So, last section for today.
integration by substitution.
So we did integration by parts, and that
was the product rule running backwards.
Integration by substitution we can think
of
as just working the chain rule backwards.
So I'm going to start out by letting
capital F of x just be the
antiderivative of little f of x.
And
then I'm going to think of x as being a
function of another variable, u.
Then by the chain rule, if I took the
derivative of capital F of g of u,
I would end up with f prime evaluated at g
of u.
And then the chain rule then says I have
to take this function
on the inside and multiply by its
derivative to, so I end up with.
A g prime of u showing up on the outside.
And then because of the way I've defined
little f of x to be the, the
derivative of capital F or rather a
capital
F to be an anti derivative of little f.
I can
write that as F of g of u times g prime of
u.
[SOUND].
And then if I integrate both sides with
respect to u.
Well, on this side, I, I just had d of u.
d by du.
So the derivative with respect to u of f
of g of u.
So that's going to be the complementary
operation to my anti derivative.
So that's just going to be f capital f of
g of u.
Is going to be equal to the integral
of little f of g of u times g prime of u
du.
And so my rule for integration by
substitution then, is if x, f of
x is an integrable function and g
of x is invertible and continuously
differentiable.
Then I can make the substitution x equals
g of u.
And it's going to change my integral in
the following way.
So if I replace the variable x in my
function f of x with g of u,
then I have to multiply by g prime of u in
order to get the same answer.
And so I think the easiest way to see how
this is going to, work is just to try an
example.
So if I look at this, I see
the top of this fraction as the
exponential function.
And then, the argument to that exponential
function is the square root of x.
So, I don't know how to integrate this
function just looking at that.
I'm not going to make any progress if I
keep it in terms of x.
But, it would be easier to integrate.
If, instead of having a square root in the
exponent here, I just had a single letter.
So if this was, say, e to the u, for
instance.
So, the way I can do that, is make a
substitution.
I'm going to define u to be the square
root of x.
And then if I do that, I could also think
of it as x being equal to u squared.
And then if I were to take the derivative
of
that, I would find that dx is equal to 2u
du.
And so let's see what happens when I
start.
Plugging in these these substitutions
into, into my integral.
So on the bottom, I can write square root
of x, so this square root of x in the
denominator just becomes a u.
In the numerator my e to the square root
of x.
That's going to be e to the u, and then my
dx term over here
this is going to be the part that sort of
picks up my bit from the chain roll.
dx is equal to 2udu So I'm going to put
that right here in place of the dx.
And then what's going to happen that's
very convenient,
is that this u here and the u
in the denominator, they're going to
cancel each other out.
And then because 2 is a constant, I can
use my,
my scaling property just to move that 2
outside of the integral.
And so I can rewrite this as 2 times the
indefinite integral of e to the u d u.
And
now since I'm integrating, this d u tells
me the variable that I'm integrating with.
This e to the u is going to be a very easy
function to integrate with respect to u,
because it's its own antiderivative.
So I end up with 2 e to the u plus c but
that's not quite finished yet,
because the answer to the question that I
asked, should be
given in terms of x because this is a
function of x.
But I know that u is equal to the square
root
of x so all I have to do in my final step.
Is just replaced this u with the square
root of x.
So this would be in, in my opinion
probably the simplest way to write down
the answers.
So to have 2 times e to the
square root of x plus an arbitrary
constant.
If you take the derivative of that, you're
going to get this e
to the square root of x divided by the
square root of x.
and here I've assumed that x is greater
than 0 just
because I want to be able to take the
square root of it.
So now let's consider trying a little bit
more complicated function.
So this, this doesn't really look very,
very much like it's a, a function
that I'm going to be able to write down as
a function of another function,
but it is something that I can still use
this rule to, to simplify quite a bit.
So, suppose I define u to be e to the x
minus e to the minus x.
So it's going to be the denominator as my
integrand.
So integrand is just a word that means the
function that I'm trying to integrate.
Well, if I take the derivative of this
function with respect to, with respect to
x, I'm going to get du is equal to e to
the x, and then this
one, by the chain rule, I'm going to have
a minus sign pop down, so this, oh, sorry.
This side, when I take its derivative, I'm
going to get e to the minus x times
the derivative of the exponent.
The derivative of minus x is negative 1,
so that's going to have that's going
to have the effect of just changing this
minus sign into a plus sign.
So that's just a little application of the
chain rule here.
And
now if I go ahead and make these
substitutions,
I end up with e to the x plus e to the
minus x times dx.
But that's just what I've said, that's my
du.
And I have 1 over e to the x minus e to
the minus x.
