding it below zero you don't have to put the minus zero but that's kind of what's going on okay maybe I'll write it just so for trickier scenarios you're already thinking that way okay then we're just going to set up our integral for the volume it's pi times definite integral these limits need to be for x for the region before we spun it so from zero to one right the region exists when X is zero up until one and then you have outer radius squared so that would be negative seven X Plus fourteen this gets squared minus in a radius squared separately DX okay so don't subtract them and then Square it you square each of them separately then subtract them okay this integral is not too bad you just gotta keep your wits about you so you don't multiply or do something incorrectly you don't want to multiply wrong okay negative 7x plus 14 squared that's going to be 49 x squared middle term is 2 times 7 times 14 so that's going to be 14 squared basically 196 x plus 196 minus 7x squared that's 49 x squared DX and then notice these 49 x squares cancel I can take out 196. and paste uh Take It Outside the integral so we have 196 pi times the integral from 0 to 1. Negative X Plus 1 which is 1 minus X DX oh now this is the easy integral so 196 pi anti-derivative is going to be x minus one half x squared from 0 to 1. so this is 196 pi times 1 minus one half minus zero so one minus one half that's a half a half of 96 Pi is 98 pi boom all right good let's do one more and then we'll move on to shells so find the volume of the solid generated by revolving the region about the given line we have the region bounded above by the line Y equals nine below by the curve y equals 9 minus x squared on the right by the line x equals three and we're spinning about the line Y equals nine so that means wherever y equals 9 is we're spinning around it so I'm just going to put a lot a lot in the positive y direction so here's nine we're spinning around it and then one two three the regions bounded above by y equals nine that's here below by the curve 9 minus x squared so 9 minus x squared that's a parabola opening down intercepts or plus or minus three on the right it's bounded by the line x equals three so where's the region it's right here and we're spinning this region about the line Y equals nine spinning that way so after we spin it it's gonna look something like that and if I slice perpendicular to the axis of rotation I have a nice washer I'm going to integrate with respect to X okay now here's the deal we don't have an outer radius and an inner radius okay we've got a disk here is the one and only radius always use the original region before you spun it so to figure out the radius just do top minus bottom what curve defines the boundary right here on the top it's y equals nine so the radius is 9 minus what's bounding the radius on the bottom it's the parabola 9 minus x squared so just like for area between curves when you found the height you did top minus bottom you do the same thing for the radius here okay this is top of the radius bottom of the radius what's bounding it on top y equals nine what's bounding it below the parabola so if I clean this up this is going to be 9 minus nine that'll cancel and then this gives me positive x squared okay good now to find the volume we have pi the limits are for X the original region was bounded from zero to three and then you have the radius squared oh what a harmless little integral so Pi zero to three x to the fourth DX so that's one-fifth x to the fifth so pi over five times x to the fifth from zero to three that's just going to be 3 to the fifth minus zero so 243 pi over five voila not so bad okay good now we're going to look at the other method that I mentioned earlier which is the method of shells cylindrical shells so the formula if you're using cylindrical shells to compute volumes is volume equals integral from A to B of 2 pi r h d x and you're going to peel parallel to the axis of rotation so the opposite direction from disks and washers okay so next example use the shell method to find the volume of the solid generated by revolving the Shaded region about the indicated axis oh how nice they already gave us a graph and we're spinning this region for example for about the y-axis so we're spinning this way okay and we always peel parallel to the axis of Revolution so my students always have a hard time drawing the cylindrical shells here's the trick ready okay we're spinning around the y-axis and we peel parallel so please draw for me a line segment in the original region in the Shaded region parallel to the y-axis right whatever you're spinning around Draw Something contained only within the region though so something parallel to the y-axis anywhere over here that your heart desires I just felt like putting it right there in the middle so here is the height of one of your cylinders please reflect it onto the other side okay I'm gonna reflect it over here then Loop it together draw yourself a little cylinder oh beautiful beautiful this was the height right here okay this down here this is your radius now we have to figure out what they are for this particular problem radius and height I'm going to say something profound please take note the radius when you're doing cylindrical shells never involves the function that bounds the region or functions the radius is always just plain old X or plain old y if you're spinning around the x-axis or y-axis if you spin around some other line like y equals nine x equals one then you'll have a constant in there along with X or Y but this will never involve like this function here no no no so if I'm spinning around the y-axis I'm going to peel parallel I'm going to integrate with respect to X oh and the radius is just X okay if I drew another cylinder over here like this the radius would be whatever X is right there so the radius is X it's always plain old X or plain old y the only time you would involve a number is if you're spinning around something other than x axis or y-axis now what's the height it's right here top minus bottom top curve they gave it to me here 4 sine X over X bottom is just y equals zero so you won't see the minus zero part but I'll write it so we know we've thought about it four sine X over x minus zero so just 4 sine X over X now let's set up this this volume is put the 2 pi outside please limits of integration should be for X whatever bounded the original region so that's going to be 0 to Pi and then you have radius which is x times height for sine X over X DX so lovely so X cancels out and then let's see now 2 pi and 4. I'm going to take that out so we have 8 Pi integral 0 to Pi sine X DX anti-derivative of sine X is negative cosine X let's put