Evaluating Improper Integrals (BC ONLY) - AP Calculus BC Study Guide

The Core Idea: Evaluating Improper Integrals (BC ONLY)

The definite integral, ${\int }_a^{b}f(x)dx$, is traditionally defined over a finite, closed interval $[a,b]$ where the function $f(x)$ is continuous. The concept of improper integrals extends the process of integration to scenarios that violate these conditions. There are two primary types of improper integrals: those with at least one infinite limit of integration (e.g., an interval of $[a,\infty )$), and those where the integrand $f(x)$ has an infinite discontinuity at some point within the interval of integration.

The fundamental technique for evaluating an improper integral is to express it as a limit. We first calculate a proper definite integral over a modified interval and then take the limit as the endpoint of that interval approaches either infinity or the point of discontinuity. The central question we answer is whether the integral converges to a finite numerical value, meaning the limit exists, or diverges, meaning the limit does not exist or approaches infinity.

Key Formulas

The evaluation of any improper integral relies on rewriting it using limit notation. The specific form depends on the nature of the impropriety.

Type 1: Infinite Limits of Integration

• If the upper limit is infinite:

  $${\int }_a^{\infty }f(x)dx=\underset{b\to \infty }{\lim }{\int }_a^{b}f(x)dx$$

• If the lower limit is infinite:

  $${\int }_{-\infty }^{b}f(x)dx=\underset{a\to -\infty }{\lim }{\int }_a^{b}f(x)dx$$

• If both limits are infinite, the integral must be split into two separate pieces at an arbitrary point $c$:

  $${\int }_{-\infty }^{\infty }f(x)dx={\int }_{-\infty }^{c}f(x)dx+{\int }_c^{\infty }f(x)dx$$

  For the original integral to converge, both resulting integrals must converge independently.

Type 2: Infinite Discontinuities on the Interval of Integration

• If $f(x)$ is discontinuous at the right endpoint $b$:

  $${\int }_a^{b}f(x)dx=\underset{c\to b^{-}}{\lim }{\int }_a^{c}f(x)dx$$

• If $f(x)$ is discontinuous at the left endpoint $a$:

  $${\int }_a^{b}f(x)dx=\underset{c\to a^{+}}{\lim }{\int }_c^{b}f(x)dx$$

• If $f(x)$ is discontinuous at an interior point $c$ where $a<c<b$:

  $${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$$

  Similar to the doubly infinite case, the original integral converges only if both resulting integrals converge independently.

Understanding Convergence vs. Divergence

The core concept that distinguishes improper integrals is the outcome of the limit evaluation. This outcome determines whether the integral converges or diverges.

• Convergence: An improper integral is said to converge if the limit used to define it exists and is a finite number. This means that the area represented by the integral, even though it may be over an infinitely long region or a region bounded by a vertical asymptote, is a finite value.

• Divergence: An improper integral is said to diverge if the limit used to define it does not exist or is infinite ($\infty$ or $-\infty$). This implies that the area represented by the integral is unbounded.

A critical rule applies whenever an integral is split into two parts, which occurs when both limits of integration are infinite or when there is a discontinuity inside the interval. For the original integral to be considered convergent, both of the component integrals must converge. If even one of the two pieces diverges, the entire original integral diverges. You cannot, for example, have one part diverge to $\infty$ and the other to $-\infty$ and conclude that they cancel out to zero.

Core Concepts & Rules

• An integral is classified as improper if its interval of integration is infinite (e.g., $[a,\infty )$) or if its integrand has an infinite discontinuity on the interval.

• The evaluation of any improper integral requires rewriting it as the limit of a proper definite integral.

• For an infinite limit of integration, replace the infinity symbol with a variable (e.g., $b$) and take the limit as that variable approaches infinity (e.g., ${\lim }_{b\to \infty }$).

• For an infinite discontinuity, approach the point of discontinuity from within the interval using a one-sided limit. For a discontinuity at $x=b$ on $[a,b]$, you would use ${\lim }_{c\to b^{-}}$.

