Improper integrals (BC only) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: improper integrals with infinite bounds, improper integrals with discontinuous integrands, convergence/divergence definitions, the p-test, the comparison test, and improper integral evaluation via limits.

You should already know: Definite integral evaluation via the Fundamental Theorem of Calculus. Limit computation at finite and infinite points. Basic integration rules for common functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Improper integrals (BC only)?

According to the AP Calculus BC Course and Exam Description (CED), this topic falls within Unit 6: Integration and Accumulation of Change, and accounts for approximately 6-8% of the total exam score. Improper integrals appear in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam, often combined with other topics like integration techniques or infinite series. By definition, an improper integral is a definite integral that fails one of the two core requirements for a standard proper definite integral: either the interval of integration is infinite, or the integrand has an infinite discontinuity (vertical asymptote) at some point in or at the endpoint of the integration interval. Since standard Riemann sums cannot be constructed for these cases, we rely on limits of proper definite integrals to evaluate them. If the limit exists and is finite, we say the improper integral converges; if the limit does not exist or is infinite, the improper integral diverges. This topic is unique to AP Calculus BC, and does not appear on the AP Calculus AB exam.

2. Improper Integrals with Infinite Bounds

Improper integrals with infinite bounds (sometimes called Type 1 improper integrals) are the most common form of improper integral you will encounter on the exam. These occur when at least one of the limits of integration is infinite, while the integrand is continuous on the entire interval of integration. By definition:

• If $f$ is continuous on $[a,\infty )$, then: $${\int }_a^{\infty }f(x)dx=\underset{b\to \infty }{\lim }{\int }_a^{b}f(x)dx$$
• If $f$ is continuous on $(-\infty ,b]$, then: $${\int }_{-\infty }^{b}f(x)dx=\underset{a\to -\infty }{\lim }{\int }_a^{b}f(x)dx$$
• If $f$ is continuous on $(-\infty ,\infty )$, split the integral at any finite point $c$, so ${\int }_{-\infty }^{\infty }f(x)dx={\int }_{-\infty }^{c}f(x)dx+{\int }_c^{\infty }f(x)dx$. The original integral converges only if both of these separate integrals converge.

The intuition behind this definition is simple: we calculate the area under the curve from $a$ to a moving bound $b$, then see what happens to that area as $b$ moves infinitely far away. If the area approaches a finite value, the integral converges; if the area grows without bound, it diverges.

Worked Example

Evaluate ${\int }_1^{\infty }\frac{1}{x^3}dx$ and state whether it converges or diverges.

1. By definition, rewrite the improper integral as a limit of proper integrals: $${\int }_1^{\infty }\frac{1}{x^3}dx=\underset{b\to \infty }{\lim }{\int }_1^b{x}^{-3}dx$$
2. Find the antiderivative of $x^{-3}$: $\int x^{-3}dx=-\frac{1}{2}{x}^{-2}+C=-\frac{1}{2x^2}+C$.
3. Evaluate the definite integral from 1 to $b$: $$\underset{b\to \infty }{\lim }\left[-\frac{1}{2b^2}-\left(-\frac{1}{2(1{)}^2}\right)\right]=\underset{b\to \infty }{\lim }\left(\frac{1}{2}-\frac{1}{2b^2}\right)$$
4. Evaluate the limit: as $b\to \infty$, $\frac{1}{2b^2}\to 0$, so the limit equals $\frac{1}{2}$.

The integral converges to $\frac{1}{2}$.

Exam tip: If you have an improper integral with infinite bounds on both ends, you must evaluate two separate limits. Writing ${\lim }_{b\to \infty }{\int }_{-b}^{b}f(x)dx$ (the Cauchy principal value) is not accepted as a test for convergence on the AP exam.

3. Improper Integrals with Discontinuous Integrands

Improper integrals with discontinuous integrands (sometimes called Type 2 improper integrals) have finite bounds of integration, but the integrand has an infinite discontinuity (vertical asymptote) at one or more points in the interval. The definition follows the same limit-based logic as Type 1, but we approach the discontinuity instead of moving a bound to infinity:

• If $f$ is continuous on $(a,b]$ and has an infinite discontinuity at $a$, then: $${\int }_a^{b}f(x)dx=\underset{c\to a^{+}}{\lim }{\int }_c^{b}f(x)dx$$
• If $f$ is continuous on $[a,b)$ and has an infinite discontinuity at $b$, then: $${\int }_a^{b}f(x)dx=\underset{c\to b^{-}}{\lim }{\int }_a^{c}f(x)dx$$
• If the discontinuity is at an interior point $c\in (a,b)$, split the integral into two improper integrals at $c$, and both must converge for the original integral to converge.

Intuition: We cannot integrate all the way up to a point where the function grows to infinity, so we integrate up to a point just next to the discontinuity, then take the limit as we get arbitrarily close. If the area approaches a finite value, the integral converges.

Worked Example

Evaluate ${\int }_0^8\frac{1}{\sqrt[3]{x}}dx$ and state whether it converges or diverges.

