Calculus II: Worked Example Walkthroughs for Convergence, Differentiation, and Volumes

Abstract

This article presents structured worked examples for three core Calculus II topics: convergence of improper integrals, logarithmic differentiation, and volumes of solids of revolution. Each section combines conceptual grounding with step-by-step problem solutions, designed to clarify common points of confusion and reinforce technique mastery.

Background

Calculus II extends single-variable calculus into applications and advanced techniques. Three topics frequently challenge students: determining whether improper integrals converge, applying logarithmic differentiation to complex functions, and computing volumes via rotation. These topics share a common thread—they require careful setup, systematic application of rules, and verification of results. This article provides worked examples to bridge the gap between abstract definitions and practical problem-solving.

Key Results and Concepts

Convergence of Improper Integrals

An integral converges when the limit of integration yields a finite value; it diverges otherwise ^{[convergence-of-integrals]}. For improper integrals—those with infinite limits or discontinuous integrands—direct evaluation often fails. Instead, we rewrite the integral as a limit and evaluate:

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

If this limit exists and is finite, the integral converges. Convergence tests such as the comparison test and limit comparison test allow us to determine convergence without explicit computation ^{[convergence-of-integrals]}.

Logarithmic Differentiation

Logarithmic differentiation simplifies the derivative of functions where the variable appears in both base and exponent, or in complex products and quotients ^{[logarithmic-differentiation]}. The method proceeds by taking the natural logarithm of both sides, then differentiating implicitly:

$\frac{d}{dx} \ln(y) = \frac{1}{y} \frac{dy}{dx}$

This transforms multiplicative relationships into additive ones, making the chain rule and product rule easier to apply ^{[logarithmic-differentiation]}.

Volume of Solids of Revolution

When a region bounded by curves is rotated about an axis, the resulting solid's volume can be computed via integration. For rotation about the x-axis, the disk/washer method yields ^{[volume-of-solid-of-revolution]}:

$V = \pi \int_a^b (f(x)^2 - g(x)^2) \, dx$

For rotation about the y-axis, we integrate with respect to $y$ ^{[volume-of-solid-of-revolution]}:

$V = 2\pi \int_c^d x (h(y) - k(y)) \, dy$

Worked Examples

Example 1: Convergence of an Improper Integral

Problem: Determine whether $\int_1^{\infty} \frac{1}{x^2} \, dx$ converges or diverges.

Solution:

Rewrite as a limit: $\int_1^{\infty} \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t x^{-2} \, dx$

Evaluate the antiderivative: $\int_1^t x^{-2} \, dx = \left[ -x^{-1} \right]_1^t = -\frac{1}{t} - (-1) = 1 - \frac{1}{t}$

Take the limit: $\lim_{t \to \infty} \left( 1 - \frac{1}{t} \right) = 1$

Conclusion: The integral converges to 1 ^{[convergence-of-integrals]}.

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Example 2: Logarithmic Differentiation

Problem: Find $\frac{dy}{dx}$ for $y = x^x$.

Solution:

Take the natural logarithm of both sides: $\ln(y) = \ln(x^x) = x \ln(x)$

Differentiate both sides with respect to $x$ using the chain rule on the left and the product rule on the right: $\frac{1}{y} \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)$

Verification: This result is difficult to obtain using standard rules directly, but logarithmic differentiation makes it straightforward ^{[logarithmic-differentiation]}.

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Example 3: Volume of a Solid of Revolution

Problem: Find the volume of the solid obtained by rotating the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ about the x-axis.

Solution:

The region is bounded above by $y = \sqrt{x}$ and below by $y = 0$. Using the disk method with rotation about the x-axis ^{[volume-of-solid-of-revolution]}:

$V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx$

Evaluate: $V = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \left( \frac{16}{2} - 0 \right) = 8\pi$

Conclusion: The volume is $8\pi$ cubic units.

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Example 4: Convergence with the Limit Comparison Test

Problem: Determine whether $\int_1^{\infty} \frac{1}{x^2 + 1} \, dx$ converges.

Solution:

For large $x$, $\frac{1}{x^2 + 1} \approx \frac{1}{x^2}$. We know $\int_1^{\infty} \frac{1}{x^2} \, dx$ converges (from Example 1). Apply the limit comparison test ^{[convergence-of-integrals]}:

$\lim_{x \to \infty} \frac{\frac{1}{x^2 + 1}}{\frac{1}{x^2}} = \lim_{x \to \infty} \frac{x^2}{x^2 + 1} = 1$

Since the limit is a positive finite number and the comparison function converges, the original integral converges ^{[convergence-of-integrals]}.

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References

• ^{[convergence-of-integrals]}
• ^{[logarithmic-differentiation]}
• ^{[volume-of-solid-of-revolution]}

AI Disclosure

This article was drafted with AI assistance from class notes (Zettelkasten). All mathematical statements and worked examples are grounded in cited source notes. The AI was used to organize, paraphrase, and structure the material for clarity and pedagogical flow. All factual claims are traceable to the source notes listed in the References section.