• If an integral has two improper features (e.g., ${\int }_{-\infty }^{\infty }f(x)dx$ or ${\int }_a^{b}f(x)dx$ with a discontinuity at $c\in (a,b)$), it must be split into two separate improper integrals.

• An integral converges if its corresponding limit evaluates to a finite, real number.

• An integral diverges if its corresponding limit is infinite or does not exist.

• If an integral is split, it converges only if both parts converge. If either part diverges, the original integral diverges.

Step-by-Step Example 1: Infinite Limit of Integration

Problem: Evaluate ${\int }_2^{\infty }\frac{6}{x^3}dx$.

Step 1: Identify the impropriety and rewrite as a limit.

The integral is improper because the upper limit of integration is $\infty$. We replace $\infty$ with a variable, $b$, and take the limit as $b\to \infty$.

$${\int }_2^{\infty }\frac{6}{x^3}dx=\underset{b\to \infty }{\lim }{\int }_2^{b}6x^{-3}dx$$

Step 2: Find the antiderivative of the integrand.

$$\int 6x^{-3}dx=6\left(\frac{x^{-2}}{-2}\right)=-3x^{-2}=-\frac{3}{x^2}$$

Step 3: Evaluate the definite integral within the limit.

We apply the Fundamental Theorem of Calculus to the integral from $2$ to $b$.

$$\underset{b\to \infty }{\lim }{\left[-\frac{3}{x^2}\right]}_2^b=\underset{b\to \infty }{\lim }\left(\left(-\frac{3}{b^2}\right)-\left(-\frac{3}{2^2}\right)\right)=\underset{b\to \infty }{\lim }\left(-\frac{3}{b^2}+\frac{3}{4}\right)$$

Step 4: Evaluate the limit.

As $b\to \infty$, the term $-\frac{3}{b^2}$ approaches 0.

$$\underset{b\to \infty }{\lim }\left(-\frac{3}{b^2}+\frac{3}{4}\right)=0+\frac{3}{4}=\frac{3}{4}$$

Step 5: State the conclusion.

Because the limit exists and is a finite number, the integral converges.

$${\int }_2^{\infty }\frac{6}{x^3}dx\text{ converges to }\frac{3}{4}.$$

Step-by-Step Example 2: Exam-Style Application (Internal Discontinuity)

Problem: Determine if ${\int }_{-1}^1\frac{1}{x^2}dx$ converges or diverges. If it converges, find its value.

Step 1: Identify the impropriety.

A common mistake is to not see the impropriety. The integrand $f(x)=\frac{1}{x^2}$ has an infinite discontinuity at $x=0$, which is inside the interval of integration $[-1,1]$.

Step 2: Split the integral at the point of discontinuity.

We must split the integral into two separate improper integrals at $x=0$.

$${\int }_{-1}^1\frac{1}{x^2}dx={\int }_{-1}^0\frac{1}{x^2}dx+{\int }_0^1\frac{1}{x^2}dx$$

The original integral converges only if both of these new integrals converge.

Step 3: Evaluate the first integral using a limit.

We will evaluate the first integral, ${\int }_{-1}^0\frac{1}{x^2}dx$. The discontinuity is at the right endpoint, so we use a left-sided limit.

$${\int }_{-1}^0\frac{1}{x^2}dx=\underset{c\to 0^{-}}{\lim }{\int }_{-1}^c{x}^{-2}dx$$

The antiderivative of $x^{-2}$ is $-x^{-1}=-\frac{1}{x}$.

$$\underset{c\to 0^{-}}{\lim }{\left[-\frac{1}{x}\right]}_{-1}^c=\underset{c\to 0^{-}}{\lim }\left(\left(-\frac{1}{c}\right)-\left(-\frac{1}{-1}\right)\right)=\underset{c\to 0^{-}}{\lim }\left(-\frac{1}{c}-1\right)$$

Step 4: Evaluate the limit.

