Okay, so that was my, my brief overview of
what
integration is and two or three problems
that can be used to to address.
And I'm going to start talking about
techniques of integration.
So these are techniques for finding
anti-derivatives of functions.
And the first one is called integration by
parts.
So essentially what I'm going to try and
do to get integration by
parts is remember, when I was computing
derivatives, we had certain rules.
One of them was the product rule which
was the derivative of the product of two
functions.
And so I'm just going to try and use that
rule backwards to be able to evaluate an
anti derivative.
So, let's start out with two functions, f
of x and g of x.
And they need to be continuous and
integrable.
Then if I looked at the product of their
anti derivatives,
and here it's a little bit, a little
bit trickier what I'm going to do with,
constant.
Because if I said f of x, if I
was allowed, to put in a arbitrary
constant there.
And I did the same thing with g of x then
when
I multiply these together I would have the
constant from f of x
times g of x and the constant from g of x
times f
of x, so for this to work I have to choose
my constant
to be zero
[SOUND].
So I'm going to take the product of these
two
anti derivatives, and take its derivative
using the product rule.
And what I'm going to end up with is
the derivative of the first function times
the second.
So this guy here.
Plus the derivative of the second
function.
Times the first.
So then if I take the integral of both
sides of that, well the
integral sort of by the fundamental
theorem
of calculus that's just undoing my
derivative
operation, so I have capital F of x times
capital G of x
is just going to be equal to the integrals
of the two terms on the right-hand side.
And then I'm just going to rearrange these
terms
to get something called the integration by
parts formula.
So I'm going to take
Oops,
this second term, move it to the left hand
side of the equation
and move the other two terms to the right
hand side of the equation.
And
now what this is telling me is that if I
can
find, if I can break my function up into a
function
that I can integrate, this g of x, and a
function
that I can take its, where I can take its
derivative.
Then I can use this formula to evaluate
this integral.
So generally the idea is going to be to
choose.
F of x and little g of x so that
this final term here is something that's
easy to compute.
So I'm starting out over here with a
capital F of x.
So the easiest thing that I could ever
have here would just be an
x, because if I took the derivative of
captial X, if capital, sorry capital F.
If that's
a constant, then I could just take this
little f,
the derivative of a constant, sorry, if
this was x,
little f of x is going to be a constant,
and I can take my constant outside my
integral sign.
And then I just have to integrate, oops,
that should be a little g of x.
No, sorry, capital G of x never mind.
so what I'm wanting
to do is choose these functions so that
this
integral is going to be something that's
easy to evaluate.
And so since I'm just stumbling over my
words
now I probably should just skip ahead to
an example.
So here this doesn't even look like it's
something that
I'm going to be able to write down as a
product.
But you always have a one.
So if I let my capital F of x just be this
function here log of 1 plus x.
And I let my g of x be 1 plus x, then
my little g of x, that's just going to be
the derivative of this guy here.
It's just going to be a constant it's
going to be one.
So I can write this function here as
capital F of x times little g of x.
So this was my integration by parts
formula.
And then I just go ahead and plug these
functions
that I've worked out here, into the
integration by parts formula.
So, capital F of x is just log of 1 plus
x, g
of x is equal to 1 so I didn't bother to
write that down.
The anti derivative of that is going to be
F
of x, capital F of x times capital G of x.
So, that's just
1 plus x times log of 1 plus x minus the
integral of little f of x.
So that's going to be the derivative of
this guy here.
Times capital g of x which is one plus x.
So, the capital G of x ends up on top
here.
The derivative of the logarithm is just 1
over the argument.
So in this case, the argument is 1 plus x.
So, little f of x is going to be 1 over 1
plus x.
And technically I skipped a step here,
because I also have to use the chain rule.
But the derivative of 1 plus x,
we already said, was equal to 1.
So, it's 1 over 1 plus x, times 1.
And now if I look at the term on the right
here, I have 1 plus x, divided by 1 plus
x.
So that's just going to cancel each other
out.
So I have 1 plus x times the log of 1 plus
x.
Minus the integral just of d x.
And so.
That integral is very easy to calculate, I
just end up
with x being the anti derivative and then
because I don't have
limits here I am asking for an
in-definitive and well just
a anti derivative so I am adding on a
arbitrary constant C,
so generally the bulk of the work in
using this formula Is looking at the
original function
and then trying to identify you know, how
am I going to write that as a capital F of
x and a lower case g of x so I want to
split up that function.
So that this term that I get on the
right-hand side, the term that I still
have to integrate after I use the
integration
by parts formula, is going to be easy to
integrate.
And unfortunately, there's just not a
really good way to
tell you how to do that.
