Improper Integrals

Improper Integrals

Part 1 of 7 — Infinite Limits of Integration (Type I)

An improper integral has either an infinite limit of integration or an integrand with a discontinuity. This part covers Type I — integrals to $\pm\infty$.

Definition of Type I Improper Integrals

$\boxed{\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^t f(x)\,dx}$

$\int_{-\infty}^{b} f(x)\,dx = \lim_{t \to -\infty} \int_t^b f(x)\,dx$

$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_c^{\infty} f(x)\,dx$

Key Fact: For $\int_{-\infty}^{\infty}$, split at any convenient point $c$ (often $c = 0$). BOTH halves must converge for the whole integral to converge.

Worked Example: $\int_1^{\infty} \frac{1}{x^2}\,dx$

$\boxed{\int_1^{\infty} \frac{1}{x^2}\,dx = 1 \quad (\text{converges})}$

Contrast: $\int_1^{\infty} \frac{1}{x}\,dx$

$\lim_{t \to \infty} [\ln t - \ln 1] = \lim_{t \to \infty} \ln t = \infty$

$\int_1^{\infty} \frac{1}{x}\,dx = \infty \quad (\text{diverges})$

AP Tip: This contrast ($1/x^2$ converges, $1/x$ diverges) is fundamental. The “border” between convergence and divergence is explored in Part 5 with the p-integral test.

Type I Practice

Convergence or Divergence?

Evaluate an Improper Integral

Key Takeaways — Part 1

Coming Up: Part 2 covers Type II improper integrals — when the integrand has a discontinuity within the interval.

Improper Integrals

Part 2 of 7 — Discontinuous Integrands (Type II)

Type II improper integrals have a vertical asymptote or discontinuity within $[a,b]$. The integrand “blows up” at one or more points.

Type II Definitions

Discontinuity at $x = b$ (right endpoint): $\int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx$

Discontinuity at $x = a$ (left endpoint): $\int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx$

Discontinuity at $x = c$ inside $(a,b)$: $\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$

Key Fact: If the discontinuity is INSIDE the interval, you MUST split the integral. Computing the antiderivative across the discontinuity gives a wrong answer. This is a classic AP exam trap.

Worked Example: $\int_0^1 \frac{1}{\sqrt{x}}\,dx$

The integrand $\frac{1}{\sqrt{x}}$ is undefined at $x = 0$ (left endpoint).

$\boxed{\int_0^1 \frac{1}{\sqrt{x}}\,dx = 2 \quad (\text{converges})}$

Contrast: $\int_0^1 \frac{1}{x}\,dx$

$\lim_{t \to 0^+} [\ln x]_t^1 = 0 - \ln t = -\lim_{t \to 0^+} \ln t = \infty$

$\int_0^1 \frac{1}{x}\,dx \quad \text{diverges}$

Interior Discontinuity: $\int_{-1}^{1} \frac{1}{x^2}\,dx$

$\frac{1}{x^2}$ has a vertical asymptote at $x = 0$, which is INSIDE $[-1, 1]$.

Wrong approach: Ignoring the discontinuity gives $[-\frac{1}{x}]_{-1}^1 = -1 - 1 = -2$. This is negative, but $\frac{1}{x^2} > 0$ everywhere — clearly wrong!

Correct approach: Split at $x = 0$: $\int_{-1}^{0} \frac{1}{x^2}\,dx + \int_0^1 \frac{1}{x^2}\,dx$

Both halves diverge ($\lim_{t \to 0} \frac{1}{t} = \infty$), so the integral diverges.

AP Tip: Always scan for discontinuities before integrating. If $f(x) \to \infty$ anywhere in $[a,b]$, the integral is improper.

Type II Practice

Identify the Type

Type II Computation

Key Takeaways — Part 2

Coming Up: Part 3 develops convergence vs. divergence criteria — when can you tell without computing?

Improper Integrals

Part 3 of 7 — Convergence vs. Divergence

Not every improper integral needs full computation. Key tests let you determine convergence or divergence quickly. This part builds your toolkit.

