Problem 8 - Thomas Calculus

Question

In Exercises \(1-16,\) find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{4}{3} \sqrt[3]{x} \quad\) b. \(\frac{1}{3 \sqrt[3]{x}} \quad\) c. \(\sqrt[3]{x}+\frac{1}{\sqrt[3]{x}}\)

Step-by-Step Solution

Answer

Antiderivatives: 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: Identify the Integral Formula

To find an antiderivative, use the power rule for integration. For a function of the form \(x^n\), the antiderivative is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

Step 2: Simplify and Integrate Part (a)

Rewrite the function \(\frac{4}{3} \sqrt[3]{x}\) as \(\frac{4}{3} x^{1/3}\). Using the power rule, integrate to get \(\frac{4}{3} \cdot \frac{x^{(1/3) + 1}}{1/3 + 1} + C = \frac{4}{3} \cdot \frac{x^{4/3}}{4/3} + C = x^{4/3} + C\).

Step 3: Simplify and Integrate Part (b)

Rewrite the function \(\frac{1}{3 \sqrt[3]{x}}\) as \(\frac{1}{3} x^{-1/3}\). Using the power rule, integrate to get \(\frac{1}{3} \cdot \frac{x^{(-1/3) + 1}}{-1/3 + 1} + C = \frac{1}{3} \cdot \frac{x^{2/3}}{2/3} + C = \frac{1}{2} x^{2/3} + C\).

Step 4: Simplify and Integrate Part (c)

Rewrite the function \(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\) as \(x^{1/3} + x^{-1/3}\). Integrate each part separately: the antiderivative of \(x^{1/3}\) is \(\frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3}\), and the antiderivative of \(x^{-1/3}\) is \(\frac{x^{2/3}}{2/3} = \frac{3}{2} x^{2/3}\). Combine: \(\frac{3}{4} x^{4/3} + \frac{3}{2} x^{2/3} + C\).

Step 5: Verify by Differentiation

Differentiate each found antiderivative to ensure they match the original functions from parts (a), (b), and (c). The differentiation should return you to the original functions.

Key Concepts

Integration

Integration is one of the core concepts in calculus. It is essentially the reverse process of differentiation. By integration, we look to find the original function based on its derivative, which gives us the antiderivative. An antiderivative of a function is another function whose derivative is the given function.

• For example, if you have a function that represents how fast something is changing, integration helps you find out how much it has changed over time.
• It is like piecing together the small parts to see the whole picture.

The constant of integration, typically denoted as "+ C", is added to represent all possible antiderivatives since adding any constant to a function doesn't change its derivative.
Understanding integration requires practice, as you must be comfortable with manipulating algebraic expressions, recognizing function forms, and applying proper rules.

Power Rule

The power rule is one of the fundamental tools for both differentiation and integration. It’s a shortcut that simplifies the process of finding derivatives and antiderivatives. When it comes to integration, the power rule states:

• If you have a function in the form of \(x^n\), its antiderivative is \(\frac{x^{n+1}}{n+1} + C\).
• This rule doesn’t work when \(n = -1\), as it leads to division by zero.

To properly use this rule, be sure to adjust the power by adding one, and then divide by the new exponent. Don't forget to add the constant of integration \(+ C\) to cover all possible solutions.
Let's consider an example: For \(x^{1/3}\), applying the power rule gives the antiderivative \(\frac{x^{4/3}}{4/3} + C\). This simple technique makes integrating power functions straightforward once you recognize the pattern.

Differentiation

Differentiation is the process of finding the derivative, which is essentially the rate at which a function is changing at any given point. It's the tool we use to verify our integration results. When you integrate a function to find its antiderivative, you can differentiate the result to check if you obtain the original function.

• For example, given the function \(x^{4/3}\), the derivative using the power rule for differentiation would be \(\frac{d}{dx} x^{4/3} = \frac{4}{3} x^{1/3}\), which should match the original function before integration.
• If it matches, it confirms the accuracy of your antiderivative.

By understanding differentiation, you'll see how the concepts of integration and differentiation tie together, serving as checks and balances within calculus.
Both processes complement each other, with integration allowing you to build up from a rate of change to a total, and differentiation letting you break down totals to see how they change.