Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{-\infty}^{1} \frac{d x}{(2 x-3)^{3}}\)

Step-by-Step Solution

Answer

The integral converges to \(-\frac{1}{4}\).

Step 1: Recognize the type of integral

The integral \( \int_{-\infty}^{1} \frac{dx}{(2x-3)^3} \) is an improper integral because the lower bound is negative infinity. We'll need to check whether this integral converges or diverges.

Step 2: Introduce a limit to handle the infinity

To deal with the infinity, rewrite the integral as a limit: \( \lim_{a \to -\infty} \int_{a}^{1} \frac{dx}{(2x-3)^3} \). You'll now solve the integral by evaluating this limit.

Step 3: Use substitution to simplify the integral

Let \( u = 2x - 3 \), which gives \( du = 2dx \) or \( dx = \frac{1}{2}du \). Change the limits accordingly: when \( x = a, u = 2a - 3 \) and when \( x = 1, u = -1 \). The integral becomes \( \frac{1}{2} \lim_{a \to -\infty} \int_{2a-3}^{-1} \frac{du}{u^3} \).

Step 4: Integrate the function

The integral \( \int \frac{1}{u^3} \, du = -\frac{1}{2u^2} + C \). Apply this result to the limits \( \frac{1}{2} \lim_{a \to -\infty} \left[ -\frac{1}{2} \left( \frac{1}{u^2} \right) \right]_{2a-3}^{-1} \).

Step 5: Evaluate the definite integral

Substitute the limits into the antiderivative: \[= \frac{1}{2} \lim_{a \to -\infty} \left[ -\frac{1}{2(-1)^2} + \frac{1}{2(2a-3)^2} \right] = \frac{1}{2} \lim_{a \to -\infty} \left[ -\frac{1}{2} + \frac{1}{2(2a-3)^2} \right].\]

Step 6: Determine the convergence of the limit

Observe that as \( a \to -\infty \), the term \( \frac{1}{2(2a-3)^2} \) approaches zero. Therefore, the limit becomes \( \frac{1}{2}(-\frac{1}{2}) \).

Step 7: Calculate the final result

Finally, compute \( \frac{1}{2}(-\frac{1}{2}) = -\frac{1}{4} \). Thus, the improper integral converges and equals \(-\frac{1}{4}\).

Key Concepts

Convergence

When working with improper integrals, one of the first things you'll want to determine is whether the integral converges or diverges. Convergence means that the integral evaluates to a specific, finite number. Conversely, divergence implies the integral does not settle on a finite value. In mathematical terms, for the integration to converge, the limit you compute in the evaluation process must exist and be finite.

To identify convergence, you often introduce a limit to replace the bounds that involve infinity. In our exercise, we rewrite the integral with a limit as the lower bound approaches negative infinity. This step does not solve the integral itself, but rather sets up a pathway to determine whether the solution will converge to a meaningful answer.

Working with these concepts not only helps in this specific exercise but builds your understanding of evaluating how different types of integrals behave.

Integration by Substitution

Sometimes referred to as "u-substitution," integration by substitution is a key strategy for simplifying integrals. You temporarily replace a part of the integrand with a new variable, making the integral easier to solve.

In our problem, the substitution method involved letting the variable \( u = 2x - 3 \). This means the differentials change as well, indicated by \( du = 2dx \) or equivalently \( dx = \frac{1}{2} du \).

• Convert the limits for \( x \) to limits for \( u \).
• Simplify the integrand so it is more manageable for integration.

Post substitution and limit conversion, you reduced the integral to an expression that is much more straightforward to evaluate. Substitution often acts as a bridge to turn complex integrations into standard forms you already know how to handle.

Limits

Limits play a crucial role in making sense of improper integrals. They offer a way to "approach" infinity or some other problematic points systematically. Rather than working with undefined values directly, limits provide a framework to ascertain behavior as these bounds are approached.

In the exercise, after employing substitution, we arrive at the expression \( \frac{1}{2} \lim_{a \to -\infty} \int_{2a-3}^{-1} \frac{du}{u^3} \). This expression doesn't just cover the integration but also examines the behavior of the integral as \( a \) tends to negative infinity.

• Setting limits ensures that any infinities are approached logically.
• Limits allow observation of trends within the integral's outputs.

If the limit yields a finite number, then the integral converges. If it doesn't, the integral diverges, underlining why understanding limits is essential in calculus.

Antiderivative

Calculating an antiderivative, or indefinite integral, is another crucial step in solving definite integrals, especially when indefinite bounds are involved. An antiderivative is essentially a function whose derivative gives the original function you're working us.

In this problem, after performing substitution, the task was to integrate \( \frac{1}{u^3} \) with respect to \( u \). The antiderivative of \( \frac{1}{u^3} \) is \( -\frac{1}{2u^2} \), plus the constant of integration \( C \). Though not needed directly here for definite integrals, it's essential to remember their existence.

• Transform indefinite integrals into evaluable forms.
• Use them to apply limits and solve complex integral problems.

Antiderivatives allow us to ratify and apply boundaries effectively, transforming abstract mathematical problems into concrete solutions.