 What do you want? Yeah? It was a question on the last website, website number four. The one with the cars? No, it was finding error bonds. Yeah? It was the integration of 27 codes that next went to the end. Oh, that one? Yeah. I could find a... From what? No. So this is really... Find Simpsons. Am I writing big enough in the back? Yes. Okay. Does anyone need me to actually do the problem, the trapezoidal approximation of this? Yes? No? There's an answer. It's either yes or it's no. Does anyone want me to actually write down what T8 is? Yes. She does. Okay. So T8. So what we do is we take our interval from zero to one and we divide it up into eight chunks. Three, four, five, six, seven, eight. And our x's that we're going to evaluate are the edges of each of these chunks. So this first one will be one eighth, then two eighths, which is a quarter, then three eighths, and four eighths, which is a half, then five eighths, then six eighths, which is three quarters, then seven eighths, then one. And for the midpoint, we have to use the middle of these steps. Let me skip the midpoint. Let's just do the trapezoid. I'll do the midpoint if you need one. Okay. And so now what we do, the trapezoid says we erect, we look at our function, it does whatever it does here, and we put little bars here and we connect the dots. And we find the area underneath what we get when we connect the dots. So the width of each of these things is one eighth. The area of each of these things is this side plus this side over two. So the height is however long this is plus however long this is over two. So it's going to be, since they're all over two, we might as well take a half. And now it's going to be f of zero plus f of one eighth. But in fact, when I do this one, the f of the one eighth plays a role and then when I do the next one, the f of the one eighth plays a role, so there's going to be two of those. And so on, whatever they are. Three eighths is next, blah, blah, blah. So in this case, it's one sixteenth. Oh, there's a twenty-seven out here. Let me just pull the twenty-seven out front. So it's twenty-seven sixteenths of, now f of zero is the cosine of zero squared. Two times the cosine of one eighth. But it's squared, so it's really one sixty-fourth. Two times the cosine of three eighths. Well, three squared is nine, and eight squared is sixty-four. Go all the way over to there or so much more. Keep going. Let me just keep going here. So let me not write them all down. Does anyone not understand what's next? Next one is the cosine of, I skipped a quarter. Sorry, this is a sixteenth. This is a quarter, nine, sixty-fourths, et cetera. Plus two times the cosine of seven eighths squared plus the cosine of one. So then you plug all this junk into your calculator and you get the answer. No, but on the exam, I won't ask such a nasty thing. On the exam, if you have an answer like one sixteenth of one half plus seven eighths plus nineteen sixty-fourths plus twenty-two over seven, that answer is correct assuming that those are the right numbers. You don't need to add those things together unless it's something stupid like one half of four plus three. Well, how about four plus eight? I would like you to write that as six. Okay, that wasn't your question. So for the midpoint, which I just erased, it's exactly the same except we only evaluate here in the middles which I don't want to write down and we multiply everything by one because we're just taking the height of the middles. If you think about these things in terms of the picture, there's sort of no way to go wrong. Okay, but your question was about the error. Yes. Right. So I do this, I get some number and the second part says what's the error? Let's do TA. It works the same way. Okay, so there's a formula. You don't need to memorize the formula for the exam. I will give it to you if you need it. Let's see if I remember the formula. The formula is, so the error is less than K times B minus A cubed over, I think it's 12 N squared. I think that's the formula. Okay. So B minus A, that's easy. This is 1 minus 0. That's 1, just forget that. We need to find K here. K is the maximum for X between 0 and 1 or my A and my B of the second derivative of X. So we need to take the second derivative and then say where will it be its biggest? So, okay. So here we need to take the second derivative of our function. So our function is 27 cosine X squared. So our first derivative, we have to reach the chain rule, is 27 times 2X times the sine of X squared. And it's negative. Yes? And the second derivative, which is the one we want, is 54, well, let me not reduce it yet. Just leave it out. Yeah, let me leave it at this. 54. And now I want to take the derivative of sine X squared. So I use the product rule, the derivative of X is 1, so I have sine X squared. And now I take the derivative of this piece, gives me X times 2X times the cosine of X. There's a negative sign that I lost. It doesn't matter because I'm taking an absolute value. Now we want to maximize this between 0 and 1. But the problem tells you, you know, don't worry about where the cosine and the sine exactly are the biggest. Just notice that sine of X is always less than 1 in absolute value, and so is the cosine. And since this is just an estimate, it's okay to just replace the sine and the cosine by 1 because they're never bigger than 1. So this is always less than, well, we have 54. I'm going to replace that by 1 plus 2. Now X squared is biggest when X is 1, not when X is 0, so I can replace that by 1. And I replace the cosine by 1. So the K that I want is 3 times 54, which is 162. So assuming I didn't make any arithmetic errors, I want K to be 162. And then, so now let's go back here. And so our error is less than or equal to 162 over 12 times n squared n was 8, so that's 64. So that's some number. It's not very good. Well, sorry. It's whatever it is. It's some number. You could plug it into your calculator and it's like 202 or... Does anyone know? Does anyone care? Okay, all the same. Okay, so does that answer your question? And for the midpoint, this is a 24. So the midpoint is exactly the same except it's half of that. And here we have to do something different for the midpoint. Other questions? Yeah? Can you go for a trick sub and arc sum? Okay. You want a trick sub with arc sum. Okay. Do you have one in mind? Okay. Do you want to do an easy one or a hard one? Hard one. Well, a hard one's going to take like an hour and a half. What do you want? Easy one. Okay, so I'm just going to make this up. So do you want it definite or indefinite? I know. Okay. So this one may come out icky. The numbers might be nasty, but do you want it... Okay. You want it middle hard or...? Okay. Let's just do this one. I can make it harder