Problem 13 - Calculus Concepts

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{3}{8}\).

Step 1: Identify the Improper Integral

The integral \( \int_{-\infty}^{-2} \frac{3}{x^{3}} \, dx \) is improper because it involves an infinite limit of integration. Specifically, the lower limit of integration is \(-\infty\). Our task is to evaluate this integral.

Step 2: Set up the Limit for Improper Integrals

To manage the infinite limit, we replace \(-\infty\) with a variable, say \(a\), and then take the limit as \(a\) approaches \(-\infty\). The integral becomes: \[ \lim_{a \to -\infty} \int_{a}^{-2} \frac{3}{x^{3}} \, dx \]

Step 3: Find the Antiderivative

Determine the indefinite integral \( \int \frac{3}{x^3} \, dx \). We find the antiderivative by rewriting the integrand as \( 3x^{-3} \), and then integrating to get:\[ \int 3x^{-3} \, dx = 3 \cdot \frac{x^{-2}}{-2} = -\frac{3}{2x^2} + C \]

Step 4: Evaluate the Definite Integral with Limits

Use the antiderivative to evaluate the definite integral from \(a\) to \(-2\):\[ \int_{a}^{-2} \frac{3}{x^{3}} \, dx = \left[ -\frac{3}{2x^2} \right]_{a}^{-2} = -\frac{3}{2(-2)^2} - \left(-\frac{3}{2a^2}\right) \]Simplifying gives:\[ -\frac{3}{8} + \frac{3}{2a^2} \]

Step 5: Take the Limit

We now take the limit as \(a\) approaches \(-\infty\):\[ \lim_{a \to -\infty} \left( -\frac{3}{8} + \frac{3}{2a^2} \right) \]As \(a\) goes to \(-\infty\), \( \frac{3}{2a^2} \) approaches 0. Thus, the limit is:\[ -\frac{3}{8} \]

Key Concepts

Limits of Integration

When dealing with improper integrals, the concept of limits of integration is crucial. The limits define where you start and stop when calculating an integral. Usually, these limits are finite numbers, like 0 to 5.
In this exercise, however, the lower limit is \(-\infty\), an infinite value, which makes the integral improper. To tackle this, imagine a workaround: replace the infinite limit with a variable, such as \(a\), which will later approach \(-\infty\). This transforms your integral into a manageable form. This approach changes our integral to:

• \( \int_{a}^{-2} \frac{3}{x^{3}} \, dx \)

This small adjustment allows us to apply the valuable tools of calculus to solve the problem.

Antiderivative

The term "antiderivative" refers to the opposite of differentiation. It's essentially finding a function whose derivative is the given function. To solve our integral, we first need the antiderivative of \( \frac{3}{x^3} \).
To find this, we rewrite it in a form suitable for basic integration rules:

• The integrand \( \frac{3}{x^3} \) can be rewritten as \( 3x^{-3} \).

Once rewritten, apply the power rule for integration. The power rule states that the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) (as long as \( n eq -1\)):

• \( \int 3x^{-3} \, dx = 3 \cdot \frac{x^{-2}}{-2} = -\frac{3}{2x^2} + C \)

The "+ C" symbolizes that there are infinitely many functions that can have the same derivative. However, for definite integrals, we will focus on evaluation, eliminating \(C\) in the process.

Infinite Limits

When a limit relates to infinity, like in improper integrals, we define it as an infinite limit. Our original problem specifies a lower bound of \(-\infty\). Infinite limits can seem challenging, but we handle them using limits.The expression changes into taking the limit as a variable approaches infinity. For the given exercise:

• \( \lim_{a \to -\infty} \int_{a}^{-2} \frac{3}{x^{3}} \, dx \)

This simply means integrating, then determining the function's behavior as our variable approaches infinity. After obtaining the antiderivative, we apply it over \([-\infty, -2]\) by taking:

• \( [-\frac{3}{2x^2}]_a^{-2} \)

This requires expanding and simplifying to evaluate the limit:

• \( -\frac{3}{8} + \frac{3}{2a^2} \) approaches \( -\frac{3}{8} \) as \( 1/a^2 \) goes to 0.

Thus, our improper integral evaluates to \:

• -\frac{3}{8}

The infinite limit transforms a seemingly complex problem into solvable steps, using limits as a tool.