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MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Now, to start out
today we're going to finish up what we did last time. Which has to do with
partial fractions. I told you how to do partial
fractions in several special cases and everybody was trying
to figure out what the general picture was. But I'd like to lay that out. I'll still only do it
for an example. But it will be somehow a bigger
example so that you can see what the general
pattern is. Partial fractions, remember,
is a method for breaking up so-called rational functions. Which are ratios
of polynomials. And it shows you that you can
always integrate them. That's really the theme here. And this is what's reassuring
is that it always works. That's really the bottom line. And that's good because there
are a lot of integrals that don't have formulas
and these do. It always works. But, maybe with lots of help. So maybe slowly. Now, there's a little bit of
bad news, and I have to be totally honest and tell you
what all the bad news is. Along with the good news. The first step, which maybe I
should be calling Step 0, I had a Step 1, 2 and 3 last
time, is long division. That's the step where you take
your polynomial divided by your other polynomial,
and you find the quotient plus some remainder. And you do that by
long division. And the quotient is easy to take
the antiderivative of up because it's just
a polynomial. And the key extra property here
is that the degree of the numerator now over here, this
remainder, is strictly less than the degree of
the denominator. So that you can do
the next step. Now, the next step which I
called Step 1 last time, that's great imagination, it's
right after Step 0, Step 1 was to factor the denominator. And I'm going to illustrate
by example what the setup is here. I don't know maybe,
we'll do this. Some polynomial here,
maybe cube this one. So here I've factored
the denominator. That's what I called
Step 1 last time. Now, here's the first
piece of bad news. In reality, if somebody gave you
a multiplied out degree, whatever polynomial here,
you would be very hard pressed to factor it. A lot of them are extremely
difficult to factor. And so that's something you
would have to give to a machine to do. And it's just basically
a hard problem. So obviously, we're only going
to give you ones that you can do by hand. So very low degree examples. And that's just the way it is. So this is really a hard step
in disguise, in real life. Anyway, we're just going
to take it as given. And we have this numerator,
which is of degree less than the denominator. So let's count up what
its degree has to be. This is 4 + 2 + 6. So this is degree 4 + 2 + 6. I added that up because this is
degree 4, this is degree 2 and (x ^2) ^3 is
the 6th power. So all together it's
this, which is 12. And so this thing is
of degree <= 11. All the way up to degree 11,
that's the possibilities for the numerator here. Now, the extra information that
I want to impart right now, is just this setup. Which I called Step
2 last time. And the setup is this. Now, it's going to take
us a while to do this. We have this factor here. We have another factor. We have another term,
with the square. We have another term
with the cube. We have another term with
the fourth power. So this is what's going to
happen whenever you have linear factors. You'll have a collection
of terms like this. So you have four constants
to take care of. Now, with a quadratic in the
denominator, you need a linear function in the numerator. So that's, if you like, B0
x + C0 divided by this quadratic term here. And what I didn't show you last
time was how you deal with higher powers of
quadratic terms. So when you have a quadratic
term, what's going to happen is you're going to take that
first factor here. Just the way you did
in this case. But then you're going to have to
do the same thing with the next power. Now notice, just as in the case
of this top row, I have just a constant here. And even though I increased the
degree of the denominator I'm not increasing
the numerator. It's staying just a constant. It's not linear up here. It's better than that. It's just a constant. And here it stayed a constant. And here I stayed a constant. Similarly here, even though I'm
increasing the degree of the denominator, I'm leaving the
numerator, the form of the numerator, alone. It's just a linear factor
and a linear factor. So that's the key
to this pattern. I don't have quite as jazzy
a song on mine. So this is so long that it runs
off the blackboard here. So let's continue
it on the next. We've got this B2 x + C2
- sorry, B3 x + C3 / (x ^2 + 4) ^3. I guess I have room
for it over here. I'm going to talk about
this in just a second. Alright, so here's
the pattern. Now, let me just do a count
of the number of unknowns we have here. The number of unknowns that we
have here is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. That 12 is no coincidence. That's the degree of
the polynomial. And it's the number of unknowns
that we have. And it's the number of degrees
