|Start||remainder there is a review session
for the exam
there's one on sunday
I will film the on sunday
so you can watch it on the web if you can't make it
however, it's probably in your best interest to go
because really the best way for a review session to work is
for you to ask questions
and say I really can't understand Simpson's rule
or I cannot do this integral, how do we do it?
rather than "tell me everything that I never learned because I don't know anything"
|0:36||uh, so, watching it is probably the least useful way to learn it
but, I understand people have obligations
on the webpage for the class, I put up old exams for mat 132 I think one actual exam two practice exams and an exam for mat126, which covers some of the same material
|1:04||the drawback of all of this is the 132 exams were behind so, they all covered volumes, which wont be in the exam although we covered just before the exam so they all covered volumes of revolutions we will not have volumes of revolution on the exam I think one of them covered arc length again, we won't get to arc length before the midterm so, you know, that means that|
|1:33||well, and the mat126 exam doesn't have any stuff on areas which we'll do today nor does it have anything on improper integrals, so none of the four or five, or how many I have put up there samples are perfect but it gives you an idea of what the exams are like and if you look at all of them, you will notice that they differ quite a lot|
anything else I need to cover?
the appropriate ones, how is that?
so, look on the webpage, by your recitation most of the people in this lecture are in ESS 001 but, one recitation, Ye Sle Cha's recitation is, she is not sitting there today
|2:34||seven, it's number seven Is I think in hall engineering 143, I think so recitation seven is in this room and six, eight no, there is no six|
|3:04||eight, nine, and ten are in ESS 001
and all the other ones are, if you are on the other lecture, well, look on the webpage
ok, so I wanna apologize, at the end of the last class
|3:30||I put a problem up, which
the answer was right, but you can't do that problem
so I meant to put up a different problem
so let me put up a different problem
so, here is what I put up
the question was, does this converge or diverge?
and really I meant to put this problem
I meant to put up this problem
you should be able to do both of them
you want me to ask them as clicker questions?
no? nobody likes the clicker questions?
|4:35||ok, so, I will just ask, this converge or diverge?
ok, so that's wrong see, it sounds good, but it's not right
|5:01||but, ok, it is a good guess, you have a fifty-fifty shot ok, this one actually converges and the reason it converges it's because this is like 1 over x^2 so, this is so, since sine of x over x^2|
|5:30||is certainly less than 1 over x^2 and it is greater than minus 1 over x^2 and the integral from 1 to infinity of 1 over x^2 converges so, sine is between 1 and minus 1 all the time so, this is never bigger than 1 on the top|
|6:02||and never less than minus 1 on the top
it may be zero, sometimes it's a half, sometimes .006
but it's never bigger than 1
ok, so a lot of times in calculus when you see a sine or a cosine you should think, oh maybe I can replace that with 1 you can't always, but sometimes you can so you should think about it so, here, this is true what I, ..., well, ok less than or equal to
|6:33||because, for certain values of x, the sine is actually 1 but the sine is always between plus 1 and minus 1 and x^2 is always x^2 so this, this integral, if it exists at all it's always less than this integral well, in absolute value because sometimes is negative|
|7:01||but, this integral we know converges I did that last time, I believe yes, I did, because when you integrate x^(-2) you get negative x^(-1) if you take the limit of x^(-1) as x goes to infinity, you get 0|
|7:31||so, this guy, the guy on top so here is 1/x^2 here is -1/x^2 and sine x over x^2 does something like this this area here, we know it's finite|
|8:00||it is, um
uh, you know, I can do this out more carefully if you'd like
so, your argument was wrong, but it sounded good
ok, what about this guy?
I won't laugh at you I didn't laugh at him. I like this argument. It is just wrong
|8:32||nobody is gonna tell me?
I mean, well, the other converges, I mean, geez, why would I put two converges ok, yes, why does it diverge?
1/x diverges and 1 plus since squared of x is always bigger than 1 so, by the same reasoning well, similar reasoning
|9:00||1 plus sine squared of x well, you take a number, you square it it is always positive, so this is certainly bigger than 1 for all x's and x is always the same as x so we have that and this integral, which also we did this diverges|
|9:31||because of what he said this is a log, log of infinity is infinity and we have something bigger than since this is bigger and that gets as big as you like this is as big as you like I know this comparison theorem is a little tricky for people|
|10:02||it's a good skill to learn, to be able to look at things, and compare to other things that you know of course, you don't necessarily know what you know yet but when you do it enough, you'll have a better feel but, in general, 1 over x to a power, there was a homework problem that asks about that and we're going to infinity and if the power is bigger than 1, it converges and if the power is 0 or less than 1, then it diverges|
|10:37||this guy? well, if it isn't obvious from the picture
I mean, I can't really draw the picture of how it diverges
because it looks like it should converge.
