Lesson 10: Improper Integration

Hello! This lesson will show our first application of integration. We might need to use any of the tools in our integration techniques toolbox to evaluate the integrals in this lesson.

Learning goals:

Evaluate an integral over an infinite interval.
Evaluate an integral over a closed interval containing an infinite discontinuity.
Use the comparison theorem to determine whether a definite integral is convergent.

Topic:

Volume 2, Section 3.7 Improper Integration (link to textbook section)

WeBWorK:

Integration – Improper Integrals

Motivating question

How can we use integrals to determine the area of an unbounded region?

Warmup exercise 1

Evaluate:

$\int_0^b \frac{2}{(x+2)^3}dx.$

Hint: since the upper limit of integration is the variable $b$ instead of a number, your answer will be a formula in terms of $b$ .

Show answer 1

We could evaluate this integral using direct substitution with $u=x+2$ , but since $du = dx$ , this substitution is straightforward enough that we can probably integrate directly (if it’s not obvious to you how to integrate this directly, definitely integrate by substitution; it’s not wrong, it’ll just take more writing).

$\int_0^b \frac{2}{(x+2)^3}dx = \int_0^b 2(x+2)^{-3}dx$

$= \left.\frac{2}{(-3+1)}(x+2)^{-3+1} \right|_0^b$

$=\left.\frac{-1}{(x+2)^2} \right|_0^b$

$= \frac{-1}{(b+2)^2} - \frac{-1}{(0+2)^2}$

$=\frac{-1}{(b+2)^2} + \frac{1}{4}.$

Warmup exercise 2

Evaluate the limit:

$\lim_{b \to \infty} \left(\frac{-1}{(b+2)^2} + \frac{1}{4}\right).$

Show answer 2

The important limit for us to remember is $\lim_{x \to \infty} \frac{1}{x} = 0$ .

So our limit:

$\lim_{b \to \infty} \left(\frac{-1}{(b+2)^2} + \frac{1}{4}\right) = 0 + \frac{1}{4} = \frac{1}{4}.$

Improper integrals…what do they mean?

Motivating example: You may have guessed by now that we can put Warmup exercises 1 and 2 together. Doing so looks like:

$\int_0^\infty \frac{2}{(x+2)^3}dx = \lim_{b \to \infty} \int_0^b \frac{2}{(x+2)^3}dx$

$=\lim_{b \to \infty} \frac{-1}{(b+2)^2} + \frac{1}{4}$

$=\frac{1}{4}.$

That is, we can write

$\int_0^\infty \frac{2}{(x+2)^3}dx = \frac{1}{4}$

But what does this even mean???

Recall from Lesson 2 (link here) that we defined the definite integral $\int_a^b f(x) dx$ as a signed area between the vertical lines $x=a$ and $x=b$ and between the graph $y=f(x)$ and the $x-$ axis.

In our motivating example, $f(x) = \frac{2}{(x+2)^3}$ , $a=0$ , and…well, we’d like to say that $b = \infty$ but since $\infty$ is not a real number, that doesn’t quite make sense (this is why the integral is called improper). But $\infty$ is a limit of real numbers, at the right end of the real number line. So we imagine a region bounded on the right by the vertical line $x=b$ and then taking a limit as $b$ approaches $\infty.$

Click on this interactive Desmos graph (link here) to get a picture of what this looks like for the motivating example; drag the slider $b$ to get larger and larger values of $b$ (Desmos can’t actually let $b$ go to infinity, but it can let $b$ go to 1,000,000, which is still pretty big even if it’s not infinite!).

So we can see that the symbol $\int_0^\infty \frac{2}{(x+2)^3}dx$ still represents a signed area, but now the region is unbounded (there is no finite boundary on the right side of the region).

This is remarkable!!!

We have seen that the integral $\int_0^\infty \frac{2}{(x+2)^3}dx$ represents the signed area of an unbounded region. With this in mind, our conclusion above that $\int_0^\infty \frac{2}{(x+2)^3}dx = \frac{1}{4}$ might seem alarming at first: the region is unbounded but it’s area is…finite??? Yes! It’s true! An unbounded (think: infinite) region can have finite area! Figures 1 and 2 below both show the graph of $\frac{2}{(x+2)^3}$ ; Figure 1 is zoomed in and shows a small part of the graph, and Figure 1 is zoomed out and shows a larger part of the graph. What we notice is that the blue region, whose area is represented by the improper integral above, gets thinner and thinner as we look further and further to the right (the function has the $x-$ axis as a horizontal asymptote as $x$ approaches $\infty$ ). What our calculation above shows is that the blue strip gets thin enough fast enough that its area is actually pretty small!

*Figure 2: Graph of $\frac{2}{(x+2)^3}$ , zoomed out*

More precisely, we say that the improper integral $\int_0^\infty \frac{2}{(x+2)^3}dx$ , which represents the area of the unbounded blue region, converges to $\frac{1}{4}$ . This is because the limit from Warmup exercise 2 exists; if it didn’t exist, we’d say our improper integral diverges.