of freedom in a polynomial of degree 11. If you have all these free
coefficients here, you have the coefficient x^ 0, x ^ 1,
all the way up to x^ 11. And 0 through 11 is 12 different
coefficients. And so this is a very
complicated system of equations for unknowns. This is twelve equations
for twelve unknowns. So I'll get rid of this
for a second. So we have twelve equations,
twelve unknowns. So that's the other bad news. Machines handle this very well,
but human beings have a little trouble with 12. Now, the cover-up method works
very neatly and picks out this term here. But that's it. So it reduces it
to an 11 by 11. You'll be able to evaluate
this in no time. But that's it. That's the only simplification
of your previous method. We don't have a method
for this. So I'm just showing what the
whole method looks like but really you'd have to have a
machine to implement this once it gets to be any size at all. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: It's one big
equation, but it's a polynomial equation. So there's an equation, there's
this function r ( x) = a11 x^2 11 + a10 x^ 10... and
these things are known. This is a known expression
here. And then when you cross-multiply
on the other side, what you have is, well,
it's a1 times, if you cancel this denominator with that,
you're going to get (x + 2) ^3 ( x ^2 + 2x + 3)( x^2 + 4)^3
+ the term for a2, etc. It's a monster equation. And then to separate it out into
separate equations, you take the coefficient on x^ 11th,
x ^ 10, ... all the way down to x^ 0. And all told, that means there
are a total of 12 equations. 11 through 0 is 12 equations. Yeah, another question. STUDENT: [INAUDIBLE] PROFESSOR: Should I write
down rest of this? STUDENT: [INAUDIBLE] PROFESSOR: Should you write
down all this stuff? Well, that's a good question. So you notice I didn't
write it down. Why didn't I write it down? Because it's incredibly long. In fact, you probably,
so how many pages of writing would this take? This is about a page
of writing. So just think of your machine,
how much time you want to spend on this. So the answer is that you
have to be realistic. You're a human being,
not a machine. And so there's certain things
that you can write down and other things you should
attempt to write down. So do not do this at home. So that's the first down-side
to this method. It gets more and more
complicated as time goes on. The second down-side, I want to
point out to you, is what's happening with the pieces. So the pieces still need
to be integrated. We simplified this problem, but
we didn't get rid of it. We still have the problem of
integrating the pieces. Now, some of the pieces
are very easy. This top row here, the
antiderivatives of these, you can just write down. By advanced guessing. I'm going to skip
over to the most complicated one over here. For one second here. And what is it that you'd have
to deal with for that one. You'd have to deal with, for
example, so e.g., for example, I need to deal with
the this guy. I've got to get this
antiderivative here. Now, this one you're supposed
to be able to know. So this is why I'm
mentioning this. Because this kind of ingredient
is something you already covered. And what is it? Well, you do this one by
advanced guessing, although you it as the method
of substitution. You realize that it's going to
be of the form (x^2 + 4) ^ - 2, roughly speaking. And now we're going
to fix that. Because if you differentiate it
you get 2x (-2), that's - 4 (x)(x) times this. There's an x in the
numerator here. So it's - 1/4 of that
will fix the factor. And here's the answer
for that one. So that's one you can do. The second piece is this guy. This is the other piece. Now, this was the piece
that came from B3. This is the one that
came from B3. And this is the one that's
coming from C3. This is coming from C3. We still need to get
this one out there. So C3 times that will be the
correct answer, once we've found these numbers. So how do we do this? How's this one integrated? STUDENT: Trig substitution? PROFESSOR: Trig substitution. So the trig substitution
here is x = 2 tan u. Or 2 tan theta. And when you do that, there
are a couple of simplifications. Well, I wouldn't call this
a simplification. This is just the differentiation
formula. dx = 2 sec^2 u du. And then you have to plug in,
and you're using the fact that when you plug in the tan^2,
4 tan ^2 + 4, you'll get a sec ^2. So altogether, this thing
is, 2 sec^2 u du. And then there's a (4 sec^2
u) ^3, in the denominator. So that's what happens when
you change variables here. And now look, this
keeps on going. This is not the end
of the problem. Because what does that
simplify to? That is, let's see, it's
2/64, the integral of sec^6 and sec^2. That's the same as cos^4. And now, you did a trig
substitution but you still have a trig integral. The trig integral now, there's
a method for this. The method for this is when it's
an even power, you have to use the double
angle formula. So that's this guy here. And you're still not done. You have to square
this thing out. And then you'll still
get a cos^2 2u. And it keeps on going. So this thing goes on