1. Identify the discontinuity: $\frac{1}{\sqrt[3]{x}}$ has a vertical asymptote (infinite discontinuity) at $x=0$, the left endpoint of the interval.
2. Rewrite the improper integral by definition as a right-hand limit: $${\int }_0^8\frac{1}{\sqrt[3]{x}}dx=\underset{a\to 0^{+}}{\lim }{\int }_a^8{x}^{-1/3}dx$$
3. Find the antiderivative of $x^{-1/3}$: $\int x^{-1/3}dx=\frac{3}{2}{x}^{2/3}+C$.
4. Evaluate the definite integral from $a$ to 8: $$\underset{a\to 0^{+}}{\lim }\left[\frac{3}{2}(8^{2/3})-\frac{3}{2}(a^{2/3})\right]=\underset{a\to 0^{+}}{\lim }\left(6-\frac{3}{2}{a}^{2/3}\right)$$
5. Evaluate the limit: as $a\to 0^{+}$, $a^{2/3}\to 0$, so the limit equals 6.

The integral converges to 6.

Exam tip: Always check for discontinuities in the integrand before integrating, even if the bounds are finite. AP exam questions often hide vertical asymptotes at interior points to test if you check the domain of the integrand first.

4. Convergence Testing: The p-Test and Comparison Test

On many exam questions, you do not need to calculate the exact value of an improper integral, you only need to determine if it converges or diverges. Two key tools for this are the p-test (for power function integrals) and the comparison test (for general positive integrands).

The p-test is a rule of thumb for common power function improper integrals, with two different cases for the two types of improper integrals:

1. p-test for infinite bounds ($a>0$): ${\int }_a^{\infty }\frac{1}{x^p}dx$ converges if $p>1$, and diverges if $p\le 1$. This makes sense because the function needs to decay fast enough as $x$ grows to have finite area.
2. p-test for discontinuity at 0 ($a>0$): ${\int }_0^a\frac{1}{x^p}dx$ converges if $p<1$, and diverges if $p\ge 1$. This makes sense because the function cannot blow up too fast near $x=0$ to have finite area.

The comparison test applies to positive functions: if $0\le f(x)\le g(x)$ for all $x$ in the interval of integration:

• If $\int g(x)dx$ converges, then $\int f(x)dx$ also converges.
• If $\int f(x)dx$ diverges, then $\int g(x)dx$ also diverges.

Worked Example

Without evaluating the integral, use the comparison test and p-test to determine if ${\int }_1^{\infty }\frac{3}{\sqrt{x}(x+1)}dx$ converges or diverges.

1. For all $x\ge 1$, $x+1>x$, so we can bound the integrand above: $$0<\frac{3}{\sqrt{x}(x+1)}<\frac{3}{\sqrt{x}\cdot x}=\frac{3}{x^{3/2}}$$
2. Recognize ${\int }_1^{\infty }\frac{3}{x^{3/2}}dx$ is a p-integral with $p=3/2>1$.
3. By the p-test for infinite bounds, ${\int }_1^{\infty }\frac{3}{x^{3/2}}dx$ converges.
4. By the comparison test, since our integrand is positive and smaller than a convergent integrand, the original integral converges.

Exam tip: Memorize the two opposite p-test conditions: for infinite bounds, convergence requires $p>1$; for discontinuity at 0, convergence requires $p<1$. AP exam multiple-choice distractors almost always mix these two up.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Evaluating ${\int }_{-\infty }^{\infty }xdx$ as ${\lim }_{b\to \infty }{\int }_{-b}^{b}xdx=0$ and concluding convergence. Why: Students confuse the Cauchy principal value with the formal definition of convergence for two-sided infinite improper integrals. Correct move: Always split the integral at a finite point and evaluate two separate limits. For this example, both limits diverge, so the original integral diverges.
• Wrong move: Applying the p-test to ${\int }_0^2\frac{1}{x^3}dx$ and concluding convergence because $p=3>1$. Why: Students mix up the p-test conditions for infinite bounds vs. discontinuity at 0. Correct move: Recall the p-test for discontinuity at 0: ${\int }_0^a\frac{1}{x^p}dx$ converges only if $p<1$, so this integral diverges.
• Wrong move: Integrating ${\int }_{-1}^1\frac{1}{x^2}dx$ directly with the Fundamental Theorem to get $[-1/x{]}_{-1}^1=-2$, ignoring the discontinuity at $x=0$. Why: Students forget to check for vertical asymptotes in the interior of the interval when bounds are finite. Correct move: Always check the integrand for domain restrictions and infinite discontinuities before integrating; split the integral at $x=0$ and evaluate two limits, both of which diverge here.
• Wrong move: Concluding that ${\int }_1^{\infty }\frac{1}{\sqrt{x}+1}dx$ converges because $0<\frac{1}{\sqrt{x}+1}<\frac{1}{\sqrt{x}}$ and ${\int }_1^{\infty }\frac{1}{\sqrt{x}}dx$ diverges. Why: Students misapply the comparison test rules for convergence and divergence. Correct move: Recall that if your function is smaller than a divergent integral, you can conclude nothing. For this example, bound below: $\frac{1}{\sqrt{x}+1}>\frac{1}{2\sqrt{x}}$, so compare to the divergent integral ${\int }_1^{\infty }\frac{1}{2\sqrt{x}}dx$ to conclude divergence.
• Wrong move: Adding a divergent and convergent improper integral and concluding the whole integral converges because one part converges. Why: Students incorrectly assume convergence of one part offsets divergence of another. Correct move: If any of the split limits for an improper integral does not exist or is infinite, the entire improper integral diverges, regardless of other parts.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following statements about ${\int }_0^1\frac{e^{-x}}{x^{0.8}}dx$ is true? A) The integral converges because $p=0.8<1$ by the p-test. B) The integral converges because $p=0.8>1$ by the p-test. C) The integral diverges because $p=0.8<1$ by the p-test. D) The integral diverges because $p=0.8>1$ by the p-test.

