Problem 6 - 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. \(-\frac{2}{x^{3}}\) b. \(\frac{1}{2 x^{3}}\) c. \(x^{3}-\frac{1}{x^{3}}\)

Step-by-Step Solution

Answer

a: \(x^{-2} + C\); b: \(-\frac{1}{4} x^{-2} + C\); c: \(\frac{x^4}{4} + \frac{1}{2} x^{-2} + C\).

Step 1: Understanding the Antiderivative of a Power Function

Recall that the antiderivative of a power function \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. We will use this rule to find the antiderivative of each term in the given functions.

Step 2: Finding the Antiderivative for Part (a)

Given the function \( f(x) = -\frac{2}{x^3} \), rewrite it as \( f(x) = -2x^{-3} \). The antiderivative is found by adding 1 to the exponent and dividing by the new exponent: \( -2 \cdot \frac{x^{-3+1}}{-3+1} = -2 \cdot \frac{x^{-2}}{-2} + C = x^{-2} + C \).

Step 3: Verifying the Antiderivative for Part (a)

To verify, differentiate \( x^{-2} + C \). The derivative is \( -2x^{-3} \), which matches the original function \( -\frac{2}{x^3} \). Thus, our antiderivative is correct.

Step 4: Finding the Antiderivative for Part (b)

For \( f(x) = \frac{1}{2x^3} \), rewrite it as \( f(x) = \frac{1}{2} x^{-3} \). The antiderivative is \( \frac{1}{2} \cdot \frac{x^{-3+1}}{-3+1} = \frac{x^{-2}}{-4} + C = -\frac{1}{4} x^{-2} + C \).

Step 5: Verifying the Antiderivative for Part (b)

Differentiate \( -\frac{1}{4} x^{-2} + C \). The derivative is \( \frac{1}{2} x^{-3} \), which matches \( \frac{1}{2x^3} \). This confirms the antiderivative is correct.

Step 6: Solving Two-term Antiderivative for Part (c)

For the function \( f(x) = x^3 - \frac{1}{x^3} \), find the antiderivative of each term separately. The antiderivative of \( x^3 \) is \( \frac{x^{3+1}}{3+1} = \frac{x^4}{4} \), and \( -\frac{1}{x^3} = -x^{-3} \) becomes \( -\frac{x^{-2}}{-2} = \frac{1}{2} x^{-2} \). Therefore, the antiderivative is \( \frac{x^4}{4} + \frac{1}{2} x^{-2} + C \).

Step 7: Checking the Antiderivative for Part (c)

Differentiate \( \frac{x^4}{4} + \frac{1}{2} x^{-2} + C \). The derivative is \( x^3 - x^{-3} = x^3 - \frac{1}{x^3} \), confirming the antiderivative is correct.

Key Concepts

Power Function

A power function is a type of function that has the form \( f(x) = x^n \), where \( n \) is a real number. This type of function is particularly important when working with antiderivatives. The key to understanding power functions is recognizing that each term can be independently integrated. When finding an antiderivative of a power function, you apply the power rule, which states: \[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]Here, \( C \) is the constant of integration, which accounts for any constant that might have been present in the original function before differentiation. The rule is simple: increase the exponent by one, then divide by this new exponent. Always remember to apply this process separately to each term in a function that involves multiple terms, as with polynomials.

For instance, rewriting and integrating \(-\frac{2}{x^3}\) involves reexpressing it as a power function: \(-2x^{-3}\). Then, adding 1 to the exponent gives \(-2x^{-2}\), and dividing by the new exponent modifies it to \(x^{-2}\). Finally, don't forget the constant \( C \).

Differentiation

Differentiation is the reverse process of finding an antiderivative. It's the process of determining how a function's output value changes as its input value changes. When we take the derivative of a function, we are finding the rate at which the function value is changing at any given point. For power functions, the differentiation process also involves a simple rule: bringing the exponent down as a coefficient, then decreasing the exponent by one.
For example, consider the function \( x^n \); its derivative is: \[x^n \, \frac{d}{dx} = n \cdot x^{n-1}\]Checking your work by differentiating the antiderivative is a fundamental step in ensuring that you've found a correct solution.

When we found that the antiderivative of \( -\frac{2}{x^3} \) was \( x^{-2} + C \), we differentiated it back into the original form \(-2x^{-3}\). Doing this confirms the antiderivative is correct and helps to develop intuition for the interplay between integration and differentiation.

Integration

Integration, the process of finding an antiderivative, is essentially the reverse of differentiation. It involves finding a function whose derivative is the given function. The integral or antiderivative of a function is often sought when one wants to determine the area under a curve, but it also plays a fundamental role in reverse-engineering a function from its rate of change.

• To integrate a power function, apply the reverse power rule: Increase the exponent by one and divide by the new exponent.
• Always include the constant of integration, \( C \), because indefinite integrals represent a family of functions.

For two-term functions like \( x^3 - \frac{1}{x^3} \), integrate each term separately:

• The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \).
• For \(-\frac{1}{x^3} \), rewrite and integrate as \(-x^{-3}\) to get \( \frac{1}{2} x^{-2} \).

Add the results together and don't forget the constant \( C \) for the full solution. This process shows how integration operates as the inverse of differentiation and is key to unlocking understanding of antiderivatives.