In fact, if you sit down with a calculator or a computer and you start
|11:00||numerically integrating 1/x
you are gonna get a number because it diverges so slowly.
That is you won't see it, but it does diverge because it is like the log.
The log goes to infinity as x does, and the log grows very slowly.
The log is an exponent, so it grows, but it grows slowly.
I mean, the log of a thousand is not that big of a number.
If you are gonna make base 10 the log, then log of a thousand is 3,
|11:31||that is not a very big number.
and from a thousand to a million if we're dealing base 10, base e is similar, then, thousand to a million goes from 3 to 6.
You can't just look at the graph you can look at the graph and draw some conclusions but a conclusion with no guarantee.
so, here I can look at the graph of this and compare to the graph of this I know that 1/x is too big and I know that this guy 1 + sin^2(x) over x
|12:01||does something up here
it touches every multiple of 2pi.
but, you know, it does something up here so, this is too big, certainly that's even bigger so that really, the comparison theorem is where you're looking for pictures but, to just look at the integral, and say, for sure it diverges you know, If I give you a penny every day
|12:30||for a billion years you'll have a lot of money, well, maybe in a trillion years but, as I always give you a penny, every day your amount of money is unbounded but it takes so long that you can't really touch depends on whether it's going to zero or infinity so, there is a homework problem like this, which asks|
|13:01||I forget whether the homework is to 0 or to infinity
anyone knows? anyone has done it? it's like the last one on the homework assignment.
You should've done it because it was, well it's something like find the values of p for which this converges or maybe is 1 to infinity is it infinity or 0?
|13:30||it doesn't matter
ok, this guy, for what values of p will this converge?
ok so, I mean, part of the point of homework it is that you're supposed to do and sort of remember how it works seems nobody remembers how it works so probably you didn't do it, one or the other yes, p less than 1
because when you do the integral, if p is less than 1 then the x goes to the top so that means that this at zero it's just gonna be zero rather growing up but, all of this, 1/x^p look like that and you can't tell just by looking and this
|14:30||1 to infinity dx over, let's put a q here this guy converges for q bigger than 1 for the same reason and when q, q and p equal to 1 it diverges in both places|
|15:08||and you can just, I mean you did just do this unless you didn't do the homework and then shame on you so, let me move along to the next topic which is|
|15:41||ok, so you are now supposedly experts on improper integrals and techniques of integration we'll find out next week whether that's true and then, the next topic that we're doing is returning to the idea of finding|
|16:01||so, let's say, this, this is easy suppose I have two curves here like y=x and of course since... y equals let's put that at 1 x minus 1, the whole, oops, that's not the right one|
|16:38||why, why can't I do this?
how about 4 minus x^2 let's just do this instead so, 4 minus x^2 looks like that and what I want here
|17:00||is this area so, this is not a hard problem because you just have to view it properly this is an integral and, in order to find the area between the curves you just substract the bottom curve from the top curve so, the area between|
|17:38||it's just going to be, well, I mean, let's do it a little formally if we make a little rectangle here its height because remember when we are integrating we are dividing its area in little rectangles its height it's going to be from here to here|
|18:01||so this is 4-x^2 minus x so the height of my rectangles is 4-x^2 minus x top curve minus the bottom curve and so the area and the integral from somewhere to somewhere of 4 - x^2 - x|
|18:33||is we'll, no we have to figure out where the somewhere and somewhere are there is a well-defined area here Im not telling you where they cross but you've gotta figure it out we'll have to find this point and we'll have to find this point in order to do this problem, you have to find these two points|
|19:11||so, how do we find those two points?
