Today we're going to talk about an integration
technique called substitution.
You can look at integration by substitution in two ways.
The first way would be by a pattern recognition,
and the second would be by change of variables.
The idea here for pattern recognition is we have
an integrand which is a composite function
where g of x is the inner function,
but g prime of x also occurs in the integrand.
Then we can apply this formula to find the antiderivative.
Remember, the derivative of this antiderivative would be
the integrand.
However, in practice it's going to be easier for us to use
the change of variables formula.
And the idea of this is we're going to let u equal the inner
function and then when find differential u, we want it
to match the derivative of g times dx.
Now this sounds kind of complicated,
but in practice it's not really that bad.
Let's take a look at one more idea here before we start.
We know that integration is the opposite
operation of differentiation.
So, for example, we know that to find the derivative
of a composite function, we need to apply the Chain Rule.
Therefore, if we find the antiderivative of this, we should
end up with our original function.
That is actually what is happening when we integrate
using substitution, recognizing this pattern, and we end
up finding the antiderivative.
Now just notice that this notation here is a little bit
different than the notation used on the previous slide,
but the relationship is the same.
The derivative of this function would be the integrand.
The derivative of this function here would also be the
integrand, remember big F represents the antiderivative.
Okay, so here are our guidelines.
We're going to choose a substitution u equals g of x.
Usually it's best to choose the inner part of a composite
function.
Then we're going to calculate differential u, which is
g prime times dx.
We write the integral.
Find the resulting integral.
Then replace u with g of x to obtain our antiderivative.
And then we can check by differentiating.
Here are some basic formulas that might help,
you may want to pause to review them.
Okay, so let's go ahead and get started.
First step, we need to identify our u.
We're going to let u equal x to the fourth plus two.
Now we need to find differential u, which is equal to
the derivative of this times dx.
So that would be 4x cubed dx.
Notice that in our integral we do have 4x cubed dx.
So now what we can do is to perform substitution
and rewrite this as u raised to the fourth power,
and all of this is just du.
Now integrate with respect to u, u to the fifth divided
by five plus C.
And now we just have to rewrite this in terms of x.
So replace our u with x to the fourth plus two.
Notice that when I wrote this I wrote it as 1/5.
Okay, let's try another.
Again our first step is to identify the u.
It's often best to let u equal the stuff that's raised
to the power, so our u will equal x to the third plus two.
Therefore, differential u will equal three x squared
times dx.
Now if we take a look at our integral,
we don't have three x squared dx, we just have x squared dx.
So what we can do over here is divide this by three on
both sides, so we would have 1/3 du is
equal to x squared dx.
I always like to try to make this match the integral exactly
so we can do our direct substitution.
Now I can rewrite this as, this blue part is our
u to the seventh, x squared dx is the same as 1/3 du.
Factor the 1/3 out and we integrate and have 1/3 u to the
eighth, divided by eight plus C, and lastly we write this
in terms of of x, so we have 1/24 times u to the eighth,
which is x to the third plus two, raised
to the eighth plus C.
Now this you probably may not know what to let u equal,
but notice how the denominator is degree two and
the numerator is degree one, so we're going to let u equal
the degree two part, therefore du would equal four x dx.
Okay, so here's the u, notice what's left is two x dx,
and we have four x dx.
So what I'll do is divide by two, so we have 1/2 du will
equal two x dx.
That'll make it match perfectly.
So now we can rewrite this as Integral of the red part would
be one half du, and the blue part is our u,
so we can one over u.
Remember that the anti-derivative of one over u is equal
to the natural log of u.
Now instead of writing u, I'm going to go ahead and replace
it with one plus two x squared plus C.
Let's go ahead and try another.
Let the function raised to the power equal our u.
So differential u equals one over x
don't forget times dx.
Let's color code our integral.
This is our u, so we'd have u to the fourth
and what's left is two over x dx
and we have one over x dx.
So what we can do then is multiply both sides by two.
So we'd have two du equals two over x dx, which makes this
substitution much easier, so our integral will be
u to the fourth
and then this is replaced by two du.
Factor the two out, we have the u there so
the antiderivative would be two times u to the fifth
divided by five plus C,
which is equal to 2/5 natural log of x
to the fifth power plus C.
So let's try a couple definite integrals.
Finding the antiderivative will be the same but then we'll
have to evaluate them at five and two in this case.
And if we pick the wrong u, then we'll come back
and do it again.
If you know your derivatives, it does make picking the u
much easier.
We let u equal natural log of x
du would equal one over x dx.
We look at the integral that's a perfect fit, here's our u
and whatever is left, is our du.
Now these limits of integration are in terms of x,
so I'm going write this I'm going to leave them off
temporarily, so we have the blue part would give us a
one over u and the red part would give us our du.
Remember the antiderivative of one over u which would be
natural log u, but again our u is actually natural log x.
Now this is a definite integral, so we do need to evaluate
this now at five two.
So we get an approximate value of point
eight four two eight.
Okay, our last example, again here I see a part that's
degree two, another part that's degree
On this last problem I see a part that's degree two
another part that's degree one
I'll let u equal the degree two part
This would be our u, du is two x dx.
Take a look at our integral, we have x dx, so I'll divide
both sides by two, which implies that 1/2 du is equal
to x dx.
We're writing this in terms of u, we'd have e to the u
and then a 1/2 du.
Again I'm leaving off the limits of integration
because this is in terms of u now.
If I wanted to include them, I'd have to replace x with
zero and one, in this equation for u.
This would equal 1/2 e to the u
of course u is x squared.
And evaluate it at one and zero,
which is approximately point eight five nine one.
I hope you found this video helpful.