in this lecture we're going to talk about integration by parts which is related to the product rule that you've seen in Cal one and so we're going to just start with a few reminders from Cal one okay so there are two objects given a function so let's say I'm given a function f let's call it f as we often do little F say continuous that means its graph has no breaks or holes that's an informal definition there are two mathematical objects that you spent a lot of time talking about in Cal one at the end of the semester of Interest so one of them is called an indefinite integral and the symbol for it looks like this so This is called the indefinite integral of f indefinite as opposed to definite and indefinite refers to the fact that there are no bounds here okay so this object this is the definition of this object this is a family of functions okay so this thing here is a family of functions and it's the family of all functions that look something like this I'm going to write a big f ofx plus C where big F is any function whose derivative is little F so such that big fime of x equals little F ofx so big F is called an anti-derivative of f the opposite of derivative and C is any constant so if you pick one particular anti-derivative the reason you get a family of functions is for each anti-derivative you pick you can add any constant so any constant makes this a different function but all of them have the property that the derivative of this is the inside the reason this constant doesn't matter is because if you take the derivative of this right hand side if you take the derivative of big F and you know it gives you little F and you take the derivative of the constant it's zero so really doesn't matter what constant is here in terms of the derivative of this right hand side being little F okay so that's one object of Interest the second object of interest is oh before we do that let's do an example so for example let's do it in green so for example if I take the integral we'll do a simple one you're going to do a lot harder one soon if I take the anti-derivative of this thing the indefinite integral of this thing I need to find an anti-derivative of 2 x to the 5th I'm going to two is a constant so it pulls through an anti-derivative of X 5th is x 6 over 6 or sometimes people write 1 16 x to the 6 that's the same plus C and you can see that 2 over 6 is the same as a 3 so this is the same as 1/3 x 6 + C okay so this guy would be my little F and this guy would be my big F and this is my plus C okay okay so the second object of interest is a definite integral and this is a definition this is a hard definition and we're going to spend more time talking about it again um next lecture but if f is Contin on some closed interval then the integral with bounds from A to B of f ofx DX this is actually a completely different object this is a number okay so here's the definition of this object it's a little bit Loosey Goosey definition because there's this notation it's the limit of some remon sums so it's the limit when n goes to Infinity of what's called a remon sum so you did this in Cal one so if if it's a little iffy for you you might want to just go back and look at that briefly but we will talk about a little bit more our next lecture for now let's just kind of feel like we remember what it is from Cal one so this is called a remon sum and remon sums are tightly connected to area so if I have the graph of a function and I divide the interval and I pick xks and I look at in uh I look at rectangles with height F of XK the sum of the areas of these rectangles that's what this would be okay so just for now let's just kind of take this intuitively dig a little bit into your calcan memory so this would be the sum this would be an area so this is actually a number and this limit is when I take more and more and more rectangles so they get finer and finer and finer and eventually in the end what you get is the exact area under the graph that's actually how we Define the area under the graph as long as f is positive you might remember that if f is um going below the axis you have positive and negative contributions to area Okay so sometimes we think of this as the area under the graph if f is above the xaxis but in any case this is a number and it's equal to the limit of these remon sums okay this thing here is called a definite integral or the definite integral of f from A to B okay okay so why would I want to do either of these two things by the way why are people in math doing this why do we need to integrate well there are lots of different uh reasons and applications one that you might have talked about before is um going back and forth between integrals and derivatives is really powerful when you have um position and velocity so for example let's say I knew the velocity of a particle let's say this little F was the velocity when you integrate it you take the anti-derivative you get position so these would be all partic all possible positions of the particle with different potential starting points all position functions and if I had a particular time frame that I had let's say these were times and and I knew the velocity of the particle and I looked at its definite integral that would be give me its change in position over time over this time interval which is actually what I'm going to talk about next the fundamental theorem but in any case this would give you the area under the velocity curve which is actually the change in position of the particle so that's one example in general if I have any rate of change like let's say I know how fast um a cylinder is filling with water I know the rate of change of the wa water which is sort of like saying the velocity of how fast that water is pouring into the cylinder when I integrate I'm going to get how much water I actually have in there so there are lots of reasons that I need to integrate um from a science or practical application point of view okay so the these two objects are tightly connected and you know it very well even though um you may not feel comfortable stating it form formally but you know it really well because you use it all the time every single time you try to calculate one of these and the relationship between these two objects between this object and this object is called the fundamental theorem of calculus okay so these two and that's a like surprising thing in a way and it's a theorem and this is what the theorem says there's several parts to it but I'm going to talk about the part that we use all the time when we calculate a definite integral so the fundamental theorem of calculus says sometimes this was called part two in a previous class but it says if f is continuous on a closed interval AB then if I actually want to calculate this integral so the definition of this integral is this limit of Theon sums but this is a really really bad way to calculate the integral it's really hard but what's easy is if I take any anti-derivative doesn't matter which one and I evaluate the difference of that anti-derivative at the end points I get this definite integral f is any anti-derivative of little F that means being an anti-derivative means big fime of x equals little F ofx have to be a little careful about your notation to distinguish