if you want. Is that hard enough? Or... That's fine. Okay. So say I have something like that. Yeah. How hard is it going to be on the test? I don't know how hard the one on the test is anymore because I know how to do it. I don't know how to play hard. This one is probably... Okay. So let me point out how it could be a little harder. So there's two ways that it can be hard. Number one, it can be hard because you don't recognize it as a trig substitution because maybe there's a 5 here and the constants are ugly and this isn't a 4, it's a 14. And now I'm going to have a square root of 5 and a square root of 14 floating around and it's just nasty. But conceptually, it's no harder. It's just nasty. So I don't want to drag around square root of 5s and square root of 14s. So that's one way it can be harder. Another way it can be harder is that the trigonometric integral that we wind up with can be hard. I don't know what the trigonometric integral we're going to wind up with is on this one because I just wrote it down. I don't think it's hard, but we'll see. It may be horrible, but it's doable. So this is probably about a good level of difficulty. Alright. So here I look at this and I say I hate this square root. I want it to go away. And I remember that sine squared plus cosine squared is one. So that means that sine squared is one minus the cosine squared. Or maybe I want to cosine because you want a dark sine. So we write it as cosine squared is one minus the sine squared. And so if this X were only some kind of sine then what's under the square root would be a cosine squared and I could take the square root and it would be great. So I just want to transform this into something that looks like one minus something squared. It's not a one, it's a four. But I can factor the four out which is really a two. So I can rewrite this as so I'm going to pull the four out but I'll leave it inside for now. And now I have an X squared over four which I can also write as X squared over two. Alternatively you can just let X be two sine squared. It's the same. So everybody okay with this so far? Alright. So this is an X over two. And so now I'm going to make the substitution X over two I want that to be a sine. Or if you prefer X is two sine theta. They're the same. And so that means that DX over two is a cosine. And so here or DX is two sine theta d theta and I'm running out of board space unless I want to sit on the floor so I'll come over here. I'm going to make that substitution. So X is two sine theta so this becomes so my DX becomes two cosine theta. My X becomes a two sine theta my X and my square root becomes one minus sine squared. So that's just plugging in. The four got eaten in this two. And in this two. Something? X is two sine theta. Well somehow my that's my X. This is X one minus X squared over two and this is DX. Yeah. If you like. You'll get the same answer in the end. It doesn't matter. I'm doing it with sine because he wanted sine. The cosine will work exactly the same. I'll get a negative sine. But then when I integrate it goes away. It's all the same. Yeah. I don't know where the four went actually. The four went on the floor. There's another two here. Thank you for being persistent about explaining where the four went. It went away because I found it annoying. The square root of four is two. So I can pull that out and this two becomes a four. So just to simplify the square root it's a two. These two cancel. So I get a two and now I have cosine theta and here I have a sine theta and here I have so this one is nasty. So now I have one minus sine squared which is a square root of a cosine squared which gives me a single cosine. So this one's nasty or it's not depending. This is an integral I hate. So this is one half the integral cosine's cancel d theta over sine theta which is one half the integral of the secant no the cosecant theta d theta. So this is not my favorite integral. And I forget how to do this one. It's the log of the cosecant plus the cotangent I think. It's negative log of cosecant plus cotangent. So we can work backwards. So the reason I hate this integral is it's a completely non-obvious substitution that makes it work. And I've forgotten what it is. You can probably suspect that if you see the integral of the cosecant thing. That was a hint. So this is negative one half the log of the cosecant plus the cotangent plus the constant. Now who actually got the cosine? Somebody. You did. I think if I had done the cosine I would get the secant plus the tangent and I would get the integral of the secant. They'll be the same when we substitute back what theta is in terms of x. But I think if I had done the cosine substitution it would be a secant and those cosecants and stuff would be. I'm not 100% sure and it wouldn't be negative to be positive. I'm not 100% sure but I'm pretty sure. Okay. So this is almost done but not really because the integral was expressed in terms of x and there were no thetas around. So we need to turn it back to x. If this had been a definite integral we would be done because when we did this transformation we could have plugged in in terms of theta and just evaluated and then happy. But it's not so we have to ask if x equals x over 2. So if x over 2 is sine theta then what's cosecant theta and cotangent theta? So we have to answer that question. So to do this usually I draw a triangle because we just both of these things are really just ways of writing the Pythagorean theorem. So we just draw a right triangle here where x over 2 is the sine of theta. So that would mean that the opposite over the hypotenuse this triangle has x over 2 equals the sine of theta and then we can use the Pythagorean theorem to figure out what the other side is. So this squared plus this squared equals this squared so that means x squared plus 4 sorry, x squared plus a thing equals 4. So the thing is 4 minus x squared. It often happens it doesn't always happen but it often happens that your thing that you substituted for turns out to be related to the other side. I don't want to say it always happens but it does often happen that you'll get the other side to be similar to the thing that you wanted to make go away in the first place. Okay, so now we can read off from this what the cosecant is. The cosecant is 1 over the cosine which means it's this over this and the cotangent is this over this really? Sine over cosine over sine, right? Yeah, so it's I believe so. Cosecant is 1 over sine and I wrote cosine, thank you. Cosecant is 2 over x that looks better and the cotangent is 4 minus x squared over x, that's better. So that means that this becomes negative 1 half the log of 2 over x plus square root of 4 minus x squared and had you done the substitution in terms of the cosine