Here is the graph of this region.

The area of this region will be defined by the following integral.

As we look at this integral, it is clear that we have a problem with it because the area is not finite. In fact, this integral has infinity for

one of its bounds. Any integral that has infinite limits or points within the interval integration that causes the function to go to infinity are called improper integrals. How do we determine the value of this integral? We must first find the value of this modified integral, which is now a definite integral.

Then we will evaluate the limit as b goes to infinity.

So the area under this curve is infinite.

The usual way this problem would be written is the following.

The steps that you will take in evaluating this problem are to first evaluate the modified definite integral, then take the limit of the result of the first step. If the limit fails to exist or is infinite, then we say that the improper integral diverges. If the limit is finite, then we say that the improper integral converges.

 Going back to this problem, the area under this curve is infinite, but is the volume of Gabriel's Horn finite or infinite? Let us find out. To do this, we must determine the volume of the solid of revolution. I will use the disk method where the volume is the following formula.

Therefore, the volume of Gabriel's Horn is finite.

 Let us now look at some examples on how to evaluate improper integrals.

Notice that the 4 causes the denominator to be zero. Therefore, this rational expression goes to infinity as x ® 4. So our limit will be evaluated as b approaches 4.

Let u = 4 - x, then du = - dx. When x = 0, then u = 4, and when x = b, then u = 4 - b.

Therefore, this integral converges, and its value is 4.

Notice that one of the bounds is infinity, so this is what we will let the limit approach.

Let u = x + 1, then du = dx. When x = b, then u = b + 1, and when x = -2, then u = -1.

Therefore, this integral converges.

0 is the value that causes problems in the denominator of the integrand, so I will let the limit of this improper integral approach 0.

Therefore, this improper integral does converge.

First of all, as we look at this improper integral, we should notice that both the upper and lower bounds go off to infinity. Therefore we cannot evaluate this as it stands. To evaluate this improper integral, we must divide the interval that is under consideration into two pieces. When we divide this region, we must pick a value in this interval that does not cause a problem in the integrand. What would cause a problem with this integrand? Well, anything that would make the denominator zero, but we do not have to worry about that. The denominator will always be positive. Why, you ask. x ² is positive, 3x ² is also positive, and then we are adding a positive 2 to this, so it will be positive. I will divide this interval at 0. Why did I pick zero? I picked zero because it is easy to work with. So let us start working on this integral.

Now we have two improper integrals. For the original improper integral to converge, both of these new improper integrals must converge. If one of them diverges, then the original improper integral diverges.

Let u = 3x ² + 2, then du = 6x dx. When x = b, then u = 3b ² +2, and when x = 0, then u = 2. Also, when x = c, then u = 3c ² + 2.

Since the first limit converges to -1/ 16 and the second limit converges to 1/ 16, the improper integral converges to 0.

Notice that when x = 1, the denominator of the integrand will be zero. This is a problem, so to handle this improper integral, I will divide the interval at x = 1.

Now, we have to deal with the absolute value sign. On the interval 0 £ x < 1, x - 1 is negative. So we will replace | x - 1| with 1 - x. On the interval 1 < x £ 2, x - 1 is positive. So we will replace | x - 1| with x - 1.

Since both limits go to infinity, therefore the improper integral diverges.

Let f and g be continuous on [a, ¥ ) and suppose 0 £ f (x) £ g (x) for all x ³ a. Then

1.

2.

If the new improper integral converges, then the original improper integral will converge by direct comparison, but if the new improper integral diverges, then the original improper integral will diverge. Therefore, we must evaluate the new improper integral.

Therefore, the new improper integral diverges. Therefore, we can conclude that

diverges by the direct comparison test.

Now, we have to determine if the new improper integral

converges or diverges.

The new improper integral converges, therefore the original improper integral

converges.

If the positive functions f and g are continuous on [a, ¥ ), and if

what the limit of f (x)/g (x) is.

I am now going to factor an x ² out of the denominator.

From the beginning discussion of Gabriel's Horn, we know that the improper integral of 1/ x on the interval [1, ¥) diverges, so this improper integral will also diverge on the interval

[2, ¥). Therefore the original improper integral

also diverges by the limit comparison test.

SOLUTION B: Now to determine whether this improper integral diverges or converges by integrating directly.

This integral is in the form of u ² - a ² which is a secant substitution. Let x = sec q, then dx = sec q tan q dq and x ² - 1 = sec ² q - 1 = tan ² q .

Now to construct the reference triangle for this substitution.

sec q = x

Therefore, this improper integral diverges.

Which method is easier? In my opinion, the first method was easier, but it really depends on which method you are comfortable with.

We are going to use the limit comparison test. The function that I am going to use for the comparison is the end behavior model for this function.

I used long division to simplify x /(x + 1).

Now, I need to see if the improper integral of f (x) converges or diverges.

Therefore, by the limit comparison test, the improper integral

converges.

We are going to use the limit comparison test to determine the convergence or divergence of this improper integral. Again, we will use the end behavior model for the comparison.

We know from an earlier problem that the improper integral

diverges, so if the limit of this new function and the integrand of the original improper integral exists, the original improper integral will diverge. I am letting g (x) be the integrand of the original improper integral.

Therefore, the original improper integral

diverges by the limit comparison test.

I will evaluate this improper integral directly.

Let u = 1 - sin x, then du = cos x dx. Also, when x = 0, then u = 1, and when x = b, then u = 1 - sin b.

Therefore, this improper integral converges.