We're asked to evaluate the integral
and then determine if the integral
is convergent or divergent.
We have what's called an improper integral
because the upper limit of integration is infinity,
which means we begin by writing
the improper integral as a limit.
The given improper integral is equal to the limit
as b approaches infinity of the integral from 3 to b.
Let's also rewrite the integral function
as 5x times e to the power of -x squared dx.
Just notice how we replace the upper limit of integration
with b, and then we have the limit as b approaches infinity
of the integral from 3 to b
of 5x e to the power of -x squared dx.
From here, we are going to have to perform
u-substitution to integrate.
We will let u = the exponent of -x squared and, therefore,
du is equal to the derivative of -x squared times dx,
which gives us -2x dx.
Notice we have 5x dx.
Let's go ahead and just solve for x dx
by dividing both sides by -2.
Simplifying, we have -1/2 du = x dx.
For the next step we'll write the integral
with respect to u.
So, in doing this, let's go ahead and factor out the 5,
and then x dx is equal to -1/2du.
Let's also factor out the -1/2.
And then we have the limit as b approaches
infinity of the integral.
Now need to be careful here.
The limits of integration from 3 to b
are x values, not u values.
So for the lower limit, let's write x = 3,
and for the upper limit, let's write x = b.
This is just emphasizing these are x values, not u values.
And now writing the integral with respect to u,
again, we factored out the 5,
and x dx is equal to -1/2 to u.
We already factored out the -1/2,
so now we have du and e to the power of -x squared
is just e to the u.
Well, the integral of e to the u is just e to the u,
which is really e to the power of -x squared.
So we have -5/2 times the limit
as b approaches infinity of the antiderivative,
which is e to the power of -x squared.
And, again, the limits integration
with respect to x are from 3 to b.
And now we need to determine big F of b - big F of 3
and then determine the limit.
So we have -5/2 times the limit as b approaches infinity of
big F of b is e to the power of -b squared,
then we have minus big F of 3,
which is e to the power of -3 squared or -9.
Let's rewrite this using positive exponents
before we determine the limit.
We have -5/2 times the limit as b approaches infinity of
1 divided by e to the power of b squared - 1
divided by e to the ninth.
And now let's determine the limit.
As b approaches infinity,
1 divided by e to the power of b squared approaches zero
because the denominator is increasing without bound
and the numerator stays at 1.
So, again, 1 divided by e to of the power b squared
approaches zero, and then one divided by e to the ninth
is not affected by b, and, therefore,
we have -5/2 times -1 divided by e to the ninth,
which is equal to positive 5 divided by 2e to the ninth.
Because the limit exists, and we got a real value
for the improper integral, we see the integral converges.
If the limit didn't exist or approached
positive or negative infinity,
we would say the integral diverges.
But, in this case, again, the integral converges.
I hope you found this helpful.