you would just permute the things in the triangle and you would get the same answer. Okay, is this hard enough for you but not too hard? So they don't the thing that gets hard about these things once you see the substitution to make is that sometimes these integrals come out to be horrible they come out to be something like I don't know, sine to the 8th cosine to the 4th which means you have to do this half angle thing a bunch of times okay other things you want me to go over? So the test isn't going to say you just have to figure out one of the sample tests I don't know if you've looked at the sample test some of them said do this do that do the other thing so I can tell you the test has nine questions I think six or seven of them are integrals but I don't tell you how to do any of them it's just here's six or seven integrals I don't remember exactly whether it's six or seven they're definitely nine questions you don't have to use the arc sine I mean the theta equals arc sine x over 2 so we don't have to use this problem what? you don't have to use the triangle thing well you have to know what the cosecant of arc sine is and you have to know what the cotangent of arc sine is and the only way I know how to know unless you just know them all the easiest way to understand cotangent of the arc sine is is by drawing a picture but if you hate pictures there are plenty of people that can do this without pictures so I don't care what method you use as long as A it's not cheating and B your method makes sense and it doesn't just come out come to you from extra sensory perception because it's very hard to tell extra sensory perception from cheating on all of the questions you do have to none of the questions say show your work but I want you to justify your answer that means show your work it means so I hate the phrase show your work because sometimes you can look at something and say oh I know the answer to that it's 3x squared plus 2 you just write it down but there's something that went on inside and how do I know that 3x squared plus 2 is the right answer so really I want you to justify your answer so that somebody who doesn't know it knows where it came from ok that's what people mean when they say show your work my son used to fail math test because they would say show your work and he said I didn't write anything down I just did it all in my head and they used to give him 0 anyway next topic where you all know it we're done ok see ya you had a question right ok so of the tests that are listed I only wrote the math 126 tests all the other ones were written by other people that I just grabbed so the show some work was written by professor jones he meant justify your answer some of them do say show your work or there's one of them that says I don't care if you show any work just write the right answer that's good it's not like matching and he said I don't care just give me this integral is a this integral is number d this number n whatever I don't use Google phrase show your work but so what the integral of convergence whether an integral converges good because what you were asking we're not starting for two weeks whether an integral converges or diverges so so this is related to but may not be obviously as so if we have an so the most obvious one is if we have something like let me do one that's integral 0 to 5 of 1 over x squared how about that we'll just make it easy I have something like that now this may be phrased as determine whether this integral converges or diverges or it may be phrased as evaluate the integral and if it diverges so if you look at this it's not obviously an improper integral but it is obviously an improper integral if you look at it because 1 over 0 is not 5 so we have a problem here and we have to say does it make sense so there's this trouble at 0 and so what we really want to say is this is the limit as L goes to 0 from above with the integral from L to 5 now this integral is easy the integral of x to the negative 2 is negative 1 third wrong way negative 1 over x assuming I can do my integral right which I can't always now we want to know does this limit make sense so when I plug in 5 this is good so I get negative 1 fifth but when I plug in L over L and now if I take the limit as x goes to 0 this blows up so this blows up so this diverges if I change the problem we do almost exactly the same problem for that problem then it will converge we do it the same way but we get a different answer so this will be the limit as L goes to 0 from above of so when I do that integral I get so this is x to the negative 1 half so when I integrate it I get x to the 1 half so I get 2x to the 1 half which I'm evaluating from L to 5 so now this is I'm still right with the limit you don't have to write all of these steps in between just sum them up at least one of them so this is 2 square root of 5 minus 2 square root of L and as L goes to 0 this goes to 0 so this guy converges to 2 square root of 5 this is saying this picture this area is infinite but for this picture which looks almost the same it hugs this much tighter and this area is 2 square root of 5 it's also possible that we just do one more again I'm choosing easier ones on purpose but really they're all the same it's also possible that maybe this was the integral from negative 1 to 5 of dx over x squared and if you had this one if you just did the integral and plugged in you'd get a number but this diverges because this is the same as the integral from negative 1 to 0 and this diverges because we just did it so does this but if you just were not very observant and you just did this integral and plugged in 5 and plugged in minus 1 you would get a number you would just get a nonsense answer that I forgot somebody over here asked does that answer your question it's also possible that things go to infinity it's the same you just take the limit to infinity yeah how do we tell odd functions or even functions this is a math 123 question wrong class sorry go away so f of x is even so this is the definition if f of let me use a instead of x is the same as f of minus a so if you just plug in a negative thing and the negatives all go away then it's even it's odd if f of a is minus f of minus a so if you plug in the negative sign and you can factor it out and it just changes the sign then it's odd so the reason I'm sure you're asking is because if I have a function let's say even or odd let's say I have an odd function like oh I don't know sign x x squared plus 5x to the 22nd this isn't defined it's 0 so I plus because when I plug in negative x this stays the same this stays the same this stays the same this changes sign so this is odd so an integral like negative 5 to 5 of sign x over x squared plus 5x to the 22 plus 1 this is 0 