Problem 21 - Calculus Single Variable

Question

In each of Exercises \(21-36,\) determine whether the given improper integral converges or diverges. If it converges, then evaluate it. $$ \int_{-\infty}^{-2} x^{-3} d x $$

Step-by-Step Solution

Answer

The integral converges and its value is \(-\frac{1}{8}\).

Step 1: Identify the Type of Integral

The integral given is \( \int_{-\infty}^{-2} x^{-3} \, dx \). This is an improper integral of the first kind, as the lower limit is \(-\infty\). To address this, the integral can be evaluated as \( \lim_{a \to -\infty} \int_{a}^{-2} x^{-3} \, dx \).

Step 2: Compute the Antiderivative

Find the antiderivative of \( x^{-3} \). This is given by \( \int x^{-3} \, dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2} \).

Step 3: Evaluate the Definite Integral

With the antiderivative found, we evaluate the integral from \( a \) to \( -2 \) as \( \lim_{a \to -\infty} \left[ -\frac{1}{2x^2} \right]_{a}^{-2} = \lim_{a \to -\infty} \left( -\frac{1}{2(-2)^2} - \left(-\frac{1}{2a^2}\right) \right) \).

Step 4: Simplify and Evaluate the Limit

First, simplify \( -\frac{1}{2(-2)^2} = -\frac{1}{8} \). Then, evaluate the limit: \( \lim_{a \to -\infty} \frac{1}{2a^2} = 0 \) since \( a^2 \to \infty \). Thus, \( -\frac{1}{8} - 0 = -\frac{1}{8} \).

Step 5: Conclusion: Determine Convergence and Value

The original improper integral \( \int_{-\infty}^{-2} x^{-3} \, dx \) converges, and its value is \(-\frac{1}{8} \).

Key Concepts

Improper Integral

An improper integral in calculus is an integral that has at least one infinite limit of integration or an integrand that becomes infinite within the range of integration. These integrals pose challenges that require special techniques to evaluate. To handle these, we often use limits to "bridge the gap" that infinity creates.

For instance, if you encounter an integral like \( \int_{-\infty}^{a} f(x) \, dx \), you can rewrite it using a limit like this:

• \( \lim_{b \to -\infty} \int_{b}^{a} f(x) \, dx \)

This approach allows us to work into a form that's easier to analyze and compute.

Improper integrals are essential because they allow us to explore areas under curves that extend indefinitely or approach undefined points, expanding our understanding of real-world phenomena.

Antiderivative

Antiderivatives are a crucial part of calculus. They allow us to "reverse" the operation of differentiation, turning derivatives back into functions and thereby making the calculation of integration possible.

An antiderivative of a function \( f(x) \) is another function \( F(x) \) such that \( F'(x) = f(x) \). In simpler terms, if you differentiate \( F(x) \), you get \( f(x) \).

• For example, the antiderivative of \( x^{-3} \) is \( -\frac{1}{2x^2} \), since the derivative of this expression gives us the original \( x^{-3} \).

Grasping antiderivatives is key to solving integrals effectively, especially when dealing with indefinite integrals, where no limits are provided. With definite integrals, they also help in finding areas under curves in a specific range.

Convergence

Convergence in the context of improper integrals is about determining whether an integral has a finite value as limits approach infinity or negative infinity. It helps us conclude whether the integral yields a real number or if it diverges, extending to infinity.

In evaluating convergence, limits play a central role. If the limit of the integral as it approaches infinity exists and is finite, the integral is said to converge. On the other hand, if the limit is infinite or does not exist, the integral diverges.

• In our example, the integral \( \int_{-\infty}^{-2} x^{-3} \, dx \) converges due to the limit evaluation, which results in a finite number \(-\frac{1}{8}\).

Understanding convergence allows us to make sense of functions and integrals that initially seem too complex or unbounded.

Definite Integral

A definite integral is an integral that has set upper and lower limits. It is used to find the exact area under a curve from one point to another along the x-axis. This area represents the accumulation of values of the function between these two points.

The definite integral is expressed as \( \int_{a}^{b} f(x) \, dx \), with \( a \) and \( b \) being the bounds. The fundamental theorem of calculus connects the derivative with the definite integral, showing how antiderivatives help us calculate these integrals efficiently.

• In our context, after finding the antiderivative of \( x^{-3} \), we evaluate it between specific limits to solve the integral.

Understanding definite integrals is essential for solving real-world problems involving rates of change and accumulated quantities over intervals.