We want to determine if the improper integral converges
or diverges.
And if it converges, we want to determine the value.
This is an improper integral because the upper limit
of integration is infinity.
Let's do a quick review of improper integrals.
And then we'll come back to this example.
Notices these integrals on the left
are all improper because the limits of integration
involve positive or negative infinity.
So looking at number 1, we want to replace infinity with b
and then take the limit as b approaches
infinity of the integral.
For number 2, we want to replace negative infinity with a
and then take the limit as a approaches negative infinity
of the integral.
And, of course, we don't have to use a and b.
But that's pretty normal.
And then notice on number 3, if both limits of integration
involve negative and positive infinity,
we're going to have to break it up
into two different improper integrals,
where c can be any real number.
And then we have to apply number 1 and number 2
to both of these integrals.
If the limit exists, we say that the improper integral
converges or is convergent.
If the limit doesn't exist, we say
the improper integral diverges or is divergent.
So looking at our example, we want
to write this as the limit as b approaches
infinity of the integral from 0 to b of 6 e
to the negative 3x dx.
Now we're going to evaluate this definite integral
and then take the limit as b approaches positive infinity.
Here, we need to recognize that we'll have
to apply u-substitution here.
So we'll have u equals negative 3x.
So differential u is equal to negative 3dx.
Now let's compare this to our integral.
u is equal to 3x.
But notice how we're left with 6dx and we have negative 3dx.
So if they multiply both sides of the equation
here by negative 2, we would have
negative 2du is equal to 6dx.
And now we can replace 6dx with negative 2du.
So let's go ahead and rewrite this in terms of u
so we can integrate.
We'll have the limit as b approaches infinity.
We'll leave out the limits of integration
temporarily because they're in terms of x.
So we'll have the integral of e to the u.
And then, again, dx is negative 2du.
So I'll put du here.
And we'll pull out the negative 2.
So we'll have the limit as b approaches
infinity of negative 2 times the antiderivative of e
to the u, which is just e to the u.
So this is negative 2 e to the u.
But again, u is negative 3x.
And we need to evaluate this at b and 0.
Before we evaluate this limit, I think
it's going to be helpful if we write this as the limit
as b approaches infinity of negative 2 over e
to the positive 3x power.
It's often easier to evaluate infinite limits
if it's fraction form.
So now we'll have the limit as b approaches
infinity of negative 2 all over e
to the 3b power minus negative 2 all over e to the 0 power.
Well, as b approaches infinity, notice how this denominator
is increasing without bound.
So this fraction is approaching 0.
And this fraction is not affected by b.
So we'd have minus--
this would just be negative 2 since e to the 0 is equal to 1.
And minus negative 2 would be positive 2.
So the value of this improper integral is equal to 2.
And since this limit exists, we can
say this improper integral converges.
We'll take a look at another example in the next video.