Problem 9 - 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. $$\text { a. }\frac{2}{3} x^{-1 / 3} \quad \text { b. } \frac{1}{3} x^{-2 / 3} \quad \text { c. }-\frac{1}{3} x^{-4 / 3}$$

Step-by-Step Solution

Answer

The antiderivatives are \( x^{2/3} + C \) for (a), \( x^{1/3} + C \) for (b), and \( x^{-1/3} + C \) for (c).

Step 1: Identify the Function for Antidifferentiation

We need to find antiderivatives for the given functions: - Part (a): \( \frac{2}{3} x^{-1/3} \)- Part (b): \( \frac{1}{3} x^{-2/3} \)- Part (c): \( -\frac{1}{3} x^{-4/3} \)An antiderivative is a function whose derivative is the given function.

Step 2: Use the Power Rule for Antidifferentiation

The power rule for antiderivatives states that for a function of the form \( ax^n \), the antiderivative is \( \frac{a}{n+1}x^{n+1} + C \), where \( C \) is a constant of integration, provided \( n eq -1 \). We will apply this rule to each function.

Step 3: Calculate Antiderivative for Part (a)

For \( \frac{2}{3} x^{-1/3} \): 1. Identify \( a = \frac{2}{3} \) and \( n = -1/3 \).2. Apply the power rule: \[\int \frac{2}{3} x^{-1/3} \, dx = \frac{2/3}{-1/3 + 1} x^{-1/3 + 1} + C = \frac{2/3}{2/3} x^{2/3} + C = x^{2/3} + C\]

Step 4: Calculate Antiderivative for Part (b)

For \( \frac{1}{3} x^{-2/3} \): 1. Identify \( a = \frac{1}{3} \) and \( n = -2/3 \).2. Apply the power rule: \[\int \frac{1}{3} x^{-2/3} \, dx = \frac{1/3}{-2/3 + 1} x^{-2/3 + 1} + C = \frac{1/3}{1/3} x^{1/3} + C = x^{1/3} + C\]

Step 5: Calculate Antiderivative for Part (c)

For \( -\frac{1}{3} x^{-4/3} \): 1. Identify \( a = -\frac{1}{3} \) and \( n = -4/3 \).2. Apply the power rule: \[\int -\frac{1}{3} x^{-4/3} \, dx = \frac{-1/3}{-4/3 + 1} x^{-4/3 + 1} + C = \frac{-1/3}{-1/3} x^{-1/3} + C = x^{-1/3} + C\]

Step 6: Verify by Differentiation

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

Key Concepts

Understanding the Power Rule

The power rule is a fundamental concept in calculus, especially when working with integration and differentiation. When computing antiderivatives, or the "reverse" of differentiation, the power rule provides a simple formula. This formula can be used to find the antiderivative of a power function. - Given a function of the form \( ax^n \), the antiderivative is calculated as \( \frac{a}{n+1}x^{n+1} + C \).- The variable \( n \) must not be equal to \(-1\) because the formula requires division by \( n+1 \).- The constant \( C \) is added to the antiderivative as every antiderivative family has a "constant of integration."This rule greatly simplifies finding integrals and is crucial for solving a wide range of integration problems effectively.

The Integration Process

Integration is the process of finding the antiderivative of a function. It is the reverse action of differentiation. When you integrate a function, you are determining what original function could have produced the derivative that you are given. - To integrate using the power rule, adjust the exponent of each term by adding one (changing \( n \) to \( n+1 \)) and then divide by this new exponent.- Don't forget to add the constant of integration \( C \) at the end of the integrated function.Each step of integration requires accuracy to ensure that the differentiation of your solution will check back to the original function. It's like piecing together a puzzle backward but with infinite possibilities due to that constant \( C \).

The Role of Differentiation in Verifying Answers

Differentiation is crucial for verifying the accuracy of your integration work. It is the process of finding a derivative, or the rate at which a function changes. When you find the antiderivative using integration, you should always check your work by differentiating the integrated function. - By differentiating, you ensure that the integration process was done correctly. The derivative of the antiderivative function must match the original function you started with. - This check acts like a balancing scale; if the weights on both sides are equal, you've done your work correctly. Checking your work through differentiation gives double assurance that your solution is correct and aligns with mathematical principles.

Importance of the Constant of Integration

The constant of integration, often represented as \( C \), is a vital part of the antiderivative. When you integrate a function, you need to add this constant because the derivative of a constant is zero. - Each function can have many antiderivatives that differ only by a constant. Therefore, \( C \) accounts for this infinite set of possible solutions.- Always remember to append \( C \) in your answers to cover every possible original function.Without including the constant of integration, you may miss potential solutions, which means your answer might not represent the full set of functions that could produce the same derivative. Keeping \( C \) in mind ensures your solution is complete.