automatically because the oddness means the negative part cancels out the positive part so if you have to do any work you can just look and say oh it's odd, yay, happy times and if it's even we can just do it by splitting it in 0 and doubling the answer is that why you do math? so I think there's on one or two of the practice tests there's some integral which is just completely quarreled but it happens to be an even function no an odd function so the answer is 0 0.26 other questions? yeah so the comparison test is related to this kind of thing related to these improper integrals suppose ok so I have let's take this one because it's on the board so we know this thing diverges if I have something that's bigger than this so if I have something like the integral from 0 to 5 of dx over x squared plus I want it bigger minus 4 in fact let's so if I look at this one I can say well you know what this looks a whole lot like that and I know that diverges what does a whole lot mean well for any x between 0 and 5 this number oh shoot ok now we're good no we're still screwed sorry so I don't want to blow it on these ok so this one doesn't quite work because it's got a problem in 2 ok so this is a bad example and I'm having trouble making it work ok I'm going to change this from a 5 to 1 still true ok this is now a 1 this is now a 1 this is now a 1 still diverges, still true and now we're good and maybe there's a 10 here I don't care doesn't matter so this thing is bigger than that thing this is bigger than isn't it except in absolute value so for all the guys in there that's bigger but this diverges remember diverges means gets really really big and so this has to diverge because this is bigger than that so the comparison test says if you have an integral where the integrand the thing under the integral sign is bigger than the integrand it's a divergent integral and it diverges the smaller thing is going to infinity the bigger thing has to go to infinity and then there's the other side which says that let me erase this jump now the other side says that similarly if we've got a thing which is smaller in absolute value so the other side so if S of X is bigger than G of X on my interval and really we can do it with absolute values and the integral diverges and so does S the little one the little one diverges the big one does too and then there's the other half if and B could be infinity is going to be infinity and if the integral of F converges so it's some number so does G so if something is going to infinity and you're sitting on top of it you go to infinity too if something stays at a given height and you're underneath it you don't get any taller so if I plant a beanstalk in this room and I let it grow it doesn't get any higher than the ceiling it's just sitting there a whole time I plant the beanstalk which could grow infinitely tall in this room no it can't, it can only get that high yeah you know some stuff so I mean that's the hard part of this you need to have a little arsenal of functions in your hand that you know and converge so you already know or you can't know if you did your homework you know that these guys you know all about them whether they're going to zero or to infinity there was a homework problem that said but you could just check these guys if we're going to zero so then they converge if the power is less than one so if there's a zero here then this converges for p less than one and these guys same guys and they're going to infinity they converge for p greater than one and they diverge otherwise so if the power is big and you're going to infinity the power on the bottom is big then it converges if the power is small and you're going to zero then it converges otherwise no so the good things to convert to compare to often are one over x to some power because you know about these and they're easy and they're ones that you should remember there are other ones that maybe you know like there's you know I could have also done this one by a trig substitution and blah blah blah so I can find this one exactly because it's an integral that I know it's an arc sign right it's ten times the arc sign ten times the arc sign of x over two twice sorry got factor the four out x squared minus one oops arc secant anyway I know this one I could do this one but it doesn't matter I don't have to have a comparison are you satisfied okay anyone else more comparison related things okay other stuff yeah even odd functions if it's not functioning it's not going to the same number like if a is the same as b then it's no use to you and you still can't do it well this can't do it so if it's an odd function the fact about odd functions is that I guess I can write here so the thing about odd functions is that they tell you whatever happens over here happens over here but underneath and so if I'm going to something balanced then this cancels that out but if I've got some more stuff over here well then I can only cancel out this much and I still got to know about that so odd functions and even functions are mostly only useful when your thing is symmetric about zero they're still even and odd but it doesn't tell you a whole lot if it's not symmetric about the point in the middle of your integration other stuff everybody's ready so let's see what didn't we talk about let me just remind you what we didn't talk about we didn't talk about substitution all good with that we didn't talk about integration by part oh you have a question I did that first so you want me to go over it with an example do you want me to go over Simpson's rule just remind you how it works and then do an example or do you have a specific example in mind make one up doesn't matter you have one or no so let me make one up let's do one more we know the answer so let's do the integral from 0 to 1 of 1 over 1 plus x squared this one we know the answer the answer is pi over 4 because this integral is one you know this is the arc tan 1 minus the arc tan 0 the arc tan of 1 is pi over 4 so this is pi over 4 in fact let's make it even easier we know it's pi so when we do it we should get an answer near 3 how many what n do you want it has to be even remind you of the picture of Simpson's rule so 1 over 1 plus x squared looks like this we're going from 0 we'll put 1 here to 1 and what we do with Simpson's rule we pick a number of intervals to chop it into 1, 2, 4, 3, whatever so let me just draw the picture with 2 so if I'm doing it with 2 I break it into two pieces and on these two pieces I evaluate the two ends and the middle on each piece so if I'm doing n equals 2 I'm going to use sorry this is n equals 4 and I have 0, 1, 2, 3, 4 points so I have a total of 5 points for Simpson's rule n always has to be even because we use the two sides in the middle and if