Worked Solution: First, identify the type of improper integral: the integrand has an infinite discontinuity at $x=0$, so we use the p-test for bounded intervals with a discontinuity at 0. For $0<x\le 1$, $e^{-x}\le 1$, so $\frac{e^{-x}}{x^{0.8}}\le \frac{1}{x^{0.8}}$. The p-test tells us ${\int }_0^1\frac{1}{x^p}dx$ converges if $p<1$. Here $p=0.8<1$, so ${\int }_0^1\frac{1}{x^{0.8}}dx$ converges, and by comparison the original integral converges. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=\frac{1}{(x-1)(x-3)}$. (a) Explain why ${\int }_0^{4}f(x)dx$ is an improper integral. (b) Split the integral into a sum of limits of proper integrals according to the definition of improper integrals. (c) Determine if ${\int }_0^{4}f(x)dx$ converges or diverges.

Worked Solution: (a) The integrand $f(x)=\frac{1}{(x-1)(x-3)}$ has infinite discontinuities (vertical asymptotes) at $x=1$ and $x=3$, both of which lie inside the interval of integration $[0,4]$. Any definite integral with an infinite discontinuity in the integration interval is improper by definition. (b) Split the integral at the two discontinuities, resulting in four improper integrals, each written as a limit: $${\int }_0^{4}f(x)dx=\underset{a\to 1^{-}}{\lim }{\int }_0^{a}f(x)dx+\underset{b\to 1^{+}}{\lim }{\int }_b^{2}f(x)dx+\underset{c\to 3^{-}}{\lim }{\int }_2^{c}f(x)dx+\underset{d\to 3^{+}}{\lim }{\int }_d^{4}f(x)dx$$ (c) Use partial fraction decomposition to get $f(x)=-\frac{1}{2(x-1)}+\frac{1}{2(x-3)}$. Evaluate the first limit: $$\underset{a\to 1^{-}}{\lim }{\int }_0^a\left(-\frac{1}{2(x-1)}+\frac{1}{2(x-3)}\right)dx=\underset{a\to 1^{-}}{\lim }{\left[-\frac{1}{2}\ln |x-1|+\frac{1}{2}\ln |x-3|\right]}_0^a$$ As $a\to 1^{-}$, $\ln (1-a)\to -\infty$, so $-\frac{1}{2}\ln (1-a)\to +\infty$. The first limit diverges, so the entire integral diverges.

Question 3 (Application / Real-World Style)

In astrophysics, the total work required to move a spacecraft from Earth's surface to infinitely far away is given by the improper integral: $$W={\int }_R^{\infty }\frac{GMm}{r^2}dr$$ where $G=6.67\times 10^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ (gravitational constant), $M=5.97\times 10^{24}\text{ kg}$ (Earth's mass), $m=1000\text{ kg}$ (spacecraft mass), and $R=6.37\times 10^6\text{ m}$ (Earth's radius). Calculate the total work required, giving your answer in joules (J).

Worked Solution: Rewrite the improper integral as a limit by definition: $$W=GMm\underset{b\to \infty }{\lim }{\int }_R^b{r}^{-2}dr$$ The antiderivative of $r^{-2}$ is $-r^{-1}=-\frac{1}{r}$. Evaluate the limit: $$\underset{b\to \infty }{\lim }\left(-\frac{1}{b}+\frac{1}{R}\right)=0+\frac{1}{R}=\frac{1}{R}$$ Substitute the given values: $$W=\frac{GMm}{R}=\frac{(6.67\times 10^{-11})(5.97\times 10^{24})(1000)}{6.37\times 10^6}\approx 6.25\times 10^{10}\text{ J}$$ This means the total work (escape energy) required to move a 1000 kg spacecraft infinitely far from Earth is approximately $6.25\times 10^{10}$ joules.

7. Quick Reference Cheatsheet

8. What's Next

Improper integrals are a critical prerequisite for upcoming topics in AP Calculus BC, most notably infinite series, where you will use improper integrals to apply the Integral Test for series convergence, and to analyze convergence of power series. Without understanding how to identify, test, and evaluate improper integrals, you will struggle to apply most common series convergence tests and work with power series representations of functions. This topic also connects integration and limits to solve real-world problems involving unbounded intervals or unbounded functions, which appear in applications like work, probability, and differential equations. For further study: Infinite series Integral test for series convergence Power series Applications of integration