(unintelligible) you'll wait for me to do it so, how do we find those two points?
|19:30||ok, right, we set them equal to each other so, we need to know so, that means that we write this as let me bring it over there x squared plus x minus four equals zero so, I need to solve that well, that's tricky it doesn't factor nice|
|20:00||but that is ok, I know the quadratic formula it is negative one plus or minus square root of one minus four ac over two so, it's two ... numbers but, they are numbers negative one plus or minus the square root of seventeen over two that's where they cross nothing wrong with the square root of seventeen|
|20:31||it's perfectly good. Do not use the calculator so, now we know where they cross this is where x is this one is negative one minus the square root of seventeen over two and this is negative one plus the square root of seventeen over two so that means that the integral we wanna do this one, negative one minus root seventeen over two|
|21:00||to negative one plus root seventeen over two of that and that's easy that is 4x minus one third x cube minus minus, um x squared over two evaluated from blah blah to blah blah|
um, do we have enough room on the board?
which is four times negative one plus root of seventeen over two minus one third of negative one plus root of seventeen cube
|22:13||minus four minus one minus root of seventeen so this numbers are ugly|
|22:31||is there a third somewhere?
but, there it is, it is enough because I miscopied it
|23:02||because it's hard to see around these
it's easier if you can just look at your paper
and then you just simplify this and you get a number
and I'm sorry I chose a big number
which I can simplify if you really insist but I don't think yourself would want me to do arithmetic and do it wrong
|23:32||so, that's a pretty easy idea when we have two curves that cross one another to find the area between them we just substract notice that you can crank this idea up a little bit sometimes let me just give an example that I know works|
|24:38||terrible (unintelligible) y^2 equals 2x + 6 so, I want to find the area between these two guys and here, um|
|25:01||this is y squared, so this is a parabola going sideways right, so if we look at the picture here y=x-1 looks like this and y^2 equals 2x +6 this is the same as saying|
|25:32||y^2 minus 6 over 2 equals x
ok, if you prefer if it makes you happy so either I can do that or if I prefer I can say y is plus or minus
|26:02||square root of 2x plus 6 same to me so, when 2x+6 is so, here this is a parabola this is a top, sorry either one, they're the same to me so this guy is 0 when x is negative 3|
|26:33||and so is this guy so this here at -3 looks like this and then this piece so that's this piece and then this piece looks like that and we are looking for area here now let's think about, so let me not do it this way|
|27:00||Im gonna start it this way
the way wherever he is
there you are
you suggested it even though you didn't know you suggested it
so Im gonna do it this way
this is a hard way
why is this way hard?
well, number one the square root suck but, number two let's look at this so, the area is the top curve minus the bottom curve
|27:31||the top curve here it's always this guy let me write it up here y equals plus square root of 2x minus ("plus") 6 and the bottom curve is not always x-1 because it changes right here so in this region I have a different bottom curve|
|28:00||than I have in this region so I can this up, let me not even find the points yet as the integral from negative 3 to some point here, let me just call it "a" the top curve is the positive square root the bottom curve is the negative square root it's minus a minus|
|28:32||so that's this area and then the other area plus the integral from "a" let me call this point "b"|
|29:02||which I didn't even find yet hb of the top part, still that square root and the bottom curve is x-1 so that's certainly so that's certainly represents the area the integrals aren't too bad, I can do those integrals they are both easy substitutions|
|29:30||I can find the points "a" and "b" without too much trouble
but I have to do two integrals
I have to hurt my brain to think about it
you want me to do this?
no? so you can do this, it is fine you'll get the right answer well, let's find "a" and "b" so, "a" is where
|30:01||is where the negative square root well "a" is where there is 2x plus, no negative square root of 2x plus 6 equals x-1 that for (a) and (b) is where x-1 equals plus square root|
|30:42||it's from being unable to read they are all pluses can't read they are pluses I just wrote the plus|
|31:00||so, I can find the points "a" and "b" that's the area this is the suck way to do this problem I mean, it is OK but, we can also do it in other way we can just lay down so, sometimes it's good to be lazy|
|31:40||of course, you lay out on the other side but we look at this problem bad picture|
|32:03||we look at this problem from the side notice that these guys always have the same curves from here to here if you look at just the y-values x-1 is always bigger than than the parabola|
|32:33||so if you slice it sideways instead of up and down this curve, the right curve so the right curve y is x-1|
|33:01||so x=y+1 on the right it's x=y+1 and the curve on the left is I wrote it down and then erase it x= y^2-6 all over 2|
|33:35||and so that means that this area it's the integral from y+1 minus one half y^2-6 dy and I already used "a" and "b", so let's use "c" and "d"|
|34:00||we integrate from "c" to "d" because of the y-values where they cross so we want to find the y-values for when these two curves cross so so, the "c" and the "d" solve y+1 equals one half of y^2-6 so, part of this|
|34:31||and, in fact, you should be thinking this way so for this part of y is useful to think about the little rectangles here my rectangles are laying down and they go this way it makes my life easier to cut it sideways that side of the curve so, that's the same as saying|
|35:00||this is y plus plus 2 is y^2 -6 which is the same as saying y^2 - 2y -8 = 0|
|35:34||so that guy factors so this is y-4 times y+2 equals 0 so that means that my that my c is -2|
|36:02||so that means that my integral it's from -2 to 4 of it's all there anyway let me leave it the other way|
|36:42||and then the integral is easy just multiply it out so, the off shot of this is you wanna think about how to slicing out the object and if it is more convenient to slice it horizontally, go ahead|
|37:01||if it's more convenient to slice it
the other way was more convenient to slice it
maybe the curve cross three times
does anybody want me to finish this integral?
