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calculusintegrationconvergencesolids-of-revolutionlogarithmic-differentiationpedagogyFri Apr 24

Calculus II: Worked Example Walkthroughs for Core Integration Techniques

Abstract

This article provides structured worked examples for three foundational Calculus II topics: convergence of improper integrals, volumes of solids of revolution, and logarithmic differentiation. Each section combines conceptual grounding with step-by-step problem solutions, designed to bridge the gap between theory and practice for students preparing for exams or consolidating understanding of integration-based applications.

Background

Calculus II extends single-variable calculus by introducing techniques for handling unbounded integrals, geometric applications of integration, and specialized differentiation methods. Three topics recur across most Calculus II curricula: determining whether improper integrals converge to finite values, computing volumes of three-dimensional objects formed by rotating planar regions, and differentiating complex functions using logarithmic methods. Each requires both conceptual clarity and procedural fluency.

This article assumes familiarity with basic integration, the fundamental theorem of calculus, and implicit differentiation.

Key Results

Convergence of Improper Integrals

[convergence-of-integrals] establishes that an integral abf(x)dx\int_a^b f(x) \, dx converges when the limit of the integral exists and is finite as the bounds approach their limits. Conversely, divergence occurs when the limit is infinite or does not exist. This distinction is essential because many real-world applications—probability distributions, work calculations, and area computations—require knowing whether a result is meaningful (finite) or unbounded.

The practical challenge is that not all improper integrals can be evaluated in closed form. Convergence tests such as the comparison test and limit comparison test allow us to determine convergence without explicit computation.

Volumes of Solids of Revolution

[volume-of-solid-of-revolution] provides the disk/washer method formulas. When a region bounded by y=f(x)y = f(x) and y=g(x)y = g(x) from x=ax = a to x=bx = b is rotated about the x-axis, the volume is:

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

For rotation about the y-axis, the shell method or inverse function approach yields:

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

These formulas arise from slicing the solid perpendicular to the axis of rotation and summing infinitesimal disk or washer areas.

Logarithmic Differentiation

[logarithmic-differentiation] describes a technique for differentiating functions where the variable appears in both base and exponent. Given y=f(x)y = f(x), taking the natural logarithm of both sides yields:

ln(y)=ln(f(x))\ln(y) = \ln(f(x))

Differentiating implicitly with respect to xx:

1ydydx=f(x)f(x)\frac{1}{y} \frac{dy}{dx} = \frac{f'(x)}{f(x)}

Solving for the derivative:

dydx=yf(x)f(x)\frac{dy}{dx} = y \cdot \frac{f'(x)}{f(x)}

This method transforms products and quotients into sums and differences via logarithm properties, simplifying differentiation of otherwise unwieldy expressions.

Worked Examples

Example 1: Testing Convergence of an Improper Integral

Problem: Determine whether 11x2+1dx\int_1^{\infty} \frac{1}{x^2 + 1} \, dx converges.

Solution:

We recognize this as an improper integral with an infinite upper bound. Following [convergence-of-integrals], we evaluate the limit:

11x2+1dx=limt1t1x2+1dx\int_1^{\infty} \frac{1}{x^2 + 1} \, dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2 + 1} \, dx

The antiderivative of 1x2+1\frac{1}{x^2 + 1} is arctan(x)\arctan(x), so:

limt[arctan(x)]1t=limt(arctan(t)arctan(1))\lim_{t \to \infty} \left[ \arctan(x) \right]_1^t = \lim_{t \to \infty} (\arctan(t) - \arctan(1))

Since limtarctan(t)=π2\lim_{t \to \infty} \arctan(t) = \frac{\pi}{2} and arctan(1)=π4\arctan(1) = \frac{\pi}{4}:

π2π4=π4\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}

Conclusion: The integral converges to π4\frac{\pi}{4}.


Example 2: Volume of a Solid of Revolution

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

Solution:

Using the disk method from [volume-of-solid-of-revolution], the region is bounded above by f(x)=xf(x) = \sqrt{x} and below by g(x)=0g(x) = 0, from x=0x = 0 to x=4x = 4:

V=π04(x)202dx=π04xdxV = \pi \int_0^4 (\sqrt{x})^2 - 0^2 \, dx = \pi \int_0^4 x \, dx

Evaluating:

V=π[x22]04=π162=8πV = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi

Conclusion: The volume is 8π8\pi cubic units.


Example 3: Logarithmic Differentiation

Problem: Find dydx\frac{dy}{dx} for y=xxy = x^x (where x>0x > 0).

Solution:

This function has the variable in both base and exponent, making the product rule or power rule alone insufficient. Following [logarithmic-differentiation], take the natural logarithm of both sides:

ln(y)=ln(xx)=xln(x)\ln(y) = \ln(x^x) = x \ln(x)

Differentiate both sides implicitly with respect to xx:

1ydydx=ddx(xln(x))=ln(x)+x1x=ln(x)+1\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln(x)) = \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1

Solve for dydx\frac{dy}{dx}:

dydx=y(ln(x)+1)=xx(ln(x)+1)\frac{dy}{dx} = y(\ln(x) + 1) = x^x(\ln(x) + 1)

Conclusion: dydx=xx(ln(x)+1)\frac{dy}{dx} = x^x(\ln(x) + 1).


References

AI Disclosure

This article was drafted with AI assistance. The structure, worked examples, and explanatory text were generated based on class notes provided as input. All mathematical statements and formulas are grounded in the cited notes. The worked examples were constructed to illustrate the concepts described in those notes and follow standard Calculus II pedagogy. A human author reviewed the mathematical accuracy and pedagogical clarity of all content.

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References

AI disclosure: Generated from personal class notes with AI assistance. Every factual claim cites a note. Model: claude-haiku-4-5-20251001.