Problem 32 - Thomas Calculus

Question

In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int x^{-3}(x+1) d x$$

Step-by-Step Solution

Answer

The most general antiderivative is \(-x^{-1} - \frac{x^{-2}}{2} + C\).

Step 1: Expand the Integrand

First, rewrite the integrand by distributing the term \( x^{-3} \) across the terms inside the parenthesis. This gives:\[ x^{-3}(x + 1) = x^{-3} \cdot x + x^{-3} \cdot 1 = x^{-2} + x^{-3}. \] Thus, the integral becomes: \[ \int (x^{-2} + x^{-3}) \, dx. \]

Step 2: Integrate Term by Term

To find the antiderivative, integrate each term separately. The integral of \( x^{-2} \) is:\[ \int x^{-2} \, dx = \frac{x^{-1}}{-1} = -x^{-1}. \]The integral of \( x^{-3} \) is:\[ \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{x^{-2}}{2}. \]So the antiderivative of the entire function is:\[ -x^{-1} - \frac{x^{-2}}{2} + C, \] where \( C \) is the constant of integration.

Step 3: Verify by Differentiation

Differentiate the result to ensure correctness. We have:\[ \frac{d}{dx} \left( -x^{-1} - \frac{x^{-2}}{2} + C \right) = x^{-2} + \frac{2x^{-3}}{2} = x^{-2} + x^{-3}. \]This matches the original integrand \( x^{-2} + x^{-3} \), confirming the solution is correct.

Key Concepts

Antiderivative

An antiderivative is a function whose derivative is the original function you started with. In other words, if you have a function \( f(x) \), an antiderivative \( F(x) \) is such that when you differentiate \( F(x) \), you get \( f(x) \) again. This process is also known as finding an indefinite integral.

When you seek the most general antiderivative, you include a constant of integration, typically denoted as \( C \). This is because the differentiation of a constant is zero, and thus any constant could be added to an antiderivative without affecting its derivative.

For example, in the solution we found:

• The antiderivative \( -x^{-1} - \frac{x^{-2}}{2} + C \) corresponds to the original function \( x^{-2} + x^{-3} \).
• To confirm this, differentiate this antiderivative, and you should retrieve the original function \( x^{-2} + x^{-3} \).

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function is changing at any given point and is foundational in the study of calculus.

Differentiation is used as a verification method to check if the antiderivative obtained is correct. This involves reverse-engineering the antiderivative by differentiating it to see if you end up with the original function.

In the exercise above, after differentiating the antiderivative \( -x^{-1} - \frac{x^{-2}}{2} + C \), we retrieved the original expression \( x^{-2} + x^{-3} \). This confirms that the integration was performed correctly.

The process of differentiation provides:

• Instant confirmation of the antiderivative's validity in solving integration problems.
• A method to understand how functions behave and change over intervals.

Polynomial Integration

Polynomial integration involves finding the antiderivative of polynomial expressions, which can be simplified into a sum of individual terms where each term is integrated separately.

With polynomial integration, each term of the polynomial is raised to a power in the form of \( x^n \), where the integral follows a standard rule: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] when \( n eq -1 \).

In the given exercise:

• The polynomial expression \( x^{-3}(x+1) \) was expanded to \( x^{-2} + x^{-3} \).
• Each term was integrated separately using the integral power rule, resulting in \( -x^{-1} - \frac{x^{-2}}{2} + C \).

This method exemplifies the step-by-step breakdown characteristic of polynomial integrations, making the entire calculation more manageable and easily verifiable.