for a long time. But I'm not even going to finish
this, but I just want to show you. The point is, we're not showing
you how to do any complicated problem. We're just showing you all
the little ingredients. And you have to string them
together a long, long, long process to get to the
final answer of one of these questions. So it always works,
but maybe slowly. By the way, there's
even another horrible thing that happens. Which is, if you
handle this guy here, what's the technique. This is another technique that
you learned, supposedly within the last few days. Completing the square. So this, it turns out, you have
to rewrite it this way. And then the evaluation is going
to be expressed in terms of, I'm going to jump
to the end. It's going to turn out to be
expressed in terms of this. That's what will eventually
show up in the formula. And not only that, but if you
deal with ones involving x as well, you'll also need to deal
with something like ln of this denominator here. So all of these things
will be involved. So now, the last message that
I have for you is just this. This thing is very
complicated. We're certainly never going
to ask you to do it. But you should just be aware
that this level of complexity, we are absolutely stuck
with in this problem. And the reason why we're stuck
with it is that this is what the formulas look
like in the end. If the answers look like this,
the formulas have to be this complicated. If you differentiate this, you
get your polynomial, your ratio of polynomials. If you differentiate this, you
get some ratio of polynomials. These are the things that
come up when you take antiderivatives of those
rational functions. So we're just stuck
with these guys. And so don't let it get
to you too much. I mean, it's not so bad. In fact, there are computer
programs that will do this for you anytime you want. And you just have to be not
intimidated by them. They're like other functions. OK, that's it for the general
comments on partial fractions. Now we're going to change subjects to our last technique. This is one more technical
thing to get you familiar with functions. And this technique is called
integration by parts. Please, just because its name
sort of sounds like partial fractions, don't think
it's the same thing. It's totally different. It's not the same. So this one is called
integration by parts. Now, unlike the previous case,
where I couldn't actually justify to you that the linear
algebra always works. I claimed it worked, but I
wasn't able to prove it. That's a complicated theorem
which I'm not able to do in this class. Here I can explain to you
what's going on with integration by parts. It's just the fundamental
theorem of calculus, if you like, coupled with the
product formula. Sort of unwound and
read in reverse. And here's how that works. If you take the product of
two functions and you differentiate them, then we
know that the product rule says that this is u' v + uv'. And now I'm just going
to rearrange in the following way. I'm going to solve for uv'. That is, this term here. So what is this term? It's this other term, (uv)'. Minus the other piece. So I just rewrote
this equation. And now I'm going
to integrate it. So here's the formula. The integral of the left-hand
side is equal to the integral of the right-hand side. Well when I integrate a
derivative, of I get back the function itself. That's the fundamental
theorem. So this it. Sorry, I missed the dx,
which is important. I apologize. Let's put that in there. So this is the integration
by parts formula. I'm going to write it one more
time with the limits stuck in. It's also written this
way, when you have a definite integral. Just the same formula,
written twice. Alright, now I'm going
to show you how it works on a few examples. And I have to give you a flavor
for how it works. But it'll grow as we get more
and more experience. The first example that I'm
going to take is one that looks intractable on
the face of it. Which is the integral
of ln x dx. Now, it looks like there's
sort of nothing we can do with this. And we don't know what
the solution is. However, I claim that if we fit
it into this form, we can figure out what the integral
is relatively easily. By some little magic of
cancellation it happens. The idea is the following. If I consider this function to
be u, then what's going to appear on the other side in
the integrated form is the function u', which is -- so,
if you like, u = ln x. So u' = 1 / x. Now, 1 / x is a more manageable
function than ln x. What we're using is that when we
differentiate the function, it's getting nicer. It's getting more tractable
for us. In order for this to fit into
this pattern, however, I need a function v. So what in
the world am I going to put here for v? The answer is, well, dx is
almost the right answer. The answer turns out to be x. And the reason is that
that makes v' = 1. It makes v' = 1. So that means that this is
u, but it's also uv'. Which was what I had on
the left-hand side. So it's both u and uv'. So this is the setup. And now all I'm going
to do is read off what the formula says. What it says is, this is