the negative all the way outside and then we'll do cosine X evaluated from 0 to Pi so I have negative 8 pi times cosine of pi minus cosine of zero cosine of pi that's negative one minus cosine of zero is one so this is negative 8 pi times negative 2. which gives us 16 pi okay very nice let's do one more a little more spicy okay that one was too easy huh I would probably never put that on an exam so use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves and lines about the x-axis so we've got x equals 18 minus y squared x equals y squared and Y equals zero bounding the region now I can tell these are two parabolas opening horizontally so let's figure out where they intersect so I'm going to set 18 minus y squared equal to Y squared that means 18 equals 2y squared so y squared is nine so Y is plus or minus 3 and I can just look back here if Y is 3 or negative 3x is nine so these two parabolas intersect at nine plus or minus 3. I'm also bounding the region with y equals zero so that's going to be x-axis okay good and then we're spinning the region about the x-axis as well okay fabulous so let's just draw the portion of the graph that we're going to spin so we've got intersection at nine and three and negative three okay they intersect here and here so x equals y squared is just going to open up like this okay x equals 18 minus y squared oh you know what I need a little more x-axis I do because we have to go all the way to 18 for that guy oh heavens okay so that's nine that looks like about 18 right there so 18 minus y squared if um Y is 0 x is 18 and then it'll intersect right here so this Parabola is going this way okay and then we're also bounding the region with y equals zero so you might say well do I use the top half or the bottom half to spin around the x-axis I mean it doesn't matter just pick a half you're only spinning half of it though let's just take the top half okay pick a half and then we're gonna spin it around the x-axis and so it's just gonna reflect you know over here it's gonna look like a little football like a deformed football okay so here's the thing we need to use the method of cylindrical shells because we were instructed to do so and I need to figure out the radius and the height what did we spin around we spun around the x-axis please draw for me a line segment parallel to the x-axis in the original region that we spun so in here can you please draw a line segment for me in this direction why yes we can right here boom okay this is the height that line segment is the height mirror it on the other side then draw in the rest to make it look like a cylinder okay and there you have your radius so the radius is going to go from the center out I'll just draw it like this that's the radius and then there's the height now the height in this case we're not going to do top minus bottom because we're going to integrate with respect to Y so you're going to think of right minus left which curves bound the height of the cylinder well on the right is this Parabola which one was that guy that was 18 minus y squared so we have 18 minus y squared minus what's bounding it on the left y squared so this is 18 minus 2y squared what's the radius the radius is just y okay we only spun around the x-axis nothing fancy schmancy so now we're ready to go volume is going to be 2 pi limits should be for y on the original region that we spun so the original region we chose to spin the top half from 0 to 3. then you're going to have radius which is y times height which is 18 minus 2y squared d y the worst is over the setup is always the gnarliest part so then we've got 2 pi integral 0 to 3. this is 18y minus 2y cubed d y and then we can go ahead integrate term break term 2 pi this is going to be 9y squared minus one half y to the fourth from 0 to 3. and then we have 2 pi times 9 times 3 squared minus one half times three to the fourth minus zero so that's two pi times 81 minus 81 halves so that's 2 pi times 81 halves so this is just 81 pi woo okay it can definitely get trickier I have more examples in my full-length video lectures but this is just enough to give you like a refresher if you feel like it's something that you're shaky on which a lot of students are then go back do more practice on volumes for sure for sure I work this is just a general recap work is based first on the idea of force force is a push or a pull on an object and force is equal to it can be equal to mass times acceleration and instead of acceleration I have second derivative of position function with respect to time if you are in the metric system then your units are Newtons Newtons are kilograms times meters per second squared that's all one newton if we're in the U.S customary system we use pounds or imperial system and then if acceleration is constant then force is also constant and so the work done can be given by force times distance or displacement and the units for work are joules if you're in the metric system or if you're in the U.S customary system it's a foot pound kind of weird I know and then the most common application that we're going to cover is when force is not constant but variable and then you integrate your Force function times DX okay and so I'm going to do the trickiest kind of problem that my students all hate um the work problems where you're pumping stuff out of Tanks all right so a conical tank is resting on its apex so what does that mean that means it looks like this guys it looks like ice cream cone the height of the tank is eight feet and the radius at the top is seven feet so this is eight and then across here is seven fabulous the tank is full of gasoline weighing 45 pounds per feet cubed how much work will it take to pump the gasoline to the top give your answer to the nearest foot pound okay so I'm gonna need to integrate Force required to pump the gasoline out of the tank Force if we're in U.S customary units is measured in pounds and they've given me here the weight density of gasoline pounds per feet cubed so I need to cancel out basically the feet cubed so that I can have some sort of representation of the force required to pump out the gasoline from the tank so feet cubed is volume so you always start these problems by figuring out what's the volume not of the whole tank no the volume of just one slice and I know it's impossible to actually like slice up a tank of gasoline but we pretend because that's how we Define integration we always take a region and we chop it up right when you first learned integrals you chopped up the region into a bunch of rectangles so here I have this region that I want to figure out the work for I'm going to chop it up into a bunch of hypothetical slices of gasoline okay so there's a slice