The $p$-Integral (Most Important Test)

$\boxed{\int_1^{\infty} \frac{1}{x^p}\,dx \quad \begin{cases} \text{converges to } \frac{1}{p-1} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \le 1 \end{cases}}$

Memory Device: $p = 1$ is the boundary — converges above, diverges at and below. Think “$p > 1$ wins.”

Type II Version of the $p$-Test

$\boxed{\int_0^1 \frac{1}{x^p}\,dx \quad \begin{cases} \text{converges to } \frac{1}{1-p} & \text{if } p < 1 \\ \text{diverges} & \text{if } p \ge 1 \end{cases}}$

The rule flips! For Type I ($\int_1^{\infty}$): converges if $p > 1$. For Type II ($\int_0^1$): converges if $p < 1$.

AP Tip: The AP exam loves testing whether students confuse the two $p$-test directions. Always identify the type first.

Convergence Quick Checks

Behavior Near the Boundary

The case $p = 1$ is special:

• $\int_1^{\infty} \frac{1}{x}\,dx = \lim_{t \to \infty} \ln t = \infty$ (diverges, but very slowly)
• $\int_1^{\infty} \frac{1}{x^{1.001}}\,dx = \frac{1}{0.001} = 1000$ (converges, but large)

This shows the $p$-test has a sharp threshold: the tiniest push above $p = 1$ makes the integral converge.

Exponential vs. Polynomial Decay

$\boxed{\text{Exponential decay} \gg \text{polynomial decay} \gg \text{logarithmic decay}}$

Classify Each Integral

$p$-Test Application

Key Takeaways — Part 3

Coming Up: Part 4 introduces the Comparison Test — compare unknown integrals to known ones.

Improper Integrals

Part 4 of 7 — Comparison Tests

When you can’t find an antiderivative, comparison tests let you determine convergence by comparing to a simpler integral whose behavior you already know.

Direct Comparison Test (DCT)

For $f(x) \ge 0$ and $g(x) \ge 0$ on $[a, \infty)$:

$\boxed{\text{If } 0 \le f(x) \le g(x) \text{ and } \int_a^{\infty} g(x)\,dx \text{ converges, then } \int_a^{\infty} f(x)\,dx \text{ converges.}}$

$\boxed{\text{If } f(x) \ge g(x) \ge 0 \text{ and } \int_a^{\infty} g(x)\,dx \text{ diverges, then } \int_a^{\infty} f(x)\,dx \text{ diverges.}}$

Think of it as:

• Smaller than a convergent → convergent (“trapped below a ceiling”)
• Bigger than a divergent → divergent (“pushed above a floor”)

Example: $\int_1^{\infty} \frac{1}{x^2 + 1}\,dx$

We know $x^2 + 1 > x^2$ for all $x$, so: $\frac{1}{x^2 + 1} < \frac{1}{x^2}$

Since $\int_1^{\infty} \frac{1}{x^2}\,dx$ converges ($p = 2 > 1$):

$\int_1^{\infty} \frac{1}{x^2 + 1}\,dx \quad \text{converges by DCT.}$

Example: $\int_1^{\infty} \frac{1}{\sqrt{x} - 0.5}\,dx$

For $x \ge 1$: $\sqrt{x} - 0.5 \le \sqrt{x}$, so $\frac{1}{\sqrt{x} - 0.5} \ge \frac{1}{\sqrt{x}}$.

Since $\int_1^{\infty} \frac{1}{\sqrt{x}}\,dx$ diverges ($p = 1/2 \le 1$):

$\int_1^{\infty} \frac{1}{\sqrt{x} - 0.5}\,dx \quad \text{diverges by DCT.}$

Limit Comparison Test (LCT)

When direct inequality is hard to establish, use:

$\boxed{\text{If } \lim_{x \to \infty} \frac{f(x)}{g(x)} = L, \quad 0 < L < \infty, \text{ then } \int f \text{ and } \int g \text{ both converge or both diverge.}}$

Why this works: If the ratio approaches a finite nonzero constant, the functions decay at the same rate.