And that's what I've called my, that's the
variable I've chosen fort my substitution.
So this whole thing simplifies into just
finding the anti-derivative of 1
over u du.
And that again was in my little dictionary
of anti-derivatives.
So that's the log of u.
And then to be finished I have to
substitute back
so u is equal to e to the x minus x.
So the value of this anti-derivative here
is going to be log of e
to the x minus e to the minus x plus an
arbitrary constant.
And then there's one more trick we can use
with integration by substitution when
we're working with definite integrals.
So by the Fundamental Theorem of Calculus,
let's see, if I get to where I used that.
When
I integrate, oops.
When I
integrate f of x, I should just have
capital f of x evaluated at its upper and
lower endpoints.
But if I'm going to change from x into u,
so I'm going to make some
sort of substitution, the endpoints have
to
change, too, because what I'm really
saying.
When I write these limits, oops, the
limits of integration,
so from a to b, I'm talking about values
of x.
So I'm talking about x equals a to x
equals b.
When there's only one variable generally
you don't write
this notation just because there's no, no
possibility of ambiguity.
Because there's only one variable that
you're talking about.
But when I make my substitution, so when I
want to use the
integration by substitution formula.
Here my
intergram is now a function of u.
So the limits on
my integral also need to be in terms of u,
rather than in terms of x.
So I need to take g inverse.
So that's the, if I have u
is equal to g of x.
then I need to use that function backwards
to get what my limits should
be once I've changed from sort of
integrating in x space into u space.
And so if we just follow this calculation
all the way through, what
we're eventually going to end up with, so
I used my integration by substitution
formula.
Then once I have this guy here, I can
run the chain rule backwards which gives
me this.
The integral and the derivative.
So I, if I take the antiderivative of a
derivative, I just end up with the same
function.
So I have
f of g of u.
I've defined my limits.
To be g inverse b and g inverse of a so
when I evaluate
this g of u at g inverse of b and g
inverse of a.
So that it's getting kind of sloppy here,
but g and g
inverse their functions that sort of undo
whatever the first function did.
So g of g inverse of x is just equal to x.
So all this is really going to end up
being is f of b minus f of a.
Which is what we were after in the first
place.
And so, this ends up being the
integration by substitution rule, for a
definite integral.
And then I'll go through a quick example
of, of using this guy.
[INAUDIBLE]
.
>>
[INAUDIBLE].
>> Yeah.
So it's, it's essentially a shortcut that
avoids you having to go back.
So you could just compute the anti
derivative.
>> Right.
>> And once you have the
anti-derivative.
Derivative.
That's all you need.
It's just a trick to get you the
anti-derivative.
But if you change the limits, you don't
have to substitute
backwards cause you already have limits
in, in the right numbers.
Or in, in the right scale I suppose for
the right variable.
>> Right.
>> So suppose I
want to integrate
and suppose I want to do this by
substitution.
Because if I looked at this, I would
probably just do the multiplication.
And then use the anti power rule.
But,
what I'm going to do is recognize that the
derivative
of this, is going to have an x squared in
it.
So I could make some sort of substitution
for this x cubed minus a constant piece.
So I'm going to let u equals, equal x
cubed minus 1.
Then du is going to be 3x squared dx.
So I already have an x squared d x here,
so to get a three I just need
to multiply by one, use the very clever
choice of one of being
one third times three, and that's going to
give me one
third, oops.
There are supposed to be a fourth power up
here which made it a little bit more
complicated.
so I'm going to put my 3x squared dx over
on the right hand side.
And that's going to be just equal to the
derivative of my substitution.
In order to put this 3 here I had to
introduce a 1 3rd,
but because that's a constant value I can
move that out of the integral sign.
And then I'll go ahead and make my
substitution,
but I'm also going to change the, the
limits when I
do this too, so, when I look at the lower
limit.
When I put negative 1 into here, I have
negative 1 cubed.
Is minus 1 times minus 1 is 1 times minus
1 again is negative 1.
Minus 1.
So that becomes negative 2.
And then 0 cubed is obviously going to be
0.
Minus
1.
So that gets me at negative one up here.
Because the 3 X squared
is what I can replace
by the, by the chain rule backwards.
Becomes my D U term.
And so I can use my anti-power rule to
figure
out what an anti-derivative for u to the
fourth is.
And then I just end up with 1 15th times
negative 1 to the fifth power.
Oops.
What have I done here?
Yeah, so this should be a 1 15th too, boy.
Lots of typos on these slides.
I don't know what happened this time
around.
at the bottom limit squared.
And if you do the math, you end up with 31
over 15.
Sorry, minus 31 over 15.