of gasoline do you like it the volume is going to be area of the base area of the base times the height times the thickness okay well area of the base it's a circle so it's going to be pi r squared and then the thickness of one slice I'm going to call Delta X some people call it Delta y it's up to you I'm going to count it where like this is the slice at x equals zero and since this tank is uh eight feet high this is the slice at x equals eight so I don't know this is the slice at x sub I okay now I want to see can I express the volume in terms of x pi r squared here's the thing depending where you slice the radius is changing isn't it isn't the radius of this slice up here different than the radius of the slice I drew lower so how can I express radius in terms of x well here's the radius of an arbitrary slice right there and then this is the height that it's at some arbitrary x sub I so you have similar triangles yes you do this is the radius this is X and this is nestled inside the big cone with radius 7 and height 8. so we similar triangles are in proportion to each other I can set up common ratio 7 over 8 is equal to R over X yes good and then I'm trying to solve for R in terms of X so that means 7 8 x is equal to R so now I'm going to just basically plug that in here to get the volume of one slice in terms of x and you know what instead of calling it volume of one slice I'll call it Delta V less writing same thing so it's pi times 7 8 x squared Delta X and the units on this would be feet cubed right because this is representing a volume so this is 49 pi over 64 x squared Delta X Phi Q now I want the force required to lift one slice so Force to lift one slice of gasoline is going to be the weight density which they gave us here 45 pounds per feet cubed so you just take 45 pounds per feet cubed and you multiply by your Delta V which is in feet cubed those are the units on it see this will cancel with this and then now I have 45 times 49 pi over 64 x squared Delta X pounds okay great 45 times 49 Pi in case you're dying to know it's 2205 pi over 64 x squared Delta X okay the last thing I need is the displacement of each slice or the distance that each slice moves so let's go back to the picture really quickly okay so work is force times distance the the layer of gasoline whose height is at x equals zero needs to be moved or move a total distance of eight feet to be out of the tank right what about the gasoline that's already all the way at the top how many feet does it have to move zero right what about the layer at one foot high it moves seven feet so at X height it moves 8 minus X to be out of the tank so that's the displacement of each slice eight minus X for how I set up my picture you could set it up to frame okay so now let's write our integral out the work is going to be definite integral you're only doing work on wherever there's gasoline in the tank and they told me it was all the way full right yes they did it's full so we're going to be doing work all the way from zero to eight those are my limits zero to eight I have Force here so that's two two o five pi over 64 x squared Delta X becomes DX times this displacement 8 minus X okay don't get scared by the integral take the constant outside immediately 2205 pi over 64. then you have integrals 0 to 8 x squared times eight minus X DX and then we can distribute the x squared to 205 pi over 64 integral 0 to 8 this is 8x squared minus X cubed DX and see now the rest is not so bad so you have 2 2 0 5 pi over 64. this is going to be eight thirds x cubed minus 1 4 x to the fourth evaluated from zero to eight so what is that 2205 pi over 64. if I plug in this 8 for X I'm going to have 8 cubed times another eight that's eight to the fourth over 3 minus eight to the fourth over four lower limit zero I can factor out this 8 to the fourth so then I'm going to have 8 to the fourth times two two o five pi over 64. times 1 3 minus 1 4. 64 cancels with 8 to the fourth and this just becomes 8 squared which is 64. so I have 64. times 2 2 0 5 Pi 1 3 minus 1 4 is 1 12. so this is over 12. 64 and 12 can cancel this I can divide a 4 out from each so this will be 16 and 3. and then now I know everything's reduced I can't cancel out any further so I'll multiply 16 by 2205 and you get thirty five thousand 280 pi over 3 what are our units the upper lovely foot pound okay that's it notice I did not reach for a calculator until this very last step I don't let my students have one at all I give them slightly nicer numbers so they won't need it here but you know learn to use your brain the calculator you were born with and clean up as much as possible only use the calculator if your teacher allows you when absolutely necessary okay okay good glad we had that talk I feel better um all right next topic is average value if you're we know about averages right if you just want to average something you take all of the some things add them up divide by how many some things you had fabulous if you want the average value of a function same idea you add up all of the values of the function how do we add up infinitely many things in math we take an integral and we do it on the interval that's given from A to B and then you want to divide by however many there are well there's infinitely many on the interval from A to B you just divide by the length of the interval B minus a okay so here let's look at a common application find the average value of the function over the given interval find C such that f equals the average value of the function over the interval and then sketch the graph of F and a rectangle whose area is the same as the area under the graph of f okay so first things first find the average value so you just have to know the formula part A F average is 1 over B minus a so that would be 2 minus negative three times the definite integral from A to B negative 3 to 2. of f of x DX okay this is a relaxing little integral one-fifth times the definite integral from negative three to two six minus x squared DX let's see this is one-fifth times six x minus one third X cubed from negative three to two of one-fifth times 12 minus eight thirds minus negative eighteen plus twenty-seven thirds that's just nine so this is going to be one-fifth times I'm already doing this 12 plus 18 that's 30. and then I have negative eight thirds minus twenty-seven thirds so that's minus 35 thirds right yes and then 30 is 90 thirds minus 35 thirds that's 55 thirds so I have one-fifth times 55 thirds and that reduces to 11 thirds so that's the average value that's the average height of the function if you want to think of it that way that's the average y value of y equals 6X squared on the interval from negative three to