Example: $\int_1^{\infty} \frac{x}{x^3 + 5}\,dx$

Compare with $g(x) = \frac{1}{x^2}$ (since $\frac{x}{x^3} = \frac{1}{x^2}$ for large $x$):

$\lim_{x \to \infty} \frac{x/(x^3+5)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3 + 5} = 1$

Since $L = 1 \in (0, \infty)$ and $\int_1^{\infty} \frac{1}{x^2}\,dx$ converges, so does $\int_1^{\infty} \frac{x}{x^3+5}\,dx$.

AP Tip: The LCT is often the fastest approach on the AP exam. Identify the dominant terms and compare to $1/x^p$.

Comparison Test Practice

Choose the Right Test

Limit Comparison Computation

Key Takeaways — Part 4

Strategy: Identify the dominant terms as $x \to \infty$, form $\frac{1}{x^p}$, and apply LCT. If $p > 1$, converges. If $p \le 1$, diverges.

Coming Up: Part 5 covers special convergence results and the interplay between $p$-integrals and comparison tests on the AP exam.

Improper Integrals

Part 5 of 7 — Special Results & AP Exam Patterns

This part consolidates key results that appear repeatedly on the AP Calculus BC exam and prepares you for free-response improper integral problems.

Gallery of Important Improper Integrals

$\boxed{\int_0^{\infty} xe^{-x}\,dx = 1 \quad \text{and} \quad \int_{-\infty}^{\infty} \frac{dx}{1+x^2} = \pi}$

AP Tip: Memorize the arctan integral: $\int_0^{\infty} \frac{1}{1+x^2}\,dx = \frac{\pi}{2}$. It appears in multiple-choice questions regularly.

The Gamma Function Connection

$\Gamma(n+1) = \int_0^{\infty} x^n e^{-x}\,dx = n!$

While $\Gamma$ is beyond the AP exam, knowing $\int_0^{\infty} xe^{-x}\,dx = 1$ via IBP is testable.

Derivation: $u = x$, $dv = e^{-x}dx$: $\int_0^{\infty} xe^{-x}\,dx = [-xe^{-x}]_0^{\infty} + \int_0^{\infty} e^{-x}\,dx = 0 + 1 = 1$

AP-Style Quick Checks

AP Free-Response Pattern

A typical AP FRQ might say:

Let $R$ be the unbounded region between $y = \frac{1}{x^2}$ and the $x$-axis for $x \ge 1$. (a) Find the area of $R$. (b) Find the volume when $R$ is revolved about the $x$-axis. (c) Set up, but do not evaluate, the volume when $R$ is revolved about the $y$-axis.

Solutions:

• (a) $A = \int_1^{\infty} \frac{1}{x^2}\,dx = 1$ (finite area)
• (b) $V = \pi \int_1^{\infty} \frac{1}{x^4}\,dx = \pi \cdot \frac{1}{3}$ (finite volume)
• (c) $V = 2\pi \int_1^{\infty} x \cdot \frac{1}{x^2}\,dx = 2\pi \int_1^{\infty} \frac{1}{x}\,dx$ (diverges!)

$\boxed{\text{Gabriel’s Horn: finite volume but infinite surface area!}}$

FRQ-Style Classification

Integration by Parts with Improper Integral

Key Takeaways — Part 5

Must-Know Integrals:

• $\int_0^{\infty} e^{-kx}\,dx = 1/k$
• $\int_0^{\infty} xe^{-x}\,dx = 1$
• $\int_0^{\infty} \frac{dx}{1+x^2} = \pi/2$

Gabriel’s Horn: $y = 1/x$ for $x \ge 1$ revolved about the $x$-axis has finite volume ($\pi$) but infinite surface area.

Coming Up: Part 6 is a Problem-Solving Workshop with mixed practice.

Improper Integrals

Part 6 of 7 — Problem-Solving Workshop

This part is a mixed-practice workshop. Every problem integrates concepts from Parts 1–5. Work through each carefully before checking answers.

