Problem 14 - Calculus Concepts

Question

For Activities 5 through \(16,\) evaluate the improper integral. $$ \int_{-\infty}^{-2}\left(\frac{3}{x^{3}}+1\right) d x $$

Step-by-Step Solution

Answer

The integral diverges.

Step 1: Split the Integral

The integral can be separated into two simpler integrals: \( \int_{-\infty}^{-2} \frac{3}{x^3} \, dx \) and \( \int_{-\infty}^{-2} 1 \, dx \). We will evaluate each one separately.

Step 2: Evaluate the Integral of \( \frac{3}{x^3} \)

The antiderivative of \( \frac{3}{x^3} \) is \(-\frac{3}{2x^2}\). Therefore, the integral from \(-\infty\) to \(-2\) can be written as \( \lim_{a \to -\infty} \int_{a}^{-2} \frac{3}{x^3} \, dx = \lim_{a \to -\infty} \left[ -\frac{3}{2x^2} \right]_a^{-2} \). This evaluates to \( -\frac{3}{2(-2)^2} + \lim_{a \to -\infty}\frac{3}{2a^2} \). Since \( \frac{1}{a^2} \to 0 \) as \( a \to -\infty \), the value is \( -\frac{3}{8} \).

Step 3: Evaluate the Integral of \(1\)

\( \int_{-\infty}^{-2} 1 \, dx \) has the antiderivative \(x\). Thus, \(\lim_{a \to -\infty} \left[ x \right]_a^{-2} = -2 - a\). Since \( a \to -\infty \) implies \(a\) is very large in magnitude, this limit does not converge; the integral is divergent.

Step 4: Combine Results

Since one of the integrals diverges, the whole integral \( \int_{-\infty}^{-2} \left( \frac{3}{x^3} + 1 \right) \, dx \) is divergent. Even though \( \int_{-\infty}^{-2} \frac{3}{x^3} \, dx \) converges separately, the overall improper integral does not have a finite value due to the divergence of \( \int_{-\infty}^{-2} 1 \, dx \).

Key Concepts

Antiderivative

When working with integrals, finding the antiderivative is a key step. The antiderivative of a function is a new function whose derivative is the original function. For the exercise above, the antiderivative for the term \( \frac{3}{x^3} \) was determined to be \( -\frac{3}{2x^2} \). This means that if you differentiate \( -\frac{3}{2x^2} \), you will obtain \( \frac{3}{x^3} \). It feels a bit like 'undoing' the differentiation process.
For the constant function inside the integral, the antiderivative is simply \( x \), since differentiating \( x \) gives back the constant 1. By recognizing and utilizing antiderivatives, we gained the ability to compute the integrals over certain intervals.

Convergence

Convergence is an essential concept when evaluating improper integrals. An integral converges when its limits yield a specific, finite result. In the step-by-step solution, we managed to find a convergent result for the sub-integral \( \int_{-\infty}^{-2} \frac{3}{x^3} \, dx \). This means its result, \( -\frac{3}{8} \), is a well-defined and finite number.
Convergence is crucial because it tells us whether a particular part of our mathematical model is capable of producing meaningful values. Much of calculus concerns itself with exploring which functions and intervals lead to convergence.

• Helps determine if a specific integral can yield a meaningful number.
• Is a necessary check especially when dealing with infinite limits.
• Guides us to manipulate partial solutions where some parts are convergent.

Divergence

Divergence is when an integral does not result in a finite value, often due to the infinite limits. In the given exercise, the integral \( \int_{-\infty}^{-2} 1 \, dx \) diverges. This is because as \( a \to -\infty \), the antiderivative expression \( -2 - a \) does not approach a finite number, but rather becomes infinitely negative.
For mathematicians and engineers, divergence is critical because it indicates that a certain condition or function cannot produce a usable numerical answer over the given interval. Identifying divergence helps in deciding alternative mathematical methods to employ, or intervals to consider.

• Divergence shows where integrals cannot remain finite.
• Often leads mathematicians to reevaluate their model or approach.
• Understanding divergence can prevent inaccurate interpretation of infinite intervals.

Limit of Integration

The limit of integration is the boundary or extreme values applied during the integration process. In this exercise, we tackled limits \(-\infty\) to \(-2\), meaning one of our bounds extends to negative infinity. In practice, limits like these require careful analysis. When limits extend toward infinity, they demand the use of limits in calculus, borrowing concepts from sequences and series, to evaluate them properly.
Recognizing how the limit of integration impacts the calculation is imperative. In earlier steps, we computed limits by analyzing how each part behaves as \( a \to -\infty \). This process is essential in understanding both convergence and divergence, serving as preparation for recognizing finite or infinite final outcomes.