two that's the average height okay now Part B said find C such that F of C equals the average value well F of C would be 6 minus c squared 6 minus c squared and they want that to equal 11 thirds okay so that means 6 minus 11 thirds is c squared 6 is 18 thirds 18 minus 11 that's seven thirds equals c squared so C equals plus or minus rad seven thirds now C needs to be on the interval from negative three to two so you would need a little calculator just to kind of eyeball I know well I know seven Birds you know that's more than two so the square root is going to be more than one it's going to be less than two though because seven thirds is less than four right so whatever this square root is it's between one and two it's about 1.53 and both are in the interval so we have two values of c I'll say C1 is negative Rod seven thirds and C2 is positive we're at seven thirds okay the graph of same as the area under the graph of f so we're going to graph F on the interval from negative three to two and then a rectangle whose area is the same well the height of the rectangle is going to be the average value that's the whole point of this okay I need an extra page I didn't know this problem would take so long here we go so I'm gonna graph f of x equals 6 minus x squared on negative 3 to 2. okay very good so let's hop to it um let's see I'm just quickly in my head if I plug in negative 3 for X what does that give me 9 6 minus nine negative three okay so I don't need to go too low in the y direction I need to go up to six though don't I something like this should be good here's my x-axis here's my y-axis one two three four five six one two three like that okay one two three one two that's as far as we're going okay so 6 minus x squared goes through zero six negative three negative three okay and then x-intercept is rad six I don't know it's between two and three who cares where pretend that's rad six right there where I crossed and then at two if I plug in 2 for x 6 minus 4 is 2. so it's going to go to here okay don't go past that don't get crazy so area under the Curve it's gonna be this portion this is positive area and then this little slipper baby slipper here is negative area okay good now we need a rectangle that has the same area well the height of the rectangle that's why we found the average value it's 11 thirds eleven thirds three and two-thirds wee bit more than three so say maybe it's like right here I'm gonna say that's 11 thirds okay and it's gonna go all the way from negative three to two okay I think I overshot it there hold on let's fix that okay that much better beautiful that is a rectangle whose area is the same there you go so the height of the rectangle maybe you can't see it so the height right here is 11 thirds the base is the same the base is going to be from negative three to two perfect eleven thirds times five gives you 55 thirds which should be the total area under the curve of 6 minus x squared from negative 3 to 2. um what was the whole point of C1 and C2 I'll show you right now where they intersect right here that's our C2 that's positive rad seven thirds and then over here that's C1 that's negative rad seven thirds okay that should be good for average value just remember the formula and the idea now we're moving on to the next topic which is probably one of my favorites techniques of integration and we're going to start with a review of integration by parts so the formula when you're applying integration by parts to evaluate an integral is that you start off with the integral of U DV you've got to figure out what to let you be and what to let DV be and then that is equal to UV minus the integral of v d u so when to use integration by parts when you have a product of two functions and you need to evaluate their antiderivative and U sub will not work okay so I always default to U sub as like my first go-to what I'm trying to solve an integral and I'm looking here at this example y times tan inverse of Y nope no U sub will save me here and I have a obvious product of two functions Y is one of them and the other one is tan inverse of Y so now it just comes down to deciding who's going to be you and who's going to be DB we're going to pick that keep in mind you have to be able to figure out d u and v before you proceed so a lot of the times only pick DV to be something that you can anti-differentiate meaning I know there's no way tan inverse of Y is going to be DV because we don't know the antiderivative off the top of our heads for Tan inverse of Y that means tan inverse of Y is going to be U and then DV is obviously going to have to be what's left over which is y d y all right and then from here let's figure out u d u and v excuse me so d u derivative of tan inverse of y That's 1 over 1 plus y squared d y how do I know that I memorized it a long time ago so you do need to just memorize all your bread and butter derivatives from calc 1. No Way Around It if you haven't pull it together quick as a bunny memorize it okay V is antiderivative of Y so that's going to be one half y squared we don't put a plus c when we do integration by parts also we have limits of integration here so this is going to be equal to look back here we have UV I just always think it's this diagonal product because I always set it up the same way so U times V that's one half y squared times tan inverse of Y I'm gonna list that this is to be evaluated from zero to one that part does not get integrated again it's done minus integral from 0 to 1 and then what's here integral of V times d u v d u is right here people so that's going to be one half y squared times 1 over 1 plus y squared d y are you all right fabulous okay so we've got here one half y squared tan inverse of Y from zero to one I'm not going to evaluate it now I'll do it all at the end when I'm done integrating minus I want to take this one half out okay get in the habit of taking your constants outside of the integral it'll make life easier then we have zero to one I'm going to rewrite this as y squared over 1 plus y squared okay so y squared over 1 plus y squared d y now how to integrate y squared over 1 plus y squared well look degree of the numerator is 2. degree of the denominator is also 2. so we need to do long division anytime the degree of the numerator is greater than or equal to degree of the denominator do long division I'm going to show you a shortcuts you don't have to do long division oh so I notice I have 1 plus y squared in the denominator I wish I had 1 plus y squared in the numerator it would make life easier so I'm going to add one and subtract one so really I just did nothing right don't freak out and then watch what's going to happen so I'm just going to rewrite