Applications and improper integrals

Imagine that we turn on a faucet and never turn it off again. The flow of water out of the faucet, measured in some unit of volume per unit time, can be given as a function.

In this case, we let the flow be described by:

The thing here is, after an infinite amount of time, we won't have an infinite amount of water that have come out of the faucet. And here is why.

At a small time interval, amount of water has run out of the faucet. If we sum up all of these small amounts over the infinite time interval, we get the integral:

If we let this faucet run forever, we would only end up with 1 unit of water!

Recall the form of a definite integral:

where and are the limits of the integration, and is an integrable function on the interval .

Now is an open interval, and the definite integral will calculate the area under the curve of from to . Should we not need to include the points and ? Not necessarily, but we must be careful, because the end points may make it an improper integral.

Integrals fulfilling the first condition are universally regarded as improper integrals of type 1. Likewise, those fulfilling the second one are called improper integrals of type 2.

It might seem counter intuitive at first, since improper integrals always contain infinities, that the area calculated by this type of integrals can sometimes be a finite number. If it is, we say that the integral converges.

On the other hand, if the area under the curve obtained by the integral goes to positive or negative infinity, we say that it diverges.

The term convergence is not unique to improper integrals, as we will see in future chapters, but wherever it appears it is used to describe a quantity that tend to approach some value.

In the case of improper integrals, this quantity is the area under the curve. As opposed to convergence, divergence means to grow or shrink without bound.

Considering that we are interested in whether an improper integral approach some finite value or not, it should come as no surprise that the way we evaluate it will be through limits:

Say the following integral is improper:

Let's look at how we would treat it depending on what makes it improper.

Type 1

If , then:

If , then:

Type 2

If as , then:

If as , then:

If the limit we equate the improper integral with exists as a finite number, we say it converges to that number. Otherwise, we say that it diverges.

In the case that the limit tend toward , we say that the improper integral diverges to positive or negative infinity.

As an alternative method to calculating limits, we can in some cases use the -test to determine whether an improper integral converges or diverges.

Hence, if we know that converges, must necessarily do so too. Similarly, if diverges, will diverge.

Now it turns out that the following is true about improper integrals of powers of :

For example, it seems reasonable when looking at the picture below that converges when integrated around , but not from some to the infinity:

Conversely, looking at below, it could kind of make sense that integrating from some to infinity, we get a nice convergence, whereas integration around result in divergence.

In the -test, we make use of these known results to determine whether an improper integral is convergent or divergent, given that we know the function to be greater or smaller than between the limits of integration.

This section is about integrals of functions with holes, disobediently deviant points and vertical asymptotes between the integration limits. We'll see that we can integrate them all, with some reservations.

We have seen when looking at integrability and properties of integrals that we can separate an integral into two without that it changes anything:

if

Say now that has a vertical asymptote at . To calculate the integral, we then need to split it in two, and write:

As an example, let's calculate the integral of on .

We write:

Now, the integral is perfectly symmetric around , so we can rewrite it as:

and from the p-test for integrals we know that the integral converges, as the exponent . Thus, we calculate the integral and get:

In front of you stands a function:

How do you do to integrate on ? Somewhat surprisingly, you can actually integrate as usual without caring about the points of trouble:

This is valid as long as there is a finite number of points on the finite limits of integration where is either undefined or has a function value which breaks out of the continuous curve.

So let´s go on, remove as many points as you want from any continuous function, we can still integrate as if nothing happened.

The difference between these functions and the vertical asymptote example is that there, the whole continuous curve went off towards infinity.

Here, we only have infinitely small points going there own way. They don't actually contribute with anything to the integrals: the area under one infinitely small point is indeed very small, compared to the total area.

Have a look at this funny-looking thing:

With some values of , mapping to multiple values of , the curve is certainly not a function, and it is important that we do not treat it as such.

So how do go about treating it instead? The answer is parametrization.

In mathematics, an arclength is a smooth curve connecting two points. For an arclength in two dimensions, such as the one above, we need two numbers to describe the points along it. We represent a point by , which is not an interval, but coordinates.

