Problem 8 - Thomas Calculus in SI Units

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{4}{3} \sqrt[3]{x}\) b. \(\frac{1}{3 \sqrt[3]{x}}\) c. \(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\)

Step-by-Step Solution

Answer

a. \(x^{4/3} + C\), b. \(\frac{1}{2}x^{2/3} + C\), c. \(\frac{3}{4}x^{4/3} + \frac{3}{2}x^{2/3} + C\).

Step 1: Rewrite the Functions

For each of the functions, rewrite them using exponents: a. \( \frac{4}{3} \sqrt[3]{x} = \frac{4}{3} x^{1/3} \) b. \( \frac{1}{3 \sqrt[3]{x}} = \frac{1}{3} x^{-1/3} \) c. \( \sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} = x^{1/3} + x^{-1/3} \)

Step 2: Apply the Power Rule for Integration

For each term, use the power rule for integration \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.**a.** \( \int \frac{4}{3} x^{1/3} \, dx = \frac{4}{3} \cdot \frac{x^{4/3}}{4/3} + C = x^{4/3} + C \)**b.** \( \int \frac{1}{3} x^{-1/3} \, dx = \frac{1}{3} \cdot \frac{x^{2/3}}{2/3} + C = \frac{1}{2} x^{2/3} + C \)**c.** \( \int \left( x^{1/3} + x^{-1/3} \right) \, dx = \frac{x^{4/3}}{4/3} + \frac{x^{2/3}}{2/3} + C = \frac{3}{4} x^{4/3} + \frac{3}{2} x^{2/3} + C \)

Step 3: Differentiate to Check Each Antiderivative

Differentiate each result to ensure it gives the original function.**a.** Differentiating \( x^{4/3} + C \) results in \( \frac{4}{3} x^{1/3} \), which matches the original. **b.** Differentiating \( \frac{1}{2} x^{2/3} + C \) results in \( \frac{1}{3} x^{-1/3} \), which matches the original. **c.** Differentiating \( \frac{3}{4} x^{4/3} + \frac{3}{2} x^{2/3} + C \) results in \( x^{1/3} + x^{-1/3} \), which matches the original.

Key Concepts

Power Rule

The Power Rule is an essential tool in calculus, especially when dealing with integration and differentiation. When we talk about integrating polynomials, the power rule makes this process simpler and more straightforward.
To integrate a function like \( x^n \), the Power Rule states that you can use the formula:

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

where \( C \) is the constant of integration. The constant \( C \) is crucial because it represents any vertical shift in the function, accounting for all possible antiderivatives.
When applying this rule, increment the exponent by one and then divide by this new exponent to get your antiderivative. This makes it an easy method to work through problems involving basic polynomials.

Integration

Integration is the process of finding the antiderivative of a function. In other words, it is about determining the original function from its derivative.
It is an important calculus operation, and its notation is \( \int \), often followed by a function and \( dx \), which reads as "the integral of."
Integration can be definite or indefinite:

• Indefinite Integrals: Involve finding a general antiderivative of a function, with the result including an arbitrary constant \( C \).
• Definite Integrals: Concerned with the area under the curve of a function between two specific points, providing a numerical result.

The power rule discussed in the previous section is a particular technique used in integration, especially for polynomial-type functions.
Understanding integration involves constant practice of applying the rules to different functions to become comfortable with the arithmetic and reasoning behind it.

Exponents

Exponents are a compact way of expressing repeated multiplication of a number by itself. They are fundamental to simplifying expressions and understanding calculus operations. In integration, rewriting functions involving roots means expressing them as exponents.
Consider \( \sqrt[3]{x} \) which is equivalent to \( x^{1/3} \). This transformation allows us to apply the Power Rule to functions that might otherwise seem complex. Similarly, reciprocal powers like \( \frac{1}{\sqrt[3]{x}} \) can be rewritten as \( x^{-1/3} \).
Remember these key ideas:

• \( x^{a/b} \) denotes the \( b \)-th root of \( x \) raised to the power of \( a \).
• Negative exponents indicate reciprocals, meaning \( x^{-n} = \frac{1}{x^n} \).

By converting radicals and fractions into exponential form, complex functions become eligible for integration and differentiation using the Power Rule, providing a powerful method to solve algebraic and calculus problems efficiently.

Differentiation

Differentiation is the process of finding the derivative of a function, which is essentially the rate at which the function's value changes. If integration is about finding the original form of a function, differentiation is about determining its rate of change.
The derivative of a function is denoted as \( f'(x) \) or \( \frac{dy}{dx} \), and it's calculated using rules like:

• The derivative of \( x^n \) is \( nx^{n-1} \).

This process is the inverse of integration.
For example, if you find the antiderivative of \( x^{1/3} \) using the Power Rule, then differentiate it to get back to possibly the original function. This is crucial for checking solutions in calculus.
Differentiation, like integration, is a fundamental concept in calculus that allows us to understand how functions behave and change. Both operations are tools that should be understood in relation to each other, providing comprehensive insights into the mechanics of functions.