this first term here minus one half okay this is the best part right here integral zero to one now we have these two I'm going to write over the denominator so we've got y squared plus 1 over 1 plus y squared minus and then this little guy's all on his own okay look at him so brave minus 1 over 1 plus y squared d y and what was the point of that I don't have to do long division this is my little shortcut way so I don't do long division does it always work now but when you have just one little term in the numerator that matches one of the terms in the denominator then you can just add and subtract whatever constant you need to make it happen if you hate this then just do the long division you'll get the same thing okay one half y squared tan inverse of Y from zero to one minus one half integral zero to one this right here this first term is just going to reduce to one isn't it I have same thing in the numerator and denominator minus and then just leave this alone one over one plus y squared d y fabulous you should be able to integrate this now I'm serious so one half y squared tan inverse of Y from zero to one minus one half anti-derivative of one is going to be y minus we should know anti-derivative of one over one plus y squared yes it's tan inverse of Y and you know what this is getting evaluated from zero to one so I'm just gonna go ahead and add another little bracket zero to one get rid of this one and say all of this is getting evaluated from zero to one okay good before I move there though let's distribute everything and make it as simple as possible so we've got one half y squared tan inverse of Y minus one half y plus one half tan inverse of Y and this is evaluated from zero to one and then oh looks like I need more room okay here we go so we have I'm gonna plug in one now the upper limit one half times one squared times tan inverse of 1 is pi over four yes you should know that minus one half times one plus one half times tan inverse of one again pi over four that's the upper limit minus if I plug in zero um first term is going to be zero minus zero plus one half times tan inverse of zero is also zero so what are we left with this is pi over eight minus a half plus pi over eight so I have 2 pi over eight that's pi over four minus a half very good I love it how was that yeah if you need like more basic examples then you can go back to the video lecture on integration by parts I figured we're reviewing for the final let's keep the spice level intense all right so here's another favorite of mine we have integral of e to the 2x cosine seven X DX this one you need to do integration by parts twice because it boomerangs boomerangs meaning the original integral comes back after your second iteration of integration by parts and how do I know that that's going to happen just by looking at it well I know the derivative of e to the X always involves e to the X it's cyclical and same thing for cosine 7x after four derivatives you get back to basically the original function uh albeit with some adjustments for for the constants but that's the idea so whenever you have a mix a product of two functions whose derivatives are both cyclical then it works out that you have this little Boomerang situation okay and the cool thing is it actually doesn't matter which function you let bu and which you let B DV as long as you're consistent through both iterations so I'll show you what I mean so let's just pick U to be e to the 2x simply because it was first and then DV is going to be cosine 7x DX all right then let's find d u so d u would be 2 e to the 2x DX and V antiderivative of cosine 7x is going to be a positive 1 7. sine 7X all right so this original integral I'm gonna write it e to the 2x cosine 7x DX is equal to u v what's that going to be that's going to be 1 7 e to the 2x sine 7x minus integral of vdu vdu is this product right here so that's going to be 2 over 7. e to the 2x sine 7X DX okay I'm gonna repetition by Parts again on this integral right here again notice I have product of E and trig function okay um you have to just match up if if U was the exponential function and DV was the trig function then choose it the same way here at this step okay you can take the constant out or you can put it in here it's kind of up to you um call it something else you can't call it just you again and DV again U Bar and DV bar is usually the appropriate choice for round two so U Bar I will call 2 7 e to the 2X and then DV bar is sine 7x DX then D U Bar would be derivative so that's 4 7 e to the 2x DX and then V bar would be a negative 1 7 cosine seven X DX all right good so let's see what we're left with now I don't feel like writing out this whole integral on the left hand side I'm just going to call it I okay so I represents all of this that's I and that equals 1 7 e to the 2x sine 7x minus and then now we have U Bar V bar so that's 2 7 times negative 1 7 so that's negative 2 over 49 e to the 2x there's no DX here what was I thinking shame on me um cosine 7X minus integral of V Bar D U Bar so that would be 4 over 49 negative so I'll make this a plus 4 over 49 e to the 2x cosine 7x d close it up okay let's clean this up a bit more so I have on the left eye my original integral equals 1 7 e to the 2x sine 7x plus 2 over 49 e to the two x cosine 7 x I'm Distributing this minus sign minus let's take the 4 over 49 out integral e to the 2X cosine 7x DX and then this is where the boomerang happens so notice can you notice that this integral that I still have here is the same as the original one that we started with who we have now called I all right so we're solving for I just like you solve for x in basic equations back in the day when you were in algebra and life was relaxing and you didn't even realize it so I'm going to replace this whole integral with I so I have 4 over 49 I negative and then here I have I equals 1 7 e to the 2x sine 7x plus 2 over 49 e to the 2x cosine 7x yada yada okay if I'm trying to solve for I then I want all the I's on the same side of the equation so I'm going to add 4 over 49 I here add 4 over 49i here technically this one I is 49 over 49i so on the left hand side I now have 53 over 49 I equals 1 7 e to the 2x sine 7x plus 2 over 49 e to the 2x cosine 7X and then the last thing that we need to do is just multiply everything by 49 over 53 all right to isolate I and then we have I equals I'm Distributing this through 49 and 7 are going to cancel so I have 7 over 53 e to the 2X sine 7x Plus boom this is going to be well the 49th cancels so we have 2 over 53 e to the 2x cosine 7x and then here's the part that's just tricky because you have to remember to put plus C at the very end since this was an indefinite integral we