Welcome to a presentation on improper integrals.
The goals of this video are to determine
whether an improper integral is convergent or divergent.
And we'll also solve an application problem
involving improper integrals.
Let's start off by considering the area under the function y
equals 2 divided by x squared on the interval from 1
to infinity.
Now what's new about this problem
is the interval of integration.
Notice how we're approaching positive infinity
for our upper limit.
So if we set this up, as we normally would,
we would have the integral of our function 2 divide
it by x squared, which I'll write
so we can apply the power rule as 2x
to the power of negative 2.
And our lower limit of integration would be 1.
Our upper limit would be positive infinity.
Having positive infinity in our upper limit of integration
is what makes this an improper integral.
And if you go back and look at the definition
of a definite integral, the interval of integration
must be a closed interval.
And you can see that this is not.
What we do is you write this improper integral as a limit
as b approaches infinity of this definition.
And notice what we've done is replace the upper limit
of infinity with a variable b.
And then we take the limit of this integral
as b approaches positive infinity.
So what we have to do is evaluate this definite integral
and then take the limit of it as b approaches positive infinity.
So let's go and give it a try.
So applying the power rule, here we'd
have 2 times x to the negative divided by a negative 1.
And we need to evaluate this at b and 1.
Let's clean this up a little bit.
This would be negative 2 divided by x evaluated it at b and 1.
So first, we replace x with b.
We have negative 2 divided by b minus negative 2/1
or negative 2.
So now we take the limit of this, negative 2
divided by b as b approaches infinity.
This approaches 0.
This is not affected by b.
So this is equal to positive 2.
Now this is actually quite amazing.
What we've done is we found the area of this unbounded region,
and it is equal to 2.
Now that is a little bit strange.
We can see that the area is getting smaller and smaller,
but, in fact, it is decreasing enough so
that we can say the area is equal to positive 2.
An integral is called improper if there are number one,
upper and lower limits of integration are infinite.
And number two, f of x has a finite number
of infinite discontinuities, this video
will only address number one.
So here are some definitions that formalize
what we did in example one.
Notice how this improper integral
has an upper limit of infinity.
We replace that with b.
And then we take the limit of that integral
as b approaches positive infinity.
We could also have a lower limit of negative infinity.
And we could also have both negative infinity and positive
infinity as a lower and upper limit.
If the limit exists, we say that the improper integral
converges or is convergent.
If the limit does not exist, then we
say the improper integral diverges or is divergent.
Let's go and take a look at another example.
Now in this example, I've done a lot of the work already for us.
And the reason why, if you take a look at the function,
we're considering it's f of x equals x squared.
This is an increasing function on the interval from one
to infinity.
So right away, if we're trying to find
the area of the region bounded by the function in the x-axis,
it's increasing without bound.
So hopefully, it makes sense that we would not
be able to find this area.
And therefore, this improper integral would be divergent.
Formally, if we took the improper integral,
wrote it as a limit of a definite integral,
applied the power rule, did our substitution with b and 1,
and then if we look at this limit
as b approaches positive infinity,
this approaches infinity, and this is unaffected by b.
Therefore, this limit would be infinity, which does not exist,
which verifies, again, that this improper integral is divergent,
which geometrically means we cannot find the area of this
unbounded region.
Let's go and take a look at another.
Now this one is a little more interesting.
The function is decreasing.
So the area is getting smaller and smaller.
So let's see what happens here.
First step, we write this as a limit.
Again, notice what we've done, we replace infinity
with the variable b.
And then we find the limit as b approaches infinity
of this definite integral.
Now the antiderivative 1/x is natural log x.
So we have the limit as b approaches positive infinity,
a natural log x evaluated at b and 1.
So this would give us the limit as b
approaches infinity of natural log b minus natural log 1.
Well, natural log 1 is equal to 0,
but natural log b is an increasing function.
Therefore, this approaches positive infinity,
which means this limit does not exist,
which means this improper integral is divergent.
So even though the function is decreasing,
that does not mean that we can always
find the area over an unbounded region.
Let's take a look at one more.
Let's rewrite this as a limit.
Now we'll evaluate the integral.
Now on this one, we do have to perform u substitution.
U is equal to negative 2x.
So the du equals negative 2dx.
Dx is equal to the u divided by negative 2.
So we'd have the limit of the integral of 4 times e to the u,
but then dx is equal to du divided by negative 2.
Let's rewrite this one more time.
4 divided by negative 2 would be negative 2 u to the u du.
Notice I'm leaving the upper and lower limits of integration
off right now, because those are 4x.
And we're in terms of u right now.
Next we'd have the limit of the antiderivative of this.
The antiderivative of e to the u is e to the u.
So we'd have negative 2e to the u,
but remember u is actually negative 2x.
And we have to evaluate this at b and 0.
So we'd have negative 2e to the negative 2b minus negative 2
times e to the 0.
So now we need to find this limit.
Looking at this first term, this is the same
as having negative 2 divided by e to the power of positive 2b.
So as b approaches infinity, the dominant grows without bounds.
So this approach is 0.
This is minus negative 2 or positive 2.
So this limit is equal to positive 2.
Therefore, this is convergent.
And the improper integral is equal to 2.
So we have found the area of this unbounded region, which
is equal to 2.
We'll end by taking a look at an application involving
an improper integral.
That's called accumulated present value
of a continuous money flow into an investment at a rate of P
dollars per year perpetually is given
by this definite integral, which is equal to P divided by k,
where k is the rate and interest compounded continuously.
Now this is very similar to the accumulated present value
formula from the previous section.
The only difference here is we have an improper integral,
because it's a perpetual money flow or money
flow that is forever.
Let's take a quick look at why this
simplifies to P divided by k.
I don't have enough time to go over all of this now.
So you may want to pause the video.
But essentially, what we have to do
is what we've been doing on the previous examples.
Evaluate this improper integral.
Perform a u substitution.
Evaluate at the upper and lower limit.
Find the limit.
And we'll find that it is equal to P divided by k.
So if I want to find the accumulated present value
of an investment for which there is a perpetual continuous money
flow of $3,200 per year at an interest rate of 6% compounded
continuously, we can just take P $3,200 and divide
by k, which is 6% as a decimal.
And this is approximately $53,333.33.
Since we are now considering a perpetual money flow,
we can think of this amount as today's value
of the perpetual continuous money flow at 6%.
Thank you for watching.
I hope you found this video helpful.