6.13 Evaluating Improper Integrals

TLDR

An improper integral has a limit of integration at infinity or an integrand that blows up somewhere on the interval. You evaluate it by rewriting the bad boundary as a limit, integrating normally, then checking what the limit does: if it gives a finite number the integral converges, and if it goes to infinity or does not exist it diverges. This is an AP Calculus BC only topic.

Why This Matters for the AP Calculus Exam

Improper integrals show up when you need to find areas or accumulated values over an infinite interval or near a point where a function is undefined. On the AP Calculus BC exam, you may be asked to evaluate an improper integral or show that it diverges, often pulling in earlier skills like u-substitution, integration by parts, or partial fractions. Knowing how to set up the limit correctly and explain convergence or divergence is what separates full credit from partial credit on these problems.

Key Takeaways

• An improper integral has at least one infinite limit of integration or an integrand that is unbounded somewhere on the interval.
• Always rewrite the problematic boundary as a limit before integrating. Do not plug infinity in directly.
• Integrate using whatever technique fits (basic rules, substitution, partial fractions), then evaluate the limit.
• If the limit gives a finite number, the integral converges to that number. If it goes to infinity or does not exist, it diverges.
• When both limits are infinite, split the integral at a convenient point and evaluate each piece as its own limit.
• If the function is unbounded at an interior point, split the integral there too, because the trouble spot must be a boundary of a limit.

Evaluating Improper Integrals

An improper integral happens when the limits of integration involve infinity or when the function being integrated becomes unbounded within the interval. In other words, the integral cannot be evaluated directly because one of the boundaries is infinite or the function shoots off near a point.

To handle these, you go back to limits from Unit 1. Limits let you evaluate integrals where one or both boundaries are unbounded. Replace the problematic boundary with a variable, then take the limit:

$$∫^∞_0f(x)dx=\lim_{b\to∞}∫^b_0f(x)dx$$

Steps for Evaluating Improper Integrals

1. Identify the issue: Determine whether the integral has one or both limits extending to infinity, or whether the integrand is unbounded somewhere on the interval.
2. Express as a limit: Replace the infinite or unbounded boundary with a variable. For example, integrating from a to infinity becomes the limit as b approaches infinity of the integral from a to b.
3. Evaluate the integral: Integrate the function as usual and apply the Fundamental Theorem of Calculus. See The Fundamental Theorem of Calculus and Definite Integrals for a refresher.
4. Evaluate the limit: Work out what happens to the result as the boundary approaches its infinite or unbounded value.
5. Check for convergence: If the limit approaches a finite value, the integral converges. If it goes to infinity or does not exist, it diverges.

Worked Examples

Example 1: Unbounded Integrand

$$∫^1_0\frac{1}{\sqrt{x}}dx$$

Step 1) Identify the issue: The graph is unbounded at $x=0$.

Step 2) Express as a limit:

$$\lim_{a\to0}∫^1_a\frac{1}{\sqrt{x}}dx$$

Step 3) Evaluate the integral:

$$\lim_{a\to0}∫^1_ax^{-\frac{1}{2}}dx$$ $$\lim_{a\to0}[2\sqrt{x}|^1_a]$$ $$\lim_{a\to0}[2-2\sqrt{a}]$$

Step 4) Evaluate the limit:

$$\lim_{a\to0}[2-2\sqrt{a}]=2$$

Step 5) Check for convergence:

The limit approaches 2, a finite number, so the improper integral converges to 2.

Example 2: Both Limits Infinite

Find the area under $y=\frac{e^x}{4+e^{2x}}$ from $-∞$ to $∞$, or state whether it converges.

$$A=∫^∞_{-∞}\frac{e^x}{4+e^{2x}}dx$$

Since both boundaries are unbounded, split the integral into two pieces at a chosen point. Here $\ln2$ is the global maximum, and the function is symmetric about it, so the areas on each side are equal and you only need to evaluate one side and double it:

$$=2\lim_{b\to∞}∫^b_{ln2}\frac{e^x}{4+e^{2x}}dx$$

Use the substitution:

$$u=e^x$$ $$u^2=e^{2x}$$ $$2udu=2e^{2x}dx$$ $$dx=\frac{du}{e^x}$$ $$=2\lim_{b\to∞}∫^b_{ln2}\frac{1}{4+u^2}du$$ $$=2\lim_{b\to∞}[\frac{1}{2}\arctan(\frac{e^x}{2})|^b_{\ln2}]$$ $$=\lim_{b\to∞}[\arctan(\frac{e^b}{2})-\arctan(\frac{e^{\ln2}}{2})]$$ $$=\frac{π}{2}-\frac{π}{4}$$ $$=\frac{π}{4}$$

So the area of the unbounded region is $\frac{π}{4}$.

How to Use This on the AP Calculus Exam

Free Response

Improper integral questions on the BC exam often combine techniques. Set up the limit clearly, then integrate using whatever method fits the integrand. Here is a walkthrough based on a released BC free-response question.

Let $f$ be defined by

$$f(x)=\frac{3}{2x^2-7x+5}$$

Using the given identity, evaluate $∫^∞_5f(x)dx$ or show that the integral diverges.

$$\frac{3}{2x^2-7x+5}=\frac{2}{2x-5}-\frac{1}{x-1}$$

Start by rewriting the integral as a limit:

$$\lim_{b\to∞}∫^b_5f(x)dx$$

The partial fraction form is already given, so substitute it in and integrate:

$$\lim_{b\to∞}∫^b_5[\frac{2}{2x-5}-\frac{1}{x-1}]dx$$ $$\lim_{b\to\infin}[\ln|2x-5|-\ln|x-1|]^b_5$$ $$\lim_{b\to∞}[\ln|2b-5|-\ln|b-1|-\ln5+\ln4]$$

As $b\to∞$, the difference $\ln|2b-5|-\ln|b-1|$ approaches $\ln 2$, so the limit is finite and the integral converges:

$$\ln2-\ln(\frac{5}{4})=\ln(\frac{8}{5})$$

Common Trap

Watch how you combine the two log terms as $b\to∞$. Each $\ln$ term alone goes to infinity, but their difference is finite because the inside ratio approaches a constant. Treat the divergent pieces separately and you will wrongly conclude the integral diverges. Combine them into a single log before taking the limit.

Clear Exam Work

When you evaluate a definite integral by hand, show the antiderivative and include the limit notation that turns the improper integral into a proper one. Writing the limit explicitly is important for clear exam work, since it shows you understand why you cannot just plug in infinity.

Common Misconceptions

• Plugging infinity in directly. You cannot substitute infinity into an antiderivative. You must use a limit. Infinity is not a number you can evaluate.
• Convergence means the area is small. Convergence just means the limit is a finite number. A convergent improper integral can still equal a large value.
• Forgetting interior trouble spots. If the integrand is unbounded at a point inside the interval, not just at an endpoint, you must split the integral at that point and take a separate limit for each piece. Skipping this can give a wrong answer that looks fine.
• Assuming every infinite-interval integral diverges. Some converge and some diverge. For example, $∫^∞_0e^{-x}dx$ converges, while $∫^∞_1\frac{1}{x}dx$ diverges. You have to evaluate the limit to know.
• Treating divergent log pieces as the final answer. When two logarithmic terms each blow up, their difference can still be finite. Combine them before deciding whether the integral diverges.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions
• 6.5 Interpreting the Behavior of Accumulation Functions Involving Area