have no limits of integration we need a plus c all right that's it I really do love these if they're obnoxious to you that just means you need to practice them more okay very good One More by parts um and I like this one because you need to actually use multiple integration techniques um notice I have a composition of functions I have cosine of natural log of x so it's not really set up just yet I don't have a product of two functions to do integration by parts but I'm feeling like maybe it's gonna go there I don't know so let's just try doing a substitution I'm not going to do a u sub because I want to save the variable u in case I have to do integration by part so I'm going to just substitute out with t what should I let T be uh just take a guess a lot of the times if you have a composition of functions let it be the inside function okay I'm just playing around guys I'm just playing around so let t equal natural log of x um you could find DT now but it's easier let's rewrite this equation in exponential form so that means e to the t is equal to X right that's the same thing all right now if I differentiate both sides that means e to the T DT equals DX so let's rewrite this integral now in terms of t so I have anti-derivative cosine instead of Ln of X I'm going to write t and then DX gets replaced with e to the T DT oh and then this is going to be another little boomerangi one yes it is okay so we're gonna pick U and we're gonna pick DV can you see how we have a product of two functions there's one function there's the other one so u d v do you want to pause the video give this one a try on your own I think you can handle it yes you should try Okay U is e to the T DV will be cosine T DT and then d u e to the t d t v is sine t so already I know I'm calling this integral I it equals u v so that's e to the T sine T minus integral vdu so that's going to be e to the T sine T DT now with the boomerang ones you gotta go two rounds okay so we need a U Bar and a DV bar pick them uh the same way that you picked round one otherwise you undo what you did you'll see what I mean I've done I've done it wrong plenty of times back in the day when I was first learning so I'm giving you tips so you don't make the mistakes I made so you bar should again be e to the T the exponential function and then DV bar sine T DT D U Bar will be e to the t d t v bar ooh anti-derivative of sine T is negative cosine t so now my integral is equal to sorry that eye looks so ugly e to the T sine T minus all right We've Got U Bar V bar so that's negative e to the T cosine T minus integral V Bar D U Bar which is going to be a negative e to the T cosine t d t so I'll switch this to Plus e to the T cosine T DT huh look this is my original integral I and here it is the boomerang came back I'll Circle actually the whole thing right there's my boomerang wow isn't this our lucky day okay so this is I this whole thing so we have I equals e to the T sine t plus e to the T cosine T minus I right this is this negative is Distributing and then add I to both sides we're trying to solve for it so now we have two I equals e to the T sine t plus e to the T cosine t last thing divide by two so I equals one half e to the T sine t plus one half e to the T cosine T and then don't forget plus C should we box it I feel so good about life no remember we did a u sub at first the original integral was in terms of x and T is equal to Ln of X which also means e to the t is X so we gotta replace everything so we're gonna have one half X sine of Ln of x plus one half X cosine of Ln of X plus c I you could leave it like that but you know what it would look so fabulous let's factor out the one-half X and then we have sine Ln of X plus cosine Elena backs yes plus c so stunning do you love it yes good the more you get excited about doing stuff like this it'll be less torturous you know so even if you feel kind of confused by my excitement hopefully it's contagious and it'll make the process more enjoyable for you okay now let's look at trigonometric integrals you just have to kind of play around with them get used to patterns the biggest thing when you have a mix of Sines and cosines and if one of them is odd it is your lucky day because you just want to take the odd man out and then do a u-sub you'll see what I mean in just a hot second so to start I notice I have an odd power of sine so I'm going to break this up 0 to pi over 2. cosine squared 6X times I'm going to write this as sine squared 6X times sine of 6X DX so I took the odd power the odd powered trig function and took one of them out all right why am I doing that because I want this quantity right here sine 6X DX I want this to be my du or d u ish off by a constant is fine by me so just think backwards then okay if d u is going to be sine 6X then what would you need to be well that means you should have been cosine 6X and then d u would be negative 6 sine 6 x DX but that's no problem I can deal with like the negative six okay so can I replace everything in the integral all of my cosine six X's with u well this would be U squared um but what am I going to do with this sine squared I'm going to bust out one of my pythagorean identities right sine squared 6X is 1 minus cosine squared 6X isn't it okay so let's rewrite it we have 0 to pi over 2 cosine squared 6X times 1 minus cosine squared 6 x times sine 6X DX and then now we're pretty much ready to roll why do I say that so we've got this is U squared this is going to be 1 minus U squared and this is almost d u negative 1 6 d u is sine 6X DX so call it d u put the negative 1 6 outside okay it need not harass you inside the integral and then we're going to change the limits of integration these limits of integration are for the variable X now my integral is in terms of U so I need to change them so that they match don't put X limits if your integral is in terms of U change them immediately like a dirty diaper don't let it faster okay how do you change them you go back to right here where you decided what you was okay so U is cosine of 6x so that means U of pi over 2 the upper limit that's going to be cosine of 6 times pi over 2. that's cosine of 3 Pi cosine of 3 Pi is negative 1. and then U of 0 is cosine of 6 times 0 well that's the same as cosine of 0 which is positive one so my new upper limit is negative one my new lower limit is positive one okay it is what it is right off the bat I'm just gonna flip the limits of integration and make this a positive 1 6 and then we're going to go from negative one to one don't you just feel better already I certainly do and then we'll distribute the U squared so we've got U squared minus U to the fourth du H love it and then now we're ready to