As $c$ approaches $0$ from the left (i.e., through small negative numbers), the term $-\frac{1}{c}$ approaches $+\infty$.

$$\underset{c\to 0^{-}}{\lim }\left(-\frac{1}{c}-1\right)=\infty$$

Step 5: State the conclusion.

Because the first integral, ${\int }_{-1}^0\frac{1}{x^2}dx$, diverges to $\infty$, we can immediately conclude that the original integral, ${\int }_{-1}^1\frac{1}{x^2}dx$, diverges. There is no need to evaluate the second integral.

Using Your Calculator

The evaluation of improper integrals is an analytical process that requires you to show the correct limit setup on the AP Exam. A graphing calculator cannot perform this symbolic step, but it can be an excellent tool for checking your final answer.

To approximate an improper integral, you can use the numerical integration function (e.g., fnInt on a TI-84).

• For an infinite limit of integration: You cannot enter \infty into the calculator. Instead, use a very large number as a substitute for the infinite bound. For example, to check \int_{2}^{\infty} \frac{6}{x^3} \, dx, you could calculate $fnInt(6/X^3,X,2,10000)$. The result should be very close to the analytical answer of $0.75$.

• For an infinite discontinuity: You cannot enter the point of discontinuity as a bound. Instead, use a number extremely close to it, staying within the interval. For example, to check the convergence of ${\int }_0^1\frac{1}{x^2}dx$ (from the second part of Example 2), you could calculate fnInt(1/X^2, X, 0.0001, 1). The large result would suggest divergence.

Important: This method only provides a numerical approximation. It is not a substitute for the required analytical work on a free-response question. It is best used to confirm if your answer is reasonable or to quickly determine convergence/divergence on a multiple-choice question.

AP Exam Quick Hit

Common Question Types

• Direct Evaluation with an Infinite Bound: You will be asked to evaluate an integral like ${\int }_1^{\infty }{e}^{-x}dx$ or show that it diverges. This requires a clean setup of the limit, correct antidifferentiation, and proper limit evaluation.

• Direct Evaluation with a Discontinuity: You will be asked to evaluate an integral like ${\int }_0^8\frac{1}{\sqrt[3]{x}}dx$, where the impropriety is a vertical asymptote at one of the endpoints.

• Evaluation Requiring Splitting: You will be given an integral like ${\int }_0^4\frac{1}{x-2}dx$ where the discontinuity is in the middle of the interval, or an integral like ${\int }_{-\infty }^{\infty }\frac{1}{1+x^2}dx$. Success requires recognizing the need to split the integral into two parts and evaluating each one separately.

Common Mistakes

• Forgetting Limit Notation: Students often write ${\left[F(x)\right]}_a^{\infty }$ and attempt to "plug in" infinity. This is mathematically incorrect and will lose credit. You must use the formal ${\lim }_{b\to \infty }$ notation.

• Ignoring Internal Discontinuities: The most common trap is failing to notice a vertical asymptote within the interval of integration. Evaluating ${\int }_{-2}^2\frac{1}{x}dx$ directly without splitting it at $x=0$ will lead to an incorrect answer. Always check the integrand for discontinuities between the limits of integration.

• Incorrect Limit Evaluation: Errors in evaluating limits as a variable approaches infinity, especially for logarithmic, exponential, or rational functions. For example, incorrectly stating that ${\lim }_{b\to \infty }\arctan (b)=\infty$ (it is $\frac{\pi }{2}$).

• Errors in Antidifferentiation: Simple mistakes in finding the antiderivative before the limit process even begins will lead to a completely incorrect result.

• Handling Split Integrals Incorrectly: If an integral is split into two parts, and one part evaluates to $\infty$ while the other evaluates to $-\infty$, concluding that the integral converges to 0. This is incorrect; if any part of a split integral diverges, the entire integral diverges.