This'll be the second half of my lectures
on integration.
So last time we just talked a little bit
about
what integration was, the problem it was
trying to solve.
we talked about the fundamental theorem of
calculus, which
is a tool that lets us actually evaluate
definite integrals.
one over a little bit of application of
integrals and then I started talking about
techniques
of integration.
So I cover something called integration by
parts which was just the product all
backwards and then something called
integration by
substitution or the substitution method
which was
just the general kind of running backwards
Today I'm going to do one more technique
of integration called completing the
square for
those of you who are taking 461.
You're going to use this when we try to
transform a normal distribution into
something called the log normal
distribution.
And then I'm going to talk about
differentiating definite integrals so a
function can
have, could be defined with integral and
having the variable in the limit.
So in that case you n actually think of
that as a, a function you can
[INAUDIBLE]
the derivative off.
I'll talk about improper integrals so
these are either integrals of functions
that take infinite values in either the x
or, or y direction.
And the, differentiating improper
integrals.
Okay, So, I'm going to get started with my
completing the square.
And I'll, I'll introduce this just by
trying
to remind everybody of something called
the Quadratic Formula.
And so, a quadratic function.
Is just a function, it's a polynomial of
degree two, so it's a times x
squared plus b times x plus c, or a b and
c are real valued constants.
And we have to add the extra condition
that a is not equal to zero.
Because if it were equal to zero this
would just be a linear function, bx plus
c.
And suppose I'm interested in finding the
value or values
of x that make my quadratic function equal
to zero.
I think everybody remembers from high
school algebra the
quadratic formula, so it's x is equal to
minus b plus or minus
the square root of b squared minus 4 ac
all over 2 a, and that's
going to give you the, the values of x
that will make this function be equal to
zero.
And the way we actually come up with this
formula is the same technique
that we will use in integration for
completing the square.
So suppose I want to solve my quadratic
function, I use
this notation to say I'm setting it equal
to zero so its
not defined to be equal to zero, its me
I'm choosing a certain level and now
once I've done that these x's aren't
vairable anymore, they are sort of
fixed numbers.
And I want to find out what they are.
So the first thing I'm going to do is just
divide every term by a.
So I multiply both sides of this by one
over a,
and I'm allowed to do that because one of
my conditions
on my quadratic function was the a is not
equal to zero.
So as long as a is not equal to zero, I
can divide by a.
And so I get this new form.
So it doesn't look too exciting yet.
And then what I'm going to try and do.
So this is sort of the, the meat and
potatoes of completing the square.
I want to add zero to this.
But I want to choose a particularly
smart choice of zero.
So I'm going to, first of all,
write b over a as 2 times b over 2 a.
So, that should be still b over a.
And then I want to add something here, and
I want to subtract something over here.
And I want to choose the thing that I'm
going to add over here.
So that the first
three terms can be written as x plus
something squared.
So lets just look really quickly at, what
I would want to have so that the
something is going to be b over 2 a and so
if I look at x plus b over 2 a squared.
Well on the first term.
So I I use a trick to remind myself called
foil, which is first,
outer, inner, last.
So the first two terms are going to be x
times x, which gives me x squared.
And then I'm going to have outer, so x
times b over 2 a.
And then inner, b over 2 a times x so
that's why I get two of these guys.
And then last will just be the last term
squared.
So this is what I'm going to choose for my
blank here.
And so my intelligent value of 0 is going
to be b
squared over 4 a squared minus b squared
over 4 a squared.
So I'm going to, oops, stick that into
these blanks here.
And now these first three terms that I
have here, I can
replace that by x plus b over two a, that
quantity squared.
And I've done two things in this line, so
I'm also going to
move these last two terms to the other
side of the equals sign.
Oh, and, what, I must have done a little
bit more here.
Okay, I also multiplied the c over a by 4
a over
4 a so that's where this 4, oops 4 a c
over a
squared comes from what, what did I do.
Think.
And so then I'm going to take the square
root of both sides, so I get x
plus b over 2 a is equal to and, we're
going to end up with two
solutions now because, providing that this
value here
is greater than zero I'm going to end
up with a positive value of the square
root and a negative value of the square
root.
And then the rest of it is just isolating
the x and making it look
pretty so I'll just move the b over two a
to the other side and then.
A square root of a fraction, I can think
of that as a square root
of the top divided by the square root of
the bottom so I'll just go
ahead and do the square root of the bottom
because 4 a squared has a
nice square root and it also happens to
be the same denominator as a half-up
frontier.
And so that gives me the Quadratic
Formula, so X is equal
to minus b plus or minus the square root
of b squared minus 4 a c all over 2 a.
So now let's look at I've sort of taken
the important
part out of the integral you're going to
have to, to solve.
In 461 when I ask you to find the expected
value and the variance of a log normal
random variable.