anti-differentiate oh my goodness this is no big deal we did these back in kindergarten this is going to be 1 3 U cubed minus one-fifth U to the fifth from negative one to one and then just keep it going so we've got one sixth I'm gonna plug in positive one so this is one-third minus one-fifth minus negative one cubed is going to stay negative one so this is negative one-third and then this will be plus one-fifth so we've got 1 6 times um this is going to be a positive one-third and minus one-fifth so that means I have two thirds minus one-fifth minus one-fifth minus two-fifths all right so this is 1 6 times that's ten minus six over fifteen how did I get that I did two times five that's ten minus two times three that's six over three times five there's your fifteen okay um so this is one sixth times ten minus six is four over 15 can we reduce we sure can this is two this is three this is 2 over 45. I love it no you don't need to go back to the original variable because we changed our limits of integration along the way so we're ready to roll problem beautiful okay so that's for Sines and cosines if one of them is odd pluck it off that's going to be D U play around with it make it happen what if it gets more spicy what if they're both even so nasty so you have to use half angle identities if you have sine squared x replace it with one half times one minus cosine two x if you don't have an odd power of cosine with it to do a u sub so you're stuck you can't do what you sub this is backup plan okay if you have cosine squared x then you're going to replace it with one half times one plus cosine 2X uh this double angle I really wouldn't use it unless I was like extremely desperate okay good so let's try one out this is a nasty one we have definite integral from zero to one half of three times sine to the fourth of two pi x DX so notice it's to an even power I don't have like an odd extra factor of sine or cosine that I could pluck off and do a u sub with so be careful when you use this um half angle identity whatever the argument is here it gets doubled so what's the argument right now it's 2 pi x it's going to get doubled watch what I mean you know me I'm taking this three out right away zero to one half and I'm first going to write this as sine squared 2 pi x squared DX right isn't that the same thing assigned to the fourth okay now I'm just going to focus right in here and I'm going to replace it with one half times one minus cosine double this double 2 pi x what does it become 4 pi x doesn't it and then all of this my Heavens is squared DX then you've got integral 0 to 1 half and a three okay check us out now this squared has to distribute to the one-half and to this whole thing so we're gonna have to foil that sucker out one half squared is one-fourth I want it outside with the three now that's a three-fourths okay zero to one half and then you have one minus cosine 4 pi x quantity squared DX can you square it three-fourths integral zero to one-half this is one minus two cosine four pi x plus cosine squared 4 pi x now let's see can we integrate this yeah no problem can we integrate this sure I just have to divide by 4 Pi we got it what do I do with this cosine squared one more time half angle use this one instead I remember them just remember the pattern and then the cosine one has a plus sign because I feel like cosine cosine matchy matchy it's so happy and then sine and cosine they're not the same so you know opposites um all right so I'm gonna replace cosine squared 4 pi x with one half times one plus cosine double this double this what's double 4 pi x 8 pi x Bravo is that what you were saying over there I hope okay so now we've got three-fourths integral zero to one half one minus two cosine four pi x plus distribute distribute one half plus one half cosine eight pi x DX before I integrate I can combine like terms and you always want to do that because then you'll have less stuff to integrate the fewer the terms the better so this is three halves minus two cosine four pi x plus one half cosine eight pi x DX all right we're almost there um so we've got three fourths times antiderivative of three halves three have sex very good minus two times so whose derivative did you take to get cosine four pi x why it was sine 4 pi x but I have to divide by 4 PI right divide by whatever coefficients in front of x to undo the chain rule think back if you were to take the derivative you would multiply by another 4 Pi so we got to undo that plus one half times whose derivative did you take to get cosine 8 pi x why it was sine 8 pi x but I also have to divide by that 8 pi and then all of this is evaluated from 0 to 1 half I'm not going to evaluate it just yet let's clean up so we've got 3 4 times 3 halves x minus cancel cancel that's a 2 now so I've got 1 over 2 pi times sine 4 pi x plus oh nothing cancels here dang it so 1 over 16 pi sine 8 pi x from 0 to 1 half all right plug in upper limit of a half so this is 3 4 times 3 halves times a half that's three-fourths minus one over two pi times sine of 4 pi times a half is sine of 2 pi that's zero plus 1 over 16 pi times sine of 8 pi times a half that's sine of 4 Pi that's also zero that's the upper limit minus if I plug in 0 yeah I'm going to get zeros for everybody else is zero so all we're left with is three-fourths times three-fourths are you a little sad don't be we did a beautiful job you should be proud okay that's like the nastiest case I I feel like would come up so if you can handle that problem you're ready um what about powers of tangents and other trig functions think back to what derivatives and what anti-derivatives you easily know do we know the antiderivative of tangent yes we do anti-derivative of tangent Theta D Theta is Ln absolute value secant theta plus C but I don't have tangent I have tangent to the fourth do I know antiderivative of tan squared theta D Theta no I don't know it we use Pythagorean identity to fix that who's tangent's friend secant squared theta do I know that antiderivative yes if it's tangent Theta okay so that has to be clear in your head why because we're going to break up tangent to the fourth two Theta into tangent squared to T oh it was a t not a Theta excuse me times tangent squared 2T DT and why am I doing that because one of these I'm going to replace with secant squared to T minus one and watch what's gonna happen now you're not going to leave it like this you're going to distribute and then now I have integral tan squared to T secant squared 2 T DT minus integral tan squared 2T DT okay let's talk about these one at a time this is going to be integral number one this is going to be number two so let's look at integral number one really quickly can