I'm happy to do it if somebody wants me to so, um this is just do the algebra here, this is
|37:32||-2 + 4 - 1/2y^2 plus 3 plus 1 is 4 plus y dy which is minus 1/6 y^3 plus 4y|
y^2 evaluated at 4?
which is minus 1/6 4 cube is 64 um, 4 times 4 is 16 did I miss a minus sign somewhere?
and half of 16 is 8
|38:34||and then you substract off -2 cube is 8 so this is 8 minus 6 and then -2 times 4 is -8 -2 square is plus 4 and then you add um let me put it out this is not a class in arithmetic|
|39:01||so, if you have things like 1/2 plus 5/12 plus 3/19 on the test, you just leave it that way I don't care if you get the 1/2 + 5/12 + 3/19 and whether you can add fractions or not is not really my concern I can't I mean I can, but the little ones too so, my concern is that you understand the material well enough to get an answer|
|39:33||that does not mean that you can leave something like the sine of pi you should know that the sine of pi is 0 so, I need you to do the obvious, and I would prefer you do not leave 8/4 you should know that 8/4 is 2 but, if you do not want to add 5/12 and 47/19 I can understand why not|
|40:04||um, so this is some stuff so, I guess I have time for another one let me just think about so|
|40:32||I have to write something down, Im sorry|
|41:00||well let me just let me find it this is on the midterm and, unfortunately, I was so worry about this technology garbage that I left my notes sitting on my desk and I can't remember the problem that I worked out doesn't matter ok, so notice that sometimes maybe your curve|
|41:30||cross more than once sorry um, ok sometimes your curve crosses more than once in which case you have to split up your integrals it is always important to remember that your area is the top curve minus the bottom curve so, in this case let's just call this f and call this g|
|42:02||so this point is 0 this point is a this point is b so this area that I want it's going to be the integral of f minus g dx from 0 to a plus the integral from a to b of|
|42:30||g minus f
this is really what you did
when you were finding areas under curves
it is just that
the curve that used to be your x-axis is now pulled up
it's the same thing
maybe we'll get to this, who knows?
|43:01||let me just so let me do another example|
|43:42||ok, so say I wanna do this one
what curve is on top?
so let's call this one f and let's call this one g
one of you go for f ok, it is a good time right? you can just look at 0 at 0, this is 0, and this is -2 that's a hint so y=x^2 minus 2 looks like that
|44:31||and the absolute value looks like the V now, both of these are even functions so I can do this in two ways I can either integrate from here to here and then from here to here or I can just realize the area over here is exactly equal to the area over there so I can just do this half of it|
|45:00||either way is OK
so, when do they cross?
so I guess one thing let me point out just to remind you the absolute value of x is x if x is positive and -x if x is negative
another way to read this is if what is inside the absolute value is positive, leave it alone if it is negative, change the sign students mess this up all the time don't panic when you see absolute value just remember that they change whenever what is inside changes and so here we need to know when does x equal
that will tell me that intersection point over here
so that is
and then the square root there
|46:30||so x-2 times x+1 ok, good, that makes it easier so that happens when x is 2 or x is -1 the -1 one I don't want that's here because I only look when x is equal and also|
|47:00||If I look on the other side I will get -2
so this is the one I want
and so that means that this area
is the integral
from 0 to 2
of x minus x^2 plus 2 dx
and then I can double it
because it is the same
as the integral from -2 to 0
|47:32||probably a good place to stop before you pack up and leave let me remind you some things the paper homework because we have a holiday now the calendar may say wednesday, but today is friday now if you have a recitation on wednesday, no you don't today paper homework is due after the break there is another webassign due it only has 7 problems on this stuff|
|48:01||and then there is the review session|