Instead of the being dependent on according to some function as we are used to, both of these will depend on some other variable, a parameter, usually denoted by .

Now and can be expressed as functions of , called the parametric equations, independent of each other.

The arclength parametrization will look like a system of equations, where only the points satisfying both equations will lie on the arc:

For an arc to arise from these equations, it is necessary that and are continuous functions, so that there are no gaps in the curve.

Parametrize the curve that is described by the equation:

Here, we want to express and as functions of a single paramter , that is:

This could be done by:

and the parametrization will look like:

The unit circle is described by the equation:

Now, we can describe the unit circle using only one parameter

This can be done by:

Have a look at Snakey:

You're a biologist, so you'd like to know how long Snakey is. The issue is that Snakey is sleeping. If you pick him up and stretch him out in front of a ruler, he'll try to bite you. You better find some other way to measure his length.

Hmmm. How about modeling Snakey's body as a curve? Then you can actually figure out his length. Ain't that clever?

Let's say Snakey's body is expressed by the curve , where ranges from to, I dunno, . An tiny increase in will induce a change in and . Let's call these changes and . The overall change in length is given by the Pythagorean theorem:

Now try factoring out , so that:

If you'd like the total length, you'd add up all those :s, until you've moved from Snakey's head to tail. As we decrease , we wind up with an integral. So Snakey's length can be written as:

What if Snakey's body corresponds to a function curve? Then we could write:

Snakey's length turns out to be:

where and are the relevant integral boundaries.

A hanging bridge is suspended between two points over a river. The distance between the points is 4 meters and the hanging bridge is described by the formula:

If we want to calculate the length of the bridge, we would have to calculate the arclength:

Differentiating to get , our integral becomes:

We have a curve in the -plane, what we'll do is to grab the curve and rotate it around the or axis. It may not be revolutionary, but it's pretty. We'll see how we can calculate the surface area and the volume of the object appearing.

This is some function rotated around the -axis:

Say we want to know its surface area. We'll build it up step by step. Call an infinite small element of the curve . From the lecture note on curve length, we know that:

As the circumference of a circle is , with the radius, the surface area of the thin band appearing as we rotate around a line is:

If the curve is rotated about the -axis, the corresponds to the function value , i.e. . Now, to get the whole surface of revolution, we need to sum all the thin bands.

This gives the surface of revolution as we rotate the curve of around the -axis:

If the curve is instead rotated about the -axis, the radius is just and the rest remains the same. So the surface area is:

Note that like when dealing with curve lengths, we require that the function be continuous.

Solids of revolution more or less means the volume appearing as we fill in the surface of revolution.

However, calculating a solid of revolution follows a slightly different, and, maybe surprisingly, easier method.

We'll build up the volume from thin discs. If we want to know the solid of revolution of between and , we cut up the -axis into small segments of length , depending on . We use the index to refer to the interval .

For the method to hold, we require to be continuous. Then, by the mean value theorem for integrals, there exists some -value in each interval , so that the volume of one disc is equals:

Now, if we let the number of discs go to infinity and sum all the discs, we get:

You may wonder why we don't need to bother with for solids of revolution, when we had to for the surfaces. The reason is that when dealing with the surface area, we lose too much precision if we replace by .

Say we want to calculate the surface area of a cone made up from rotating the curve . Then, each element is . That makes every thin ring that the surface area is built from times bigger using the correct instead of , and that's a big difference.

But, when approximating the whole volume, the difference between using and is negligible. Making the curve segments infinitely small will eliminate all difference between using or for the volume.

Find the volume between of the shape that is produced by rotating and around the axis, where:

The first step is to define what the area of intersection will be in our problem. To do this it can be useful to note that in the interval , as we can see below.

The next step is to define the area that each function encloses. These become and .

We find the area of intersection to be . We can easily see this in the picture below.

By multiplying with we get the volume elements . Next, we simply integrate :

Thus, the volume is .