n is not even we have two sides in the middle if it's odd there's something funny here so what n would you like 2, 4, 6 give me an n 2 we want the easiest possible 1 so n equals 2 so n equals 2 is a little bit atypical because we're just using one interval so n equals 2 we have 3 points we need to find so for n equals 2 we're going from 0 to 1 we want to find 3 points equally spaced 0 to 1 so there's 0, 1, 1, 1 there we go that's it so since n is 2 Simpson's rule tells us that we take 1 third average in constant so Simpson's rule always has a 1 third and then the width of this thing which is a half and then we evaluate our function at each of the points I'm going to wrap around here but we do kind of a funny averaging that the middle counts more if we had been doing instead n equals 4 then this one would be a 2 and then we would have a 4 so it goes 1, 4, 2, 4, 2 4, 2, 4, 2, 4, 2, 4 1 so since we're only doing 2 there's nothing with the coefficient of 2 so in this case we get a 6 here f of x, 0 that means we plug 0 into our function that's 4 over 1 plus 0 and then we have 4 times f of x, 1 x, 1 is a half 4 over 1 plus 1 half squared is a quarter and then 1 times 4 over 1 plus 1 squared so this is arithmetic that I can actually do this is about the limit of my arithmetic abilities but okay so this is 1, 6 4 plus 16 1 plus a quarter is 5 fourths I flip it I get 4 fifths and 4 over 1 is that's 2 16 times 4 fifths is 64 over 5 so that's 1, 6 times 6 plus 64 over 5 so that's 1 plus 64 over 30 which is around 3 and change so that's about right I mean it's some number right it's 94 30 which is 3 and 4 30 which is around 5 so I know it's about right had I done n equals 4 I would have had 3 more terms contributing to the mix and I would have gotten better at it the pattern is we count the middle 4 times the edges but the interior edges since they're the edge from the right and the edge from the left count twice so it's 1 4 2 4 2 4 2 4 1 I mean this really has no sense you're talking about paper homework paper homework into problem 2 it has sense you just have to use integration by parts so if you try and do you're talking about paper homework problem where you have to do Simpsons rule I mean I told you exactly what to do but you have to read what I wrote which makes it hard 0 1 log cosine something like that so you try and plug in you get infinity it's no good you can't use Simpsons rule instead you integrate by parts gives you something where you can as long as you take limits so really remember this is an improper integral so this is the limit as L goes to 0 from above but even this limit is no good until you integrate by parts to make it better when you integrate by parts then the limit makes sense and everything's good which is what I tried to say in the words when I wrote the problem but now you can do the problem okay so let me remind you of the topics the exam covers and then you can say ooh ooh ooh one of those so substitution of course integration by parts then various techniques of integration which include powers of trig functions so you know sine to the n cosine to the n trig substitutions integration by partial fractions improper integrals which we just talked about an area between curves it's short it's an easy exam you know all of those things so I don't need to do anymore I don't know where I am anymore I started over there so this is really just observations about appropriate functions of sine to the cosine so for example and it doesn't matter if the powers are negative it's still the same sine cubed x over cosine squared x dx so if we use the fact sine squared plus cosine squared is one that means we can turn even numbers of cosines into sines and even powers of sines into cosines using this fact we can turn this into something where we have an extra sine laying around and everything else is in terms of cosines and then this gives us an easy substitution to make if this power was one we're done u equals cosine this is one over u du negative one over u squared du negative easy but it's not a one it's a three so we'll use this fact let me just do this example and then I'll write the rule we can use this fact to turn all of one of these sines into cosines so this is peel away two of those sines and now I'm going to turn this guy into cosines it's an easy substitution du is a cosine so du is minus the sine which is sitting right there oops how'd my dx get to the bottom there doesn't make any sense sorry my du is sitting right there so this becomes what does it become one minus u squared over u squared and this is a negative du and that integral is easy so now it's negative of negative of one over u u is the cosine I make a mistake one of these is minus plus minus one is good but I think that's right and in general if one of these numbers is odd so the rule with these guys if m or n is odd change all but okay this doesn't even make any sense okay I'll try it over here if m is odd I'm going to use that to change m minus one to cosines if n is odd then I change cosines to sines again using sine squared plus cosine squared is one and if they're both even this doesn't work if they're both odd pick your favorite you like sines, change the cosines so if they're both even then you use a different identity to cut the power in half you use the fact one half minus cosine and then cosine squared is one half still cosine is a plus I know it's right so this cuts the power in half and then now maybe one of the powers is odd and you use that trick if not do it again if not do it again eventually you cut an even number in half unless it's zero you get down to an odd well you don't have a sine two theta so it's not much use so so it's not the angle that's even the powers are even so if I have an integral let's do the simplest one like that then the simplest case here is I would replace it with this and now it becomes easy so this becomes using this fact one half the integral of one minus cosine two theta d theta and that's easy this is just one half theta minus sine two theta but I'm letting u equal two theta so du is two d theta so I pick up another half so that became easy because I changed the power from a two if I had sine to the fourth would have to do this trick twice the sine to the fourth would become stuffy in terms of this thing squared square it out do it again so these quickly get annoying like if you have sine to the 16th it takes a long time to get down to something normal but you can keep cutting the powers in half until one becomes odd so like if you had sine to the sixth you could do this trick and now I have cosine q's and now I can use the other trick people are right with this so it's important to remember that the angle keeps doubling this is a cosine two theta so if I had to do it