equal to u v. So u is this and v is that. So it's x ln x, minus, so that
again, this is uv. Except in the other order, vu. And then I'm integrating, and
what do I have to integrate? u ' v. So look up there. u' v with
a minus sign here. u' = 1 / x, and v = x. So it's 1 / x, that's u'. And here is x, that's v, dx. Now, that one is easy
to integrate. Because (1/x) x = 1. And the integral of 1 dx
is x + c, if you like. So the antiderivative
of 1 is x. And so here's our answer. Our answer is that this
is x ln x - x + c. I'm going to do two
more slightly more complicated examples. And then really, the main thing
is to get yourself used to this method. And there's no one way
of doing that. Just practice makes perfect. And so we'll just do a
few more examples. And illustrate them. The second example that I'm
going to use is the integral of (ln x) ^2 dx. And this is just slightly
more recalcitrant. Namely, I'm going to
let u be (ln x)^2. And again, v = to x. So that matches up here. That is, v' = 1. So this is u v'. So this thing is u v'. And then we'll just
see what happens. Now, the game that we get is
that when I differentiate the logarithm squared, I'm going to
to get something simpler. It's not going to win us the
whole battle, but it will get us started. So here we get u'. And that's 2 ln x ( 1 / x). Applying the chain rule. And so the formula is that this
is x (ln x)^2, minus the integral of, well it's u' v,
right, that's what I have to put over here. So u' = 2 ln x ( 1
/ x), and v = x. And so now, you notice
something interesting happening here. So let me just demarcate
this a little bit. And let you see what it is
that I'm doing here. So notice, this is the
same integral. So here we have x (ln x) ^2. We've already solve that part. But now know notice that the
1 / x and the x cancel. So we're back to the
previous case. We didn't win all the way, but
actually we reduced ourselves to this integral. To the integral of ln x,
which we already know. So here, I can copy that down. That's - 2 (x ln x - x), and
then I have to throw in a constant, c. And that's the end of
the problem here. That's it. So this piece, I got
from Example 1. Now, this illustrates a
principle which is a little bit more complicated than
just the one of integration by parts. Which is a sort of a general
principle which I'll call my Example 3, which is
something which is called a reduction formula. A reduction formula is a case
where we apply some rule and we figure out one of these
integrals in terms of something else. Which is a little bit simpler. And eventually we'll get down to
the end, but it may take us n steps from the beginning. So the example is
l(n x^ n) dx. . And the claim is that if I do
what I did in Example 2, to this case, I'll get a simpler
one which will involve the n - 1st power. And that way I can get
all the way back down to the final answer. So here's what happens. We take u as ln x^ n. This is the same discussion
as before, v = x. And then u' is n l(n
x) ^ n - 1( 1 / x). And v' is 1. And so the setup is similar. We have here x ( ln x)^
n minus the integral. And there's n times, it turns
out to be (ln x)^ n - 1. And then there's a 1 / x
and an x, which cancel. So I'm going to explain this
also abstractly a little bit just to show you what's
happening here. If you use the notation Fn (x)
is the integral of (ln x)^n dx, and we're going to forget
the constant here. Then the relationship that we
have here is that Fn (x) = n ln, I'm sorry, x (ln x)^ n. That's the first
term over here. Minus n times the
preceding one. This one here. And the idea is that eventually
we can get down. If we start with the nth one,
we have a formula that includes, so the reduction is
to the n - 1st. Then we can reduce to the n -
2nd, and so on. Until we reduce to the
1, the first 1. And then in fact we can even
go down to the 0th one. So this is the idea of
a reduction formula. And let me illustrate it exactly
in the context of Examples 1 and 2. So the first step would be to
evaluate the first one. Which is, if you like,
(ln x)^ 0 dx. That's very easy, that's x. And then F1 ( x) =
x ln x - F0 (x). Now, that's applying
this rule. So let me just put
it in a box here. This is the method
of induction. Here's the rule. And I'm applying it for n = 1. I plugged in n = 1 here. So here, I have x ln x
^ 1 - 1 ( F0 ( x). And that's what I put right
here, on the right-hand side. And that's going to generate
for me the formula that I want, which is x ln x - x. That's the answer to this
problem over here. This was Example 1. Notice I dropped the constants
because I can add them in at the end. So I'll put in parentheses
here, (+ c). That's what would happen at
the end of the problem. The next step, so that was
Example 1, and now Example 2 works more or less
the same way. I'm just summarizing
what I did on that blackboard right up here. The same thing, but in much
more compact notation. If I take F2 ( x), that's going
to be equal to x (ln x)^2 - 2 F1 ( x). Again, this is box for n = 2. And if I plug it in, what I'm
getting here is x (ln x) ^2 - twice this stuff here. Which is right here.
