# How to Solve Improper Integrals Video

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• 0:12 How Do We Get an…
• 2:18 What an Improper…
• 6:14 Solving an Improper Integral
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Lesson Transcript
Instructor: Kelly Sjol
What does it mean when an integral has limits at infinity? These integrals are 'improper!' In this lesson, learn how to treat infinity as we study the so-called improper integrals.

## How Do We Get an Improper Integral?

Have you ever thrown darts at a dartboard? Maybe you're good and you get really close to the bull's-eye. Or maybe you're more like me and throw them way off course. It's almost like you let go of the dart too soon, and it ends up behind you.

If you look carefully at all of the holes created by people's throws at a dartboard, you might see a distribution where there are a lot more close to the center. But there are also dart holes really far away from the center, perhaps even in the bartender's rear end!

Now let's say you graph the number of darts thrown with respect to how far away to the left or right they are from the center of the bull's-eye. You might end up with a graph that looks a little something like this. So you've got f(x), where x is the horizontal distance to the center of the dartboard, and y, or f(x) here, is the number of darts thrown at that particular location.

So you've got y=f(x) and f(x)= 1 / (1+x^2). You've got a lot of darts that are thrown close to the center of the dartboard and fewer and fewer darts the further out you go.

If you want to get an idea of how many darts were thrown in total, you want to know the area underneath this curve. And we know, from all the calculus that we have learned, that the area under the curve is equal to the integral of that region.

Here, the integral has bounds of minus infinity to infinity. If you're throwing at a dartboard, sure, you're going to get really close to the dartboard because you're probably much better at darts than I am. But if I throw at the dartboard, I might end up an infinite distance to the left of the dartboard or an infinite distance to the right of the dartboard. I mean, I've got a pretty strong arm; I'm just not very accurate.

## What an Improper Integral Looks Like

So the area under this curve is the integral from minus infinity to infinity of our integrand, 1 / (1+x^2) * dx. Now this is what we call an improper integral. Here, something goes to infinity. When I say something goes to infinity, that might mean that one of the limits goes to infinity, or possibly, the value of the integrand goes to plus or minus infinity somewhere in this region. Whenever you see integrals that look like this, you're going to replace the infinity with some variable like, say, b, and you're going to take the limit of this integral as b goes to infinity.

So let's actually look at our dartboard problem. Our integral goes from minus infinity to infinity of 1 / (1+x^2) * dx. The first thing I'm going to do is I'm going to divide this into two separate integrals: one from 0 to infinity, and one from minus infinity to 0. So here, I'm going to treat the right-hand side of this region and the left-hand side of this region.

Both of these integrals look like 1 / (1+x^2) * dx. So let's take a minute to consider this integral with no limits and look at the indefinite integral of 1 / (1+x^2) * dx. Because this is 1+x^2, we might want to use a trig substitution. I'll just tell you, use the substitution x equals the tangent of theta. If that's true, then dx is the secant squared of theta d theta. With these substitutions, our integral becomes 1 / (1+tan^2 of theta) (because x^2 becomes tan^2) times the sec^2 theta d theta, where sec^2 theta d theta is equal to dx based on our substitution.

Now at this point, the integral doesn't look any easier; in fact, it looks harder because now you've got secants and tangents rather than just x. But let's go back for a minute to trigonometry. If I have cosine squared of theta plus sine squared of theta equals 1, that's a very standard trig identity that you should all remember. If I take that equation and divide each term by cos^2 of theta, then the first term becomes 1. The second term - which is sin^2 / cos^2 - is the same thing as tan^2, because the tangent of an angle equals the sine of the angle divided by the cosine of the angle. The term on the right side, 1, when I divide it by cos^2 of theta, well that's 1 over the cosine of theta, all squared, which is the definition of sec^2.

So I know that 1 + tan^2 of theta=sec^2 of theta. That means that this entire bit here simplifies to just sec^2 of theta. So if I plug this in, I get 1 / (sec^2 of theta) * sec^2 of theta d theta. These sec^2 of thetas cancel out, and I end up with the integral of d theta. Well that's just theta + some constant C. I can now plug in my substitution to make this final integral theta + C, so I can put it in terms of x. To do that, let's solve x=tan(theta) for theta. So I take the inverse tangent, the arctan, of both sides and I end up with the arctan of x = theta. So my integral, 1 / (1+x^2) * dx = arctan x + C.

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