again then it would become fourth theta and so on other questions yeah, you in the back they're all the same to me but they're obviously not all the same to you so sure do you have one in mind? okay x over x over dx oh x over e to dx sure okay so this one's okay at zero because at zero we have zero over one it's just at infinity to deal with it so this becomes just to rewrite it but we still have to do the integral of x e to the minus x in fact it's probably easier to think of this this way it's the same but it's easier to think of it this way because now what technique am I going to use to do this integral? parts I'm going to use parts and I'm going to let this be my thing so by parts I let u be x so du is dx and dv v e to the minus x dx so v is minus e to the minus x so then this integral becomes the integral of uv the x is in front evaluated from zero that's an m not an l minus the integral of v du so v is negative so that becomes a plus and this is again from zero to m right so now this integral this bit this is the limit as m goes to infinity of minus m e to the minus m minus at zero so it becomes a plus zero so there's one limit we have to do and then this integral is easy to do the integral of e to the minus x we already did it once it's minus e to the minus x so minus e to the minus m minus minus zero that was a one so far now we just have to do these two limits this limit is pretty easy as m goes to infinity 1 over e goes to zero 1 over e to the m goes to zero so this is zero so this piece goes to zero this piece also goes to zero because e grows faster than x e to the x grows faster than x so the limit of x over e to the x as x goes to infinity is zero so I get zero plus zero plus one it converges to one unless I screwed up somewhere which is not unheard of those of you in my lecture know that well so in some sense all of these are the same but I can do more of them if you want it seems to have mastered integration by parts so since nobody wants me to do integration by parts because that's the one topic we haven't done anything about we also haven't done anything on area between curves which one? not parts I meant partial fractions so the two topics that we have not touched on are partial fractions and area between curves curves? okay yeah I'm trying to get people to ask me but nobody's asked me but he asked me for area between curves do you have one in mind? okay do you want me to make one up? do the ones she wanted first okay do you have one in mind? okay so I can either look for the book and hope that the book isn't mean or I can just make one up and maybe the numbers will be big or do you want me to take it out of the book? just make one up so I suppose actually you could try and do this one by a trig substitution if you completely forgot integration by partial fractions but partial fractions should be good here so in fact we're going to get two terms which involve stuff on top that isn't just a constant because I have an x squared and I have an x squared plus one I can't reduce this and I can't reduce that okay so what I use is the fact I want it to look like something over x squared so the thing that I want is when I split them up I always want the power of the top to be one less than the power of the bottom plus something else over x squared plus one so now that's what I need to solve so if I cross multiply I get just on top I get three x equals so a x plus oh that was yeah oh well so this one was messier than I can anticipate it because I wasn't thinking sd so that's what I have to solve so I guess I want to need another letter but let's not you can't do a over x plus b over x you can do a over x plus b over x squared and that's the same as this except you get different numbers for a and b with the same answer so I prefer this because see there's nothing I can do with this okay I can do x minus i x plus i so I can't factor this anymore so the rule of thumb that I prefer the book says okay write this as a over x squared plus b over x plus cx plus b it's the same I'm just going to get a different number here so I prefer to do it this way okay so let's just go ahead and do this so now I can either multiply everything out and then equate coefficients where I can pick various values of x either one I like to use a mix but if you prefer I can go with just one which way do you prefer or just let me do whatever I do do whatever I'm going to do okay so this has to be true for every value of x that means that if I multiply everything out so for example I'm going to get an ax cube term and a cx cube term that means a plus c has to be 0 if I multiply everything out but I can also just pick numbers because it has to be true for every value of x so for example if x is 0 I get 0 3 times x is 0 equals a times 0 plus b times 1 because x squared plus 1 is 1 plus 0 because x is 0 just plugging in for x equals 0 kills everything and tells me immediately b is 0 I'm done with b so b is 0 if x is well it's not obvious what else to pick here but if x is 1 if x is 1 I get 3 equals a plus b times 2 plus well b is 0 so that's gone plus x is 1 so I get c plus d so I know that a plus 2a 2a plus c plus b is 3 if x is minus 1 I get negative 3 equals negative a b is 0 so that's done times 2 plus d minus c times 1 and so now I have another equation here let me write those where should I write this? here so I have already known b is 0 and I know that 2a plus c plus d equals 3 and negative 3 equals negative 2a plus negative c plus d and if I add those two together I get 0 equals 0 plus 0 plus 2d so d is 0 so I don't need d either so d is 0 so now I can go back again and see that since d is 0 I've got 3 equals 2a plus c and negative 3 equals 2a minus c so if I add those together I get 4a is everything 0? what the heck? alright I made a mistake somewhere negative this is a negative yes so this is no information to me this is a negative so this is no good for me so I need to get another equation so what other equation can I use here? well I notice here if I multiply this out that ax cubed plus cx cubed is 0 so I know a plus c is 0 so a is negative c so now I'm down to 1 let's put I'm sorry so I'm just going to pick another number how about 2 and I already know that a and c are opposite of each other so I have picking x equals 2 this tells me that 6 a times 5 and b is 0 so that's gone c is negative a so that's negative a times 2 times 4 if I did this right is that correct? let me write it a little longer so picking x equals 2 so 6 3 times 2 equals a times 2 that's gone it wasn't 5 it was 10 5 and then c here is negative a so this is negative 2 times 4 since d is 0 so that means 10 a minus 8 a is 6 so a is 3 and c is negative so after all of that garbage we know that this equals 3 over x squared minus 3 over