between these two guys these are two different functions okay by the way it doesn't matter if I took a different anti-derivative if I took a different one with a different C when I included the C I'd have plus C here but then I'd subtract C here so the C would go away so it doesn't matter which anti-derivative I pick I always get the same thing and you guys know really well how to do this for example if I have the integral that I had before to F from 0 to one I do my 2 * x 6/ 6 I'm not going to write plus C CU I don't care it doesn't matter I'm going to write this bar on the right between 0 and 1 which let's say I simplified to 1/3 x 6 0 and 1 and this is 1/3 1 6 - 1/3 * 0 this is the simplest integral you're going to see this whole semester so be happy while you can okay so this you know really well because you've used it a million times okay it connects the definite integral and the indefinite integral so now what we're going to do is we're going to talk about integration technique and this integration technique is called integration by parts and integration by parts comes from the product rule so first I'm going to remind you what the product rule says we often write the product rule with functions f and g but I'm going to write it with functions u and v just because that's the notation people often use for integration by parts okay so what does the product rule say the product rule says two two functions u and v and I multiply them and I want to take the derivative I'll write just Prime I could write DDX of this product I'll just write Prime the product rule says that I take the derivative of the first one times the second function plus and I could write different things here I could write the first function times the derivative of the second or I could write V Prime Time U it doesn't matter in which order I write this or or this okay that's what the product rule says the product rule assumes that these u and v are differentiable which I need because I need to be able to calculate these things okay so now let's combine this with what we've just said about integrating now I'm going to integrate so I'm going to integrate both sides so let's say my u and v are differentiable instead of being continuous I need them to be differentiable they're differentiable on an interval AB then if I take the integral of the Der of this left hand side I'm going to get the integral of this okay and I'm going to just write the integral of this plus the integral of this because I know that if I have an integral of two things I can integrate them separately I know that integrals add okay so that's what I'm going to do so I'm going to write the integral of U Prime V of X DX from A to B plus the integral from a to U of x v Prime of X DX okay but what did I just see with my fundamental theorem of calculus I saw that integration and differentiation are opposite right I had that the integral of f ofx was Big F and there was this plus C okay I'm not going to worry about the plus c because here I've chosen bounds but this is saying the integral of f Prime the integral the integral of big F Prime because if F big fime is f I'm saying when I integrate a derivative I get back where I started that makes sense right because derivatives and integrals are opposite if I take the integral of 2x is 1/3 x Cub if I take the derivative of 1/3 x Cub I just get back to sorry if I take the integral of X2 it's 1/3 x Cub if I take the derivative of 1/3 x Cub I just get back to x^2 so what happens when I take the integral of a derivative here I just get this function and since I have it between A and B this is just equal to this function inside at B minus that function inside at a so this is U of B * V of B minus U of A * V of a and that's equal to this these this thing I have on the right hand side okay so what integration by parts tells me is what this integral equals if I want to solve for this integral I'm going to put this one on this side with a minus sign okay so I'm going to State the integration by parts formula and we've Bas basically just proved it so it's a theorem integration by parts and then we'll do a bunch of examples so you can practice because it's tricky to figure out how it works okay so we need u and v to be differentiable so suppose u and v are differentiable functions on some interval AB I'm not going to worry too much about details about the end points but I need them to be very good on this whole interval then what the integration by parts formula tells you is it gives you this equals this minus that one so I'm going to write it down so then the integral from A to B of U of x v Prime of X DX equals U of b v of B minus U of A V of a that's my left hand side here minus the this integral here okay so this looks so complicated why bother with this asinine formula okay why bother well we're going to see in the examples the reason is sometimes we cannot calculate this integral but we're able to calculate this one so if we're a ble to calculate this one this we can calculate directly and that gives us this integral so the whole idea of integration by parts is if this is really hard to calculate if I can't do it directly maybe I can simplify the piece that I have to integrate that's going to be really important because we're going to see It's Tricky a little bit to figure out what to choose for U and what to choose for V and one of the criteria is that you want this to be simpler to integrate than what you started with so often you want a function U whose derivative simplifies for example if U is X2 the derivative of U is 2X x is simpler than x squ so that's a good thing okay you're also going to need to find um a function V that if I know V Prime I can figure out what V is so we're going to talk about that a little bit more in a minute just a couple of more notes about notation before we do some examples sometimes people use instead of they kind of get rid of X and they just write this they write U instead of writing U of X they write U and instead of writing V Prime of X DX they write DV okay so DV is the same as V Prime of X DX it means the same thing and U of X is just I'm writing it as U I'm kind of suppressing X in the notation and sometimes people just write * V here and they write this bar from A to B and this means exactly U of x * V of X evaluated at B minus evaluated at a so that means the same as this okay and then they write minus the integral from A to B instead of U prime time um V they'll write instead of V of X they'll just write V and instead of U Prime of X DX they'll just write du okay so this is actually the most common way of writing the integration by parts formula okay it's the same thing it's just kind of a shorthand notation and by the way if I didn't have these bounds here I argued with these bounds but if I didn't have these bounds it's exactly the same formula because it's still the product rule if I have indefinite integrals everywhere I can still do this and so this same exact formula holds but without any of these bounds so it would look like this without the bounds it would look like integral U DV is U * V minus integral of VD okay that's what it would look like okay so I won't write that down because it's basically the same formula without the constants but you get the idea