So now moving on.
So
the first six lessons I was just trying to
talk about what an integral is, the
problem I'm trying
to answer and then some techniques that
we're going to
use in 461 and in later courses for
evaluating integrals.
Now I need some, some special results, eh,
in
particular I'm driving at is, is
differentiating improper integrals.
Because this is going to be something you
need to be able to do
when you want to work with the black
[UNKNOWN],
spacing formula.
Eh, and so I'm going to first
get started by just differentiating
definite integrals.
Then I'll introduce what an improper
integral is.
And then we will learn how to take their
derivatives.
So this is actually not too difficult of a
topic.
So hopefully this set will go by quickly.
So first start out by pointing out that a
definite integral, so if I
integrate a function f of x over an
interval ab, that's a real number.
So it's constant, it's not changing.
So if I try and take the derivitive of
this function, it's constant so it's
derivative is zero.
So the only way this is going to get
exciting is
if the limits so that the A and the B are
now functions of a, of another variable.
So I, I chose T.
Then when I evaluate this definite
integral
it still going to be a function of
T and so I might want to consider taking
its derivative with respect to T.
And then more generally the integrand
could also have T in it.
But since we haven't talked about
functions of more than
one variable yet, I'm going to leave that
for lesson three.
And so the question we want to consider is
how can I calculate the derivative, with
respect to T,
of the integral of, of function f of x
where the limits depend our functions of
the value t.
So it's actually pretty straightforward.
If I wanted to just evaluate this
integral.
What I would do is find an antiderivative
of my function.
And then by the fundamental theorem of
calculus.
I could evaluate this integral.
It's just going to be the anti-derivative
evaluated at the top
end point, and the anti-derivative
evaluated at the bottom end point.
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And
now, if I take the derivative of this
well,
all I have here is a function of a
function.
So I can take the derivative of that,
using the chain rule.
And so what I'm going to get is G prime of
T, so G
is a function of T, so I'm taking it's
derivative with respect to T.
That's just going to be equal to the
derivative
of the outside function, evaluated at the
inside function.
And then by the chain rule, I have to
also multiply that by the derivative of
the inside function.
And it looks a little bit complicated,
but all I'm doing is just I have to put
once
for the upper end point, and once for the
lower end point.
And then by the way we've defined this
capital F, I know its derivative is
lowercase f, so the formula for
differentiating a definite integral is
just going to be f(b(t)) times b'(t) minus
f(a(t)) times a'(t).
And so just to state it with all of its
actual conditions.
So this is the, this is the definition
that came out of the course textbook.
So F has to be a continuous function.
And a of t and b of t have to be
differentiable functions.
Then, the derivative with respect to T, of
the integral from A of
T to B of T of F of X DX is just F of
B of T times B prime of T minus F of A of
T times A prime of T.
A so just a, a quick example of this.
I think this is my simplified version of
exercise 11 from, from chapter one.
So, let's let g of x be defined as 1 over
the square root of
2 pi times the integral from 0 to b of x
of e to the minus t squared over 2.
And then, b of x is just going to be
defined by this
function here.
Which can be simplified the way I've done
it in the light grey on the right.
And so this is actually a calculation
you'll, you'll come across
These are, these are sort of pieces of the
black sholes pricing formula, so this is
the interest rate.
This is the, this is the volatility but
again since
I haven't talked about functions of
multiple variables yet I
just wanted to plug in specific numbers so
it doesn't
get too confusing about what I'm actually
trying to do.
And so the, the goal then is to compute g
prime of x,
and so is this the, the formula from the
previous slide.
But now since my lower limit is zero, so
it's constant with respect to x, if I
take its derivative, this, this a prime of
t is just going to be equal to 0.
So derivative of a constant.
So this formula will simplify a bit,
because of a being equal to 0.
So all I need to do is just evaluate my
function.
And then multiply by the
derivative of the inside.
So the first step I will do is just
evaluate my function g of x,
so this bit here.
I'm just going to evaluate that at b.
And so I get, I had 5 times log of x plus
0.06.
So when I square that, I'm going to end up
with
25 times the log of x plus 0.06, that
quantity squared.
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And
then I still need to multiply it by the
derivative, then of b of x.
So, b of x was equal to 5 times the log of
x plus 0.06.
And so that's pretty easy to see the
derivative of, that's 5 times log of x.
That's derivative is 5 divided by x and 5
times 0.06, that's just going to be
a constant, so we don't even need to
bother
doing the math, that's going to have a
derivative of 0.
So b'(x) is just going to be 5 divided by
x,
and so I just need to put all of the
pieces
together, which just means substituting in
5 over x for b'(x),
And that gives me the derivative of my
function g(x).