x squared plus 1 so my answer is I didn't leave enough room I'll write it here so this 3 over x squared integrates to negative 3 over x and this is 3 over 10 and as I said you could have done a trig substitution and you should have gotten the same answer but that was because I was an expert now this one let me just let you know of the level of difficulty here this one if I put it on the test I would expect maybe 10% of the students to be able to do it I would expect probably 40%, 50% of the students to be able to start it but only about 10% to be able to get all the way to the end on a test what you can do on a test is much less than you can do on a homework because on a homework you can make mistakes and then you can relax and say oh jeez I did a stupid thing and then you fix it and so on on a test is a very different situation and a lot can make a lot more stupid mistakes yeah probably I mean maybe I did it wrong what should be 3x nope oh a yes a you're right see I would be one of those students that didn't get full credit so on this if this were out of 10 I would probably get 7 or 8 so I know how hard the test is and so I set the test to figure it out how does the curve on the test work it's not a curve based on how people do so it's possible for everybody to get an F for everybody to get an A I've never seen it happen but in general about 30% get a C 30% get a B 10% get an A maybe it's 15% A's maybe it's 5% A's usually it's around 70% maybe 80% that do better than awful yeah how much points how many points would you lose if you left it it kind of depends on how crucial that is to the problem so like here for this problem where I already screwed it up if it's out of 10 then probably 2 or 3 of the points are being able to actually do 10 10 maybe if the points are being able to actually do the integrals now this integral was well it's not that much harder so this is a log and that's you make the substitution use so this one has no trig functions so here the integrals on this problem are probably like 20 to 30% of the points on the problem because the hard part on this problem is to be able to notice the partial fractions set up the partial fractions correctly solve the partial fractions get the integrals and now do it but if the problem is what is this integral and you say it's arc sine you don't get a whole lot of points for that so it kind of depends on what the problem is tested if the whole point of the problem is to recognize the inverse tangent as an integral then you get zero points for thinking that it's a log if the whole point of the problem is to do that's just a little piece of the problem then it doesn't count so much how much time is 90 minutes an hour and a half given that there's nine problems but a couple of the problems are really easy and so another thing about the test all of the problems are worth the same in fact they're all worth 20 points so every problem is worth 20 points so it's out of 180 some of them are really easy some of them you should just look at them and almost write down the answer maybe write one thing and then write down the answer so that's 20 points for free some of them are quite involved and hard so part of your job is to recognize what will take you a lot of time and what will take you not a lot of time and do the problems that will take you not a lot of time first so they're out of the way and done they aren't necessarily the first problem or on the first page so what you should do when you get the test is open it up and say oh that's easy, do that that one looks hard, I'll skip it for now that one's easy so go through the test do what you can do quickly then go back through the test and do what takes you longer so what you don't want to do is find some hard problem start with that spend the whole 90 minutes doing that problem and only getting halfway and then only getting 5 points out of 180 because you only got halfway through a problem that was hard and you didn't even get the free points and they're not free I don't even remember who asked that question nobody asked that question I just said it yeah well if you encountered the square root of negative 1 you did something a little bit weird or you did something unexpected that works out to give you a square root of negative 1 and if you carry on it will go away again so for example I can factor this into x minus i x plus i if I really want to and then I only have and then I only have linear factors over here but things will work out a little weirder and I have to realize what to do with that so none of the problems need to be done involving complex numbers yeah right you have one that you think does but there's no negative 1 here I don't understand why you need to take this so this one is so let me write the problem that has so this is on one of the practice exams oh by the way I would put the solutions to the practice exams up tomorrow so all of the solutions will appear on the webpage tomorrow the reason they're not there now is to encourage you to try and do them before you see the answer so again remind me of the function there has to be a t here somewhere h of t equals sine square i c plus e oh I see what you're saying arc tan and you're supposed to find that's the question you're asking about well this is use the fundamental theorem of calculus right if you take the derivative and then you're going to integrate it oh yeah I guess it will have an i it will be e to the i because so this is h of 1 minus h of minus 1 yeah so it will have an i now it doesn't have an i okay so now I understand your question so that was just I didn't know which problem that was but we have it just being silly but yeah so it's just this by the fundamental theorem of calculus alright so now I finally understand your question sorry it took me so long to realize I can guarantee you that there should be no i's on the test that I wrote I mean your i's should be on the test but not on the test other questions other things people want me to go over yeah sure I'm listening 1 to infinity wait wait wait 1 to infinity on what square what's in I'm sorry sign okay it's just harder to hear you now sign square x over x cubed done it so and the question is not what is it it is does it convert or not so the question is does this yeah we compare it to 1 over x square oh this is a cube we compare it to 1 over x cubed and if that converges then but it's a diversion right well no it only goes one way so we use the fact that sign squared of x is always less than 1 because you're taking a number less than 1 and you're squaring it you get something between 1 and 0 and so that means that if we divide it by an x cube it's between 1 over x cubed and 0 and so if this converges then happy times if this doesn't converge we know nothing