you think of something to do you're gonna do a u-sub secant squared 2T DT is the derivative sort of of tangent to T not tangent squared just plain old tangent okay don't get wild but U equal tangent to T then d u would be 2 secant squared to T DT right okay so one half d u is secant squared to T DT which is exactly what I have right here and then I have tan squared 2T remember U was just tan 2T to the first power so this is U squared this is d u but I need a one-half so I'll put it in the front okay oh we can integrate this no big deal one-half times one-third U cubed plus c one so this is 1 6 what was U it was tangent two t so tangent Cube two t plus C1 there's that first little integral now what about the second one tan squared two t d t I just told you a second ago we don't know the antiderivative of tan squared but we do know anti-derivative of secant squared and we can get there very easily tan squared 2T is equal to secant squared to T minus 1 DT and then you should be able to integrate this no big deal this is going to be no U sub needed this is just one half tan 2T not squared just tan 2T and then minus anti-derivative of 1 is going to be t plus C2 all right and then just remember we have subtraction between these two integrals so I'm going to take integral number one minus what I have from integral number two so we have 1 6 tan cubed two T minus one half tan two t plus t plus c where C is C1 minus C2 was it not it was okay it really does come down to you have to know your trig identities and your trig derivatives so solid so you could come up with a strategy otherwise you will just stare at it for a half an hour and get nowhere you know writing down a bunch of gibberish like you're in a beautiful mind part two so just memorize the basics the bread and butter because that's how you are able to knock out these problems effortlessly okay we've got another one cotangent to the fourth three T DT it's going to be so similar almost identical to the last one we did I gave you some limits of integration pi over 12 to pi over 6 just to keep it spicy why don't you pause the video and try it on your own and I'm not even joking when I say it's almost identical to the last one okay did you pause it I hope you did so yeah it's gonna start off the same I'm gonna break it into cotangent squared times cotangent squared so we've got pi over 12 to pi over six cotangent squared 3T times cotangent squared 3T DT and then this cotangent squared I'm going to replace with cosecant squared 3T minus 1. and so we'll distribute now and we have pi over 12 to pi over 6. this is going to distribute here and to here okay so I'm going to just do it all in one shot so cotangent squared 3T cosecant squared three T DT minus integral pi over 12 to pi over 6. cotangent squared 3T DT write your limits of integration every step of the way don't stop writing them just because you're lazy unless you want to lose points okay good um let's just figure out how we're going to work with integral number one so we have here we're going to do a u sub let U equal cotangent 3T then d u is negative three cosecant squared three T DT let's see here I'm going to change my limits of integration for this first one okay so U of pi over 12 is going to be cotangent of 3 times pi over 12 which is pi over 4. so cotangent of pi over 4 is just one and then U of pi over 6 is cotangent of 3 times pi over 6 so that's pi over 2 so that's cosine over sine not pi over 2 which is 0. so we're going to have a negative one-third integral from 1 to 0 U squared d u and I would right away change this to positive go zero to one U squared d u then we're gonna add one divide by the new exponent so it's going to be 1 3 U cubed so I have 1 9 U cubed from 0 to 1. which is just 1 9. okay great now let's work on integral number two here it is pi over 12 to pi over 6 cotangent squared 3T so we've got pi over 12. to pi over 6 cotangent squared 3T DT so we don't know antiderivative of cotangent squared but I do know anti-derivative of cosecant squared so I can replace that using my pythagorean identities with cosecant squared 3T minus 1 DT you guys if you're Rusty on your pythagoreans it's fine hopefully you know this sine squared theta plus cosine squared theta equals one right even on a bad day you can hopefully remember that and then you're like dang it I need what's the identity again with the cotangents and the cosecans I can't remember well I know cotangent Theta is cosine Theta divided by sine Theta so if I want to squeeze out the identity with cotangent squared divide Everybody by sine squared because cotangent has sine in the denominator and then you go hmm what's this going to turn into 1 plus oh that's cotangent squared equals cosecant squared and that took me all of 10 seconds to do in the little margin of my paper so you can do it okay if you're stuck on an exam just remember the OG Pythagorean identity and then you're in business okay you can come up with what you need all right anyways now you should be able to integrate so what's the antiderivative of cosecant squared 3T it's going to be negative one-third cotangent 3T and then anti-derivative of negative 1 would be negative T and this is from pi over 12 to pi over 6. then now let's evaluate very carefully so we've got negative one-third times that's cotangent of pi over 2 right because I'm plugging in pi over 6 for my upper limit and that's 0 minus pi over 6 minus lower limit that's negative one-third times cotangent of pi over 4 was 1 minus pi over 12. what is this minus pi over 6 plus a third plus pi over 12 all right so then we're left with a negative pi over 12 plus 1 3 and that is the result from integral number two so putting everything together what did we have integral number one gave us 1 9. and we have 1 minus two so 1 9 minus this mess okay so we have now one minus two which is 1 9 minus negative pi over 12 plus a third which is 1 9 plus pi over 12 minus the third which is minus three-ninths yes so this is pi over 12 minus two over nine woo okay I hope you liked that one anyways that concludes part one I need a break we will do more we still have trig sub partial fractions Simpsons rule midpoint rule trapezoidal rule Arc Length surface area differential equations and sequences in series amp alert and con oh my goodness so there's more parts to come give this video a thumbs up subscribe if you haven't already if you need more in-depth explanations go to the calculus to video lectures everything's in order you can see breakdown explaining the concepts and more simple examples laying the foundation should you need it also you can catch me on Instagram and Tick Tock at mathftv with Professor V stay tuned guys I'll be back keep it up