x ln x - x. If you like, + c. So I'll leave the c off. So this is how reduction
formulas work in general. I'm going to give you
one more example of a reduction formula. So I guess we have to
call this Example 4. Let's be fancy, let's
make it the sine. No no, no, let's be
fancier still. Let's make it e^ x. So this would also work for
cosine x and sine x. The same sort of thing. And I should mention that on
your homework, you have to do it for cosine of x. I decided to change my mind
on the spur of the moment. I'm not going to do it for
cosine because you have to work it out on your homework
for cosine. In a later homework you'll
even do this case. So it's fine. You need the practice. OK, so how am I going
to do it this time. This is again, a reduction
formula. And the trick here is to pick
u to be this function here. And the reason is
the following. So it's very important to pick
which function is the u and which function is the v. That's
the only decision you have to make if you're going to
apply integration by parts. When I pick this function as
the u, the advantage that I have is that u' is simpler. How is it simpler? It's simpler because it's
one degree down. So that's making progress
for us. On the other hand, this function
here is going to be what I'll use for v. And if I
differentiated that, if I did it the other way around and I
differentiated that, I would just get the same level
of complexity. Differentiating e^x just
gives you back e ^ x. So that's boring. It doesn't make any progress
in this process. And so I'm going to instead
let v = e ^ x and, sorry this is v '. Make it v ' = e ^ x. And then v = e ^x. At least it isn't any
worse when I went backwards like that. So now, I have u and v ', and
now I get (x ^ n)( e ^ x). This again is u, and this is v.
So it happens that v = t v ' so it's a little
confusing here. But this is the one
we're calling v '. And here's v. And now minus the
integral and I have here nx ^ n - 1. And I have here e ^ x. So this is u ' and
this is v dx. So this recurrence is
a new recurrence. And let me summarize it here. It's saying that Gn ( x) should
be the integral of (x ^ n)( e ^ x) dx. Again, I'm dropping the c. And then the reduction formula
is that Gn (x) = this expression here, (x ^ n)(
e ^ x) - n Gn - 1 (x). So here's our reduction
formula. And to illustrate this, if I
take G0 (x), if you think about it for a second that's
just, there's nothing here. The antiderivative of e ^ x,
that's going to be e ^ x, that getting started at the
real basement here. Again, as always, 0 is
my favorite number. Not 1. I always start with the easiest
one, if possible. And now G1, applying this
formula, is going to be equal to x e ^ x - G0 ( x). Which is just right, because
n is 1 and n - 1 is 0. And so that's just
x e ^ x - e^ x. So this is a very, very
fancy way of saying the following fact. I'll put it over on
this other board. Which is that the integral of
x e^ x dx = x e^ x - x + c. Yeah, question. STUDENT: [INAUDIBLE] PROFESSOR: The question
is, why is this true. Why is this statement true. Why is G 0 = e^ x. I did that in my head. What I did was, I first wrote
down the formula for G0. Which was G0 is equal to the
integral of e^ x dx. Because there's an x to
the 0 power in there, which is just 1. And then I know the
antiderivative of e ^ x. It's e ^x. STUDENT: [INAUDIBLE] PROFESSOR: How do you know when
this method will work? The answer is only
by experience. You must get practice
doing this. If you look in your textbook,
you'll see hints as to what to do. The other hint that I want to
say is that if you find that you have one factor in your
expression which when you differentiate it,
it gets easier. And when you antidifferentiate
the other half, it doesn't get any worse, then that's
when this method has a chance of helping. And there is there's
no general thing. The thing is, though, if you
it with x^ n (e^ x), x ^ n cosine x, especially on
sine x, those are examples where it works. This power of the ln. I'll give you er one
more example here. So this was G1 ( x), right. I'll give you one more
example in a second. Yeah. STUDENT: [INAUDIBLE] PROFESSOR: Thank you. There's a mistake here. That's bad. I was thinking in the
back of my head of the following formula. Which is another one which
we've just done. So these are the types of
formulas that you can get out of integration by parts. There's also another way of
getting these, which I'm not going to say anything about. Which is called advance
guessing. You guess in advance what the
form is, you differentiate it and you check. That does work too, with
many of these cases. I want to give you
an illustration. Just because you these formulas
are somewhat dry. So I want to give you just
at least one application. We're almost done with the
idea of these formulas. And we're going to get back now
to being able to handle lots more integrals than
we could before. And what's satisfying is that
now we can get numbers out instead of being stuck and
hamstrung with only a few techniques. Now we have all of the
techniques of integration that anybody has. And so we can do pretty much