Warm-Up: Quick Classification

Classify each integral before computing:

Key Lesson: Always check the interval boundaries carefully. A function can be discontinuous at a point, but if that point isn’t in $[a,b]$, the integral is proper (at least regarding that point).

Workshop Problem Set 1

Decision Flowchart

$\text{Is the interval infinite?} \xrightarrow{\text{Yes}} \text{Type I}$ $\text{Is }f\text{ undefined at any point in }[a,b]\text{?} \xrightarrow{\text{Yes}} \text{Type II}$ $\text{Can you find the antiderivative?} \xrightarrow{\text{Yes}} \text{Compute directly}$ $\xrightarrow{\text{No}} \text{Use Comparison (DCT or LCT)}$

Direct Computation Steps:

1. Replace the problematic bound with a limit variable
2. Compute the definite integral
3. Evaluate the limit
4. State convergent (with value) or divergent

Workshop Problem Set 2

Challenge: Mixed Type

$\int_0^{\infty} \frac{1}{\sqrt{x}(1+x)}\,dx$

This has BOTH issues: discontinuity at $x=0$ (Type II) and $\infty$ upper limit (Type I).

Split at $x = 1$:

Part A: $\int_0^1 \frac{1}{\sqrt{x}(1+x)}\,dx$ — Type II at $x=0$

Near $x = 0$: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x}}$. Since $\int_0^1 x^{-1/2}\,dx$ converges ($p = 1/2 < 1$), Part A converges.

Part B: $\int_1^{\infty} \frac{1}{\sqrt{x}(1+x)}\,dx$ — Type I

For large $x$: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{x^{3/2}}$. Since $\int_1^{\infty} x^{-3/2}\,dx$ converges ($p = 3/2 > 1$), Part B converges.

Full integral converges. (Its exact value is $\pi$, via the substitution $u = \sqrt{x}$.)

Workshop Computation

Workshop Recap

Common Mistakes to Avoid:

1. Forgetting to check for interior discontinuities before integrating
2. Confusing Type I and Type II $p$-test directions
3. Applying comparison tests with the wrong inequality direction
4. Not splitting integrals that have both Type I and Type II issues

Coming Up: Part 7 is a Comprehensive Review covering all improper integral concepts.

Improper Integrals

Part 7 of 7 — Comprehensive Review

This final part reviews every concept from the improper integrals unit. Treat it as an AP exam simulation.

Complete Reference Table

$\boxed{\text{Type I: } p > 1 \text{ converges} \qquad \text{Type II: } p < 1 \text{ converges}}$

Review Problem Set

Classification Review

AP FRQ Practice

Problem: The region $R$ is bounded by $y = \frac{1}{x^2}$, the $x$-axis, and $x = 1$.

(a) Show the area of $R$ is finite: $A = \int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{t \to \infty} [-\frac{1}{x}]_1^t = 0 + 1 = 1$

(b) Volume revolved about the $x$-axis: $V = \pi\int_1^{\infty} \frac{1}{x^4}\,dx = \pi \cdot \frac{1}{3} = \frac{\pi}{3}$

(c) Compare: $y = 1/x$ (Gabriel’s Horn) has finite volume ($\pi$) but infinite surface area. The $y = 1/x^2$ version has finite everything.

Key Insight: Whether area/volume/surface area is finite depends on the power of $x$ in the denominator.

Final Computation

Unit Summary — Improper Integrals

The Big Picture:

• Type I: infinite bounds → replace with limit
• Type II: discontinuous integrand → one-sided limit
• $p$-test: the fundamental convergence criterion
• Comparison tests: extend $p$-test to harder integrals
• Key results: $e^{-kx}$, $\arctan$, Gabriel’s Horn

AP Exam Checklist:

• Can you identify improper integrals?
• Can you set up and evaluate both types?
• Do you know both $p$-test directions?
• Can you apply DCT and LCT?
• Can you handle FRQ area/volume problems?

Congratulations! You have mastered improper integrals for AP Calculus BC.