Problem 2 - Calculus: Early Transcendentals

Question

Verify that \(\int_{0}^{3} 2 \pi x\left(x+1-(x-1)^{2}\right) d x=\frac{27}{2} \pi\).

Step-by-Step Solution

Answer

The integral evaluates correctly to \( \frac{27}{2} \pi \).

Step 1: Simplify the Integrand

First, simplify the expression inside the integral: \( 2 \pi x \left( x + 1 - (x-1)^2 \right) \). Start by expanding \((x-1)^2\) which gives \(x^2 - 2x + 1\). Substitute this back into the expression: \( 2 \pi x \left( x + 1 - x^2 + 2x - 1 \right) = 2 \pi x \left( 3x - x^2 \right) \). Thus, the integrand becomes \( 2 \pi x (3x - x^2) = 2 \pi (3x^2 - x^3) \).

Step 2: Set up the Integral

Rewrite the integral with the simplified expression: \( \int_{0}^{3} 2 \pi (3x^2 - x^3) \, dx \). This step involves ensuring the integral expression is correctly written with respect to the variable \(x\).

Step 3: Separate the Integral

The integral can be separated into two parts: \( \int_{0}^{3} 2 \pi \cdot 3x^2 \, dx - \int_{0}^{3} 2 \pi \cdot x^3 \, dx \). Write both integrals distinctly to solve each one separately.

Step 4: Calculate the First Integral

Calculate the first part: \( 2 \pi \cdot 3 \int_{0}^{3} x^2 \, dx \). This results in \( 6 \pi \left[ \frac{x^3}{3} \right]_{0}^{3} \). Evaluating this from 0 to 3 gives \( 6 \pi \left( \frac{27}{3} - 0 \right) = 6 \pi \times 9 = 54 \pi \).

Step 5: Calculate the Second Integral

Calculate the second part: \( 2 \pi \int_{0}^{3} x^3 \, dx \). This evaluates to \( 2 \pi \left[ \frac{x^4}{4} \right]_{0}^{3} \). Substituting the limits gives \( 2 \pi \left( \frac{81}{4} - 0 \right) = \frac{162 \pi}{4} = \frac{81 \pi}{2} \).

Step 6: Subtract the Results

Subtract the result from the second integral from the first: \( 54 \pi - \frac{81 \pi}{2} \). To subtract, convert \( 54 \pi \) to \( \frac{108 \pi}{2} \), resulting in \( \frac{108 \pi}{2} - \frac{81 \pi}{2} = \frac{27 \pi}{2} \).

Step 7: Verify the Final Result

Compare the result from the integration process \( \frac{27}{2} \pi \) with the target result \( \frac{27}{2} \pi \). Since they are equal, the integrals were calculated correctly.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used to integrate products of functions. It's especially useful when direct integration is complicated. The rule comes from the product rule for differentiation and is given by:

• \( \int u \, dv = uv - \int v \, du \)

To apply this, pick two parts of the integrand: one to differentiate \(u\), and one to integrate \(dv\).
Not every integral is suited for this method; however, it provides a clever trick for certain challenging cases.
This exercise doesn't require integration by parts directly, as each component can be integrated using elementary techniques.
That said, understanding integration by parts is crucial as it deals with products of functions like polynomials and trigonometric or exponential functions.

Polynomial Integration

Polynomial integration involves integrating terms like \(x^n\), where \(n\) is a non-negative integer.
The process involves using the reverse of the power rule. For a term \(x^n\), the integral is:

• \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

This basic rule is what you'll use in our solved exercise, albeit expanded into a sum of similar terms.
For a more complex polynomial, split it into simpler terms, integrate each one separately,
and then sum those results.
In the case of the original exercise,\(6\pi(3x^2)\) and \(\frac{81\pi}{2}(x^3)\) were integrated using this straightforward technique.
It showcases how easily polynomial forms can be tackled once simplified.

Calculus Proofs

Calculus proofs ensure solutions are not only correct but understood deeply.
They provide logical foundations for each step we take in integration or differentiation.
In the exercise, simplifying the expression first is a form of proof.
By breaking down and verifying step-by-step, you ensure that the simplification does not alter the function's meaning.

• First, confirm that expanding \((x-1)^2\) to \(x^2 - 2x + 1\) is consistent.
• Then, simplify everything inside the integral to \(2\pi(3x^2 - x^3)\).

Lastly, comparing \(\frac{27}{2}\pi\) as a result provides a numeric double-check that your process was correct.
Proofs like these indicate that mathematics is not just an abstract concept, but a method grounded in clear, repeatable logic.