Two types of improper integrals

There are two types of unbounded regions whose areas we can compute using improper integrals:

Type 1: one or the other (or both) of the limits of integration are infinite (like in our motivating example above).
Type 2: both limits of integration are finite, but the function is unbounded (a.k.a. has a vertical asymptote) between the limits of integration.

Improper integrals are limits

For both types, we must set up the improper integral as a limit of definite integrals:

Definition of improper integrals of Type 1:

$\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$
$\int_{-\infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x) dx$
$\int_{-\infty}^\infty f(x) dx = \int_{-\infty}^b f(x) dx + \int_b^\infty f(x)dx$

Improper integrals of Type 1 are easier to recognize because at least one limit of integration is $\pm \infty$ . Improper integrals of Type two are a bit harder to recognize because they look like regular definite integrals unless you check for vertical asymptotes between the limits of integration.

Definition of improper integrals of Type 2:

If the function $f(x)$ has a vertical asymptote at $x=a$ , then $\int_a^b f(x)dx = \lim_{t \to a^+}\int_t^b f(x)dx$
If the function $f(x)$ has a vertical asymptote at $x=b$ , then $\int_a^b f(x)dx = \lim_{t \to b^-}\int_a^t f(x)dx$
If the function $f(x)$ has a vertical asymptote at $x=c$ where $a <c< b$ , then $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$ .

Examples

Video 1 below summarizes the theory we discussed above. It also shows examples of improper integrals of Type 1 that converge or diverge for different reasons:

improper integrals that converge (like our motivating example above),
improper integrals that diverge because the function does not have a horizontal asymptote,
improper integrals that diverge even though the function has a horizontal asymptote (the thin strip doesn’t get thin enough fast enough for the integral to converge).

Video 1 also shows an application of improper integrals to finance.

]Video 1

Videos 2 to 5 below show more examples of improper integrals of Type 1.

Video 2

Video 3

Video 4

Video 5

Videos 6 and 7 below show more examples of improper integrals of Type 2.

Video 6

Video 7

Comparing with other improper integrals

If we need to know only whether an improper integral converges or diverges, but not what if it converges to (if it converges) we have another tool at our disposal. We’ll use this type of comparison again in Lesson 16.

Video 8 below provides us with a much needed musical interlude. Outkast’s 2003 classic hit Hey Ya includes one line in particular that will be relevant for this part of our lesson. Can you guess what the relevant line is? Hint: it’s after the 3 minute mark of the song.

Video 8

Click to see the relevant line from the song

Now, what’s cooler than being cool? (Ice Cold!)
I can’t hear ya
I say what’s, what’s cooler than being cool? (Ice Cold!)

Alright, alright, alright, alright, alright, alright, alright, alright, let’s see why this line matters to us now. Figure 3 below shows two blobs (it doesn’t matter at this point what the blobs represent; they’re just blobs for now). The smaller one is labeled $A$ and shaded dark grey and the larger one is labeled $B$ and is shaded light grey. The $A$ blob sits inside the $B$ blob.

All we care about is whether the blobs are big or small. We don’t have a scale, so all we have to go on is how the blobs compare to one another. Here are two claims we can make:

Let’s assume that blob $A$ is big. Well, we know that blob $B$ is even bigger than blob $A$ . So we can conclude that blob $B$ must also be big. (What’s bigger than being big? (Big!))
Let’s assume that blob $B$ is small. Well, we know that blob $A$ is even smaller than blob $B$ . So we can conclude that blob $A$ must also be small. (What’s smaller than being small? (Small!))

Notice that if we assume that blob $A$ is small, we don’t know whether blob $B$ is big or small. Similarly, if we assume that blob $B$ is big, we don’t know whether blob $A$ is big or small. So these are unhelpful comparisons.

What does this have to do with improper integrals?

We’ll translate the two relationships above into one about improper integrals of Type 1 (there is such a comparison for improper integrals of Type 2 as well). Now $A$ and $B$ will represent improper integrals of two different functions.

Theorem: Let $f(x)$ and $g(x)$ be functions defined on $[a, \infty)$ for some real number $a$ . Assume $0 \leq g(x) \leq f(x)$ (this means that $g(x)$ is playing the role of blob $B$ and $f(x)$ is playing the role of blob $A$ ).

Assume that the improper integral $\int_a^\infty g(x)dx$ diverges. Well, we know that $f(x) \geq g(x)$ . So we can conclude that the improper integral $\int_a^\infty f(x)$ also diverges. (What’s bigger than divergent? (Divergent!))
Assume that the improper integral $\int_a^\infty f(x)dx$ converges. Well, we know that $g(x) \leq f(x)$ . So we can conclude that the improper integral $\int_a^\infty g(x)$ also converges. (What’s smaller than convergent? (Convergent!))

Examples of comparison

To use comparison to determine whether an improper integral converges or diverges, the key is to finding an appropriate second integral to compare it to.

Video 9 below walks us through the ideas above and applies them to an improper integral of Type 1.

Video 9

Video 10 below walks us through a comparison for an improper integral of Type 2.

Video 10

Additional Video Resources

Improper Integration

Exit ticket

Evaluate:

$\int_0^5 \frac{5}{\sqrt[5]{x+5}}dx$