so now the question is what about anyone know wrote it on the board a little while ago nobody knows so let's check so this is so if you forget we can just do it because if we integrate this guy we get negative 2 over x squared from 1 to infinity and if we take this is the limit as m goes to infinity of negative 2 over m squared minus of minus 2 over 1 minus of 0 so this is 2 and so the answer is yes it converges great did I do something wrong could be a half I don't know anyway this converges so yes so this converges this converges by comparison with the other guy if this had diverged we would get no information so if we change this problem we change this problem to sine squared over square root x we have to work a whole lot harder because we can't compare it to 1 over x squared we have to compare it to something else what if then you couldn't use this technique that's what I just said so if we change this problem let's write a similar problem but not the same what if I want something that diverges I want this to be bigger so say we have that problem instead so now this is always bigger than 1 over square root x so square root x diverges and since this is bigger this diverges too so you have to compare it to something else here if I change this power to a square root I can't directly compare it to 1 over square root x I have to work harder so instead I change the problem other questions you're all experts you're all going to get a's wouldn't mind giving you all a's yeah so these problems are pretty easy let me just if we have two functions f and g then the area between them is just the integral of the top minus the bottom so here g is on the top and the only hard thing is to figure out where they cross but that's an algebra problem so you have to figure out what a and b are often to know where they cross so I can do a really stupid example where I can just do this or maybe the other thing to point out is sometimes it's easier to integrate these things sideways so if I have of course I don't have an example in mind say I have some curve that looks like that and I have another curve that looks like this and I want to find this area to make it look a little more obvious then I might want to integrate this way instead I might want to integrate dy so it's the same I just have to think take this minus this say this is y equals f of x and this has to be x equals g of y because it's not a function and so I would prefer probably to integrate this one this way because to integrate this way so I would do the integral from I have to give these names let's call this a b this one's at the top so this would be f so I know this is not a specific example and I can do a specific example but that's the idea it's easier to do this one dy if you prefer to do it dx so let me just draw the picture again this is the part I'm caring about if we prefer to do it dx we have to split this up and go from here cd and then here the top curve is one branch of f inverse and the other curve is the other branch of f inverse so I have to do two different integrals one over here and one over here okay so now let me rather than just making one up let me steal one from the book so that the numbers don't come out two horrendous like 5 squared 17 over 18 4 plus y 4x plus y squared equals 12 x equals y so if we're doing this we have to figure out what the region is we just have to figure out where they cross and figure out what to do here so what does the picture of this look like well this is the same thing as saying y squared equals 12 minus 4x so this is a parabola opening this way this way right when x is 0 y is 12 goes through there 12 and then so I get something like this no I did this backwards I'm sorry so this is the same so when x is 0 still 12 when y is 0 then x is 3 and as y why can't I do this is it open which way 12 minus y squared should open down still plus this right when x is 3 y is 0 yes and y equals s that's right now finally when they cross so I have to figure out when they cross so that's just when y equals x equals 3 minus y squared over 4 so that is the y values are y equals 3 minus y squared over 4 4y is 12 minus y squared so y squared y squared minus 4y minus 12 is 0 so that factors y minus 6 and y plus 2 right so this crosses at negative 2 at 6 so my picture is a little bit off but ok and now the area is pretty easy now we're going to integrate dy we integrate minus 2 at 6 of whatever the function way the top curve which is the parabola 3 minus y squared over 4 minus the bottom curve sorry yeah that's a y so this is the top we're actually in the right and so now we just get 3y minus y cubed over 12 minus y squared over 2 from minus 2 to 6 which is 18 minus whatever 6 cubed is 2 minus what we get when we plug in negative 2 so minus 6 minus plus 8 minus 12 minus 4 2 so this is 36 over 2 72 over 2 8 over 12 is 4 no 2 over 3 is a minus is some number this is minus 18 plus 6 is minus 12 minus 10 so I get that but maybe I screwed up so I mean these are all this kind of thing yeah so one of the homework problems was broken and there was a multiple choice answer with one choice well three points for you so the problem is now labeled if you go back and do it the problem is now labeled as defective maybe because some people have done it and didn't want to take away the points still a good problem it's just the last part of stupid math because it says if the answer is 5 what's the answer other questions issues if they fix the early point issue then I'll go back to it the problem is that the mechanism where early points were detected worked for some people and gave some people early points all the time and few people got jealous when their neighbors got early points for doing it two minutes before it was fixed so okay so there's another review session on Tuesday if you think of other things to ask between now and Tuesday then plan the other review session on Tuesday is shorter it's only an hour and 20 minutes because they have to get out of the room yeah that one is it's in the time slot before your exam so that slot is what 650 to 810 so the review is 650 to 810 it's in old engineering 143 there's a physics exam afterwards you seem to have run out of questions this afternoon so maybe you're all prepared or maybe you haven't even thought about it yet I don't know there's also I mean I don't know you probably received an email about a discussion board there have been two questions asked I answered them both but there's something called Piazza you might have deleted it because it looked like spam but I set up a discussion board so that you can post questions there and your other students can answer them or maybe see you I'll see a few of you tomorrow and the other one maybe normally I'll see you later see you some day