You just have to practice, and every time
you do it, you get
a little bit better at seeing how you
could break these things up.
So suppose I want to integrate, I want to
find
the anti-derivative for the function x
squared times log of x.
So this time it's going to be a little bit
simpler, I mean this is clearly.
The product of two functions.
I have x squared times log of x.
So, let's just start there by saying log
of x is going to
be the f of x and I could also choose to
make x squared the f of x.
And log of x to be.
The, the g of x, but I'm
going to have to integrate this thing.
So I want to pick that in a way where it's
going to be easy to do.
So the capital F of x, that's something
I'm going to have to take the derivative
of.
And it's going to be easier to take the
derivative of this than it is to integrate
it.
On the other hand, this is just, something
I can use my anti power rule on.
So that's going to be something that's
probably not going to
be too difficult to find an anti
derivative of.
So let's
choose that to be the part where
I'm going to have to compute the
antiderivative.
So, so if my my capital F of x is log of x
and my capital G of X is 1 3rd x cubed.
Then my little g of x is going to be x
squared.
And so now we can figure out what the anti
derivative
of this function is just by plugging these
values into the
integration by parts formula, and hoping
that we're able to integrate
the term that's going to end up on the
right-hand side.
So, let's see.
Yeah, got it right this time.
So the integral of capital F of x, g of x.
So, just plug everything in.
So, capital F of x, that was the log of x.
The g of x is the x squared.
And so my integration by parts formula
tells me that I take capital F of x times
capital G of x.
So I get one third x cubed times log of x
minus the integral.
And now here I have my lowercase x, so
here
I just have to take the derivative Of this
function
and back in my little dictionary slide I
have the
derivative of log of x is just 1 over x,
so I have 1 over x times capital g of x
which is one third x cubed and so not
only does it is it possible to integrate,
it even
gets a little simpler since this 1 over x
is going to
cancel one of my x's in my g of x
function.
So I end up with one third x cubed log x
minus the integral of one third x squared.
So I could use the anti power rule to
evaluate this guy here.
And add up with one third x cubed times
log of x, minus one ninth x cubed.
And then because I haven't put any limits
on this, I'm asking just for
an anti derivative, an indefinite
integral, so
I have to add an arbitrary constant c.
And we're also going to be able to.
Do this for definite integrals.
And in practice, really, it's not going to
change anything.
You just use the trick to find the anti
derivative.
But you can actually work though it if you
want to.
this will probably be the one and only
time you do with the limits here.
So this is exactly the same setup that I
had before.
I have the.
Derivative of capital f of x times capital
G of
x and I'm going to integrate that over an
interval ab.
So, because the, by the fundamental
theoreum of calculus, this integration and
differentiation.
They are sort of complementary operations.
This integral is going to cancel out my
derivative and I just have to
then evaluate my anti-derivative here at
its end points and take the difference.
So that's going to give me capital F of b
times capital G of b minus capital F of a
times capital G of a.
And I can then just rearrange the terms.
So I can use the linearity property to put
this integral sign
in front of this function as well, so I
can write
this integral of a sum as the sum of two
integrals.
And that gives me.
The integration by parts formula for a
definite integral.
Except I believe, oh no,
yeah, so we just end up evaluating this
term here at it's upper and lower end
point.
So I think this can be a little bit
confusing.
So if you, if you're not getting the
correct answer the first time you do this.
you might want to try and use this longer
formula.
And see if, that helps you spot your
mistake.
And so, one final example.
I'll try and integrate from one to three
the function x times e to the x.
So again I'm going to let the function
that I'm going to take the derivative of
be x, because when I take the derivative
of that I'm just going to get one.
That will be something easy to work with.
The function I'm going to have to
integrate should be oh, e to the x.
Its anti-derivative is just going to be
itself.
So that's going to be an easy choice.
And so I just have to then go ahead and
plug these.
Into the integration by parts formula.
So remember on the previous slide I'd
written this as f of
b times g of b minus f of a times g of a.
And that's exactly what I mean by this
notation here.
So here what I mean is take capital F of
three times capital G of three.
Minus capital F of 1 times capital G of 1
and the rest of
it should be clear just because the, the
limits are still on the integral signs.
And so I'm going to get
this line just by plugging in.
The functions I've worked out into this
formula here.
And so this is my capital F and my capital
G.
And then this one was easy to do, just
because it's its own anti-derivative.
So this is 3 times e to the third minus
1 times e to the 1, so this comes from the
first term here and then from this
second term I end up with e cubed minus e.
And I can do a little bit of
simplification, and get the, answer 2 e
cubed.