anything we want that's possible to do. So here's, if you like, an
application that illustrates how integration by parts
can be helpful. And we're going to find the
volume of an exponential wine glass here. Again, don't try this
at home, but. So let's see. It's going to be this
beautiful guy here. I think. OK, so what's it going to be. This graph is going
to be y = e^ x. Then we're going to rotate
it around the y axis. And this level here is
the height y = 1. And the top, let's
say, is y = e. So that the horizontal here,
coming down, is x = 1. Now, there are two ways to
set up this problem. And so there are two methods. And this is also a good review
because, of course, we did this in the last unit. The two methods are horizontal
and vertical slices. Those are the two ways
we can do this. Now, if we do it with, so
let's start out with the horizontal ones. That's this shape here. And we're going like that. And the horizontal slices mean
that this little bit here is a thickness dy. And then we're going to
wrap that around. So this is going to
become a disk. This is the method of disks. And what's this distance here? Well, this place is x. And so the disk has
area pi x ^2. And we're going to add up the
thickness of the disks and we're going to integrate
from 1 to e. So here's our volume. And now we have one last little
item of business before we can evaluate this integral. And that is that we need to know
the relationship here on the curve, that y = e ^ x. So that means x = ln y. And in order to evaluate this
integral, we have to evaluate x correctly as a
function of y. So that's the integral
from 1 to e of (ln y)^2, times pi, dy. So now you see that this
is an integral that we did calculate already. And in fact, it's sitting
right here. Except with the variable x
instead of the variable y. So the answer, which
we already had, is this F2 ( y) here. So maybe I'll write
it that way. So this is F2 (y)
between 1 and e. And now let's figure
out what it is. It's written over there. It's y (ln y) ^2 -
2(y ln y - y). The whole thing evaluated
at 1e. And that is, if I plug
in e here, I get e. Except there's a factor
of pi there, sorry. Missed the pi factor. So there's an e here. And then I subtract off, well,
at 1 this is e - e. So it cancels. There's nothing left. And then at 1, I get ln 1 is 0,
ln 1 is 0, there's only one term left, which is 2. So it's - 2. That's the answer. Now we get to compare that with
what happens if we do it the other way. So what's the vertical? So by vertical slicing,
we get shells. And that starts, that's
in the x variable. It starts at 0 and ends
at 1 and it's dx. And what are the shells? Well, the shells are, if I can
draw the picture again, they start, the top value is e. And the bottom value
is, I need a little bit of room for this. The bottom value is y. And then we have 2 pi x is
the circumference, as we sweep it around dx. So here's our new volume. Expressed in this
different way. So now I'm going to plug
in what this is. It's the integral from
0 to 1 of e - e ^ x. That's the formula for y. 2 pi x dx. And what you see is that you get
the integral from 0 to 1 of 2 pi e x dx. That's easy, right? That's just 2 pi e ( 1/2). This one is just the
area of a triangle. If I factor out the 2 pi e. And then the other piece
is the integral of 2 pi x e^ x dx. From 0 to 1. STUDENT: [INAUDIBLE] PROFESSOR: Are you asking me
whether I need an x ^2 here? I just evaluated the integral. I just did it geometrically. I said, this is the area
of a triangle. I didn't antidifferentiate and
evaluate it, I just told you the number. Because it's a definite
integral. So now, this one here, I can
read off from right up here. Above it, this is G1. So this is equal to, let's
check it out here. So this is pi e, right, - 2 pi
G1 ( x), evaluated at 0 and 1. So let's make sure that
it's the same as what we had before. It's pi e - 2 pi times
here's g1. So it's x e ^ x - e^ x. So at x = 1, that cancels. But at the bottom
end, it's e^ 0. So it's - 1 here. Is that right? Yep. So it's pi e - 2. It's the same. Question. STUDENT: [INAUDIBLE] PROFESSOR: From here to here,
is that the question? STUDENT: [INAUDIBLE] PROFESSOR: So the step here is
just the distributive law. This is e 2 pi x, that's
this term. And the other terms, the
minus sign is outside. The 2 pi I factored out. And the x and the e ^x stayed
inside the integral sign. Thank you. The correction is that
there was a missing minus sign, last time. When I integrated from
0 to 1 x e^ x dx, I had a x e^ x - e^x. Evaluated at 0 and 1. And that's equal to + 1. I was missing this minus sign. The place where it came in was
in this wineglass example. We had the integral of
2 pi x e - e ^x dx. And that was 2 pi e integral of
x dx, from 0 to 1, - 2 pi, integral from 0 to
1 of x e^x dx. And then I worked this
out and it was pi e. And then this one was - 2 pi,
and what I wrote down was - 1. But there should have been an
extra minus sign there. So it's this. The final answer was
correct, but this minus sign was missing. Right there. So just, right there.