So, I want to compute the integral from
zero to infinity of e to the, so
it's a minus fraction here, and then I
have log of y minus Mu, that whole
quantity squared Divided by two sigma
squared.
And here, mu and sigma are going to be
real valued constants.
you can think of them as just the
mean and the standard deviation of a
statistical distribution.
So, I'm going to use my integration by
substitution trick from, from last time.
And this thing, the log of y minus mu,
that
looks like something I wouldn't be
particularly happy to have.
So I'd like to get rid of that and just
replace it by a single value u.
So if I do that then I can solve for y in
terms
of u, and I get e to the u times e to the
mu.
And then if I take the derivative of y, I
just end up with
y du, which is, I'll put that as my value
of y, but in
this particular case it's going to cancel
out nicely.
So now I make my substitution.
So lat time I told you for definite
integrals, I can also change the limits
here.
So this is going to be a case where
probably
be nicer if you the limits just into u.
Rather than remember that this is u equals
something, and
then try and switch back to y at the end.
And so it turns out if the mu doesn't
really matter, since I'm going from zero
to infinity.
if I, if I have log of zero, that's
going to be minus infinity.
And if I have log of infinity, that's
going to be plus infinity.
So the limits on u are going to be minus
infinity to infinity.
And then I go ahead and substitute in
everything
that I worked out in the, the, second
bullet-point here.
And then replace
My y by e to the u e to the Mu and
now one of the, the nice things that's
going to happen remember Mu is
a constant so I can, that's just scaling
my integral so one
of my properties allows me to pull that
outside of the integral sign.
And so now I end up with e to the Mu, e to
the Mu integral
and I have e to some power of u times e to
the u.
So really whats going to end up happening
here.
If I have e to a power times e to a power
I
can also think of that as e to the sum of
those two powers.
So, I'm going to pause for a second and
say take a look at the intergrand.
So, inter-grand is just this thing.
It's the function that I'm integrating.
And so if I give them both a common
denominator then I can go ahead and do the
sum and now I'm starting to see something
sort
of like what I had, when I was looking
at the completing the square when i was,
so the same problem that
I needed the solve when I wanted to solve
a quadratic equation.
And so here, it turns out that the,
the intelligent value to add is
sigma to the fourth and then I can
write this in this form here.
And
this thing here is just going
to be u minus sigma squared.
So I have u minus.
Sorry, not e minus sigma squared.
I have to say that twice.
So it's u minus sigma squared.
And that
whole quantity is squared as well so that
if you, if you
expanded this out you just get the thing
that's in parentheses here.
And so now I have, so here when I had Two
functions raise to an exponent multiplied
together so e to one
power times e to another power I was able
to add
those powers together and just write it as
a single exponential.
And now I'm just going to do the same
thing but backwards.
So I'm integrating with respect to you.
So that means this Sigma
to the fourth over two sigma squared,
that's just going to be a constant number.
So I'm going to rewrite this as e to the
minus quantity u minus sigma squared,
squared divided by two sigma squared times
e to the sigma squared over two.
But now this is just another number that I
could
bring outside of my integral sign if I
wanted to.
So let's go back now that, now that I have
this little intermediate result.
And look at the function I'm trying to
integrate again.
So I can replace the inter-grand now with
this function that I've this
transformation of it that I've, so all
I've done is taken what used to be there
and replaced it with this value.
And now I can also bring this, this part
of the product out in
front of the integral sign because it, it
doesn't depend on you any more as,
as far as I'm concerned its just a, a
scalar quantity so I can write it like
this.
So now I have e to the Mu times sigma
squared over 2.
Integral from minus infinity to infinity.
And then, sooner or later, you're going to
learn to recognize this form.
Here, I just had to make a hint.
Sort of one of the problems with
mathematics, it's all sort of intertwined.
So when you try to put it in order, so
that I can give them
in sequential lectures, eventually you
want to
use something that you haven't talked
about yet.
And so I stuck that in a little gray box
here.
But it turns out that this integral, so if
I have e
to the u minus something, so it doesn't
matter what number is there.
In this particular case it happens to be
sigma squared, which is
the same value down here, but that's just
because I was lucky.
So if I have e to the minus, and then my
variable u minus something, that quantity
squared, divided by two sigma squared.
This is something called the normal
density function.
if I integrate that from minus infinity to
infinity,
we get the square root of two pi sigma.
And then so this is something in, in
fourth week,
we'll, we'll talk about how to actually
evaluate this integral.
But if I look back up here that's exactly
the integral I have on
the right hand side so I can just replace
that with 2 pi sigma.
So I end up with 2 pi sigma e to the u
plus sigma squared over 2.
So, that probably doesn't seem like the
most useful example, until you get to
the homework in 461, where you have to do
this again for the first time.
And then I think you'll find these two
slides very, very useful.