Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate \(\int_{-4}^{4} \frac{1}{16-x^{2}} d x\) or show that it diverges.

Step-by-Step Solution

Answer

The integral diverges due to vertical asymptotes at the endpoints.

Step 1: Identify the Function and Limits of Integration

The integral we need to evaluate is \( \int_{-4}^{4} \frac{1}{16-x^{2}} \, dx \). Notice that the function inside the integral, \( f(x) = \frac{1}{16-x^{2}} \), is defined except at points where the denominator is zero, which are \( x = -4 \) and \( x = 4 \). These are the points where the function becomes undefined and may cause the integral to diverge.

Step 2: Analyzing Points of Discontinuity

At \( x = \pm 4 \), the function \( f(x) = \frac{1}{16-x^2} \) becomes undefined because these values make the denominator zero (i.e., \( 16-x^2 = 0 \)). Because these points coincide with the limits of integration, there is a possibility that the integral is divergent.

Step 3: Determine the Type of Singularities

Since the function is undefined at the points \( x = -4 \) and \( x = 4 \), these are vertical asymptotes. We need to analyze if the function has an improper integral due to these vertical asymptotes at the boundaries.

Step 4: Break Down the Integral

To evaluate the improper integral, we split it at the point of discontinuity and take the limit. The integral should be considered as two separate integrals: \( \int_{-4}^{c} \frac{1}{16-x^2} \, dx \) and \( \int_{c}^{4} \frac{1}{16-x^2} \, dx \), where \( c \) is a point between \(-4\) and \(4\). This allows us to examine the behavior of the integral as \( c \) approaches the points of discontinuity.

Step 5: Evaluate Limit Approaches at Singularities

Let's evaluate the limits: - For \( \int_{-4}^{c} \frac{1}{16-x^2} \, dx \) as \( c \to -4^+ \), the integral becomes unbounded.- For \( \int_{c}^{4} \frac{1}{16-x^2} \, dx \) as \( c \to 4^- \), the integral also becomes unbounded.Both integrals are divergent due to the asymptotes at the endpoints.

Step 6: Combine the Results

Since both halves of the split integral diverge, the original integral \( \int_{-4}^{4} \frac{1}{16-x^{2}} \, dx \) diverges as well. An integral is said to diverge if the limit of the integrals evaluated between the points of discontinuity does not exist.

Step 7: Conclusion

After evaluating the improper integral at its points of discontinuity and noticing that it diverges, we conclude that the integral \( \int_{-4}^{4} \frac{1}{16-x^2} \, dx \) does not converge to a finite value; instead, it diverges due to the vertical asymptotes at \( x = -4 \) and \( x = 4 \).

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of calculus focused on the concept of integrals and their properties. In this context, integration is the process of finding the total accumulation of quantities, such as areas under curves. Integrals can be classified into definite and indefinite integrals.

• **Definite Integrals:** These have specific upper and lower limits of integration, like \(\int_{a}^{b} f(x) \, dx\). Definite integrals yield a specific numerical value that represents the area under the curve from \('a'\) to \( 'b'\).
• **Indefinite Integrals:** These do not have limits of integration and produce a general expression representing an antiderivative. E.g., \(\int f(x) \, dx\).

Improper integrals can arise when the function is not bounded or when the limits of integration involve infinity or points of discontinuity. As seen in this exercise, they require careful evaluation using limits at points of discontinuity. If the limits do not exist or suit a logical or finite value, the integral is deemed divergent.

Asymptotes

Asymptotes are lines that a graph approaches but never actually reaches. They are critical in understanding the behavior of functions, especially around points of discontinuity or infinity.

• **Vertical Asymptotes:** These occur when the function f(x) increases or decreases without bound near a particular x-value. In this exercise, vertical asymptotes appear at \(x = -4\) and \(x = 4\) since the function \( f(x) = \frac{1}{16-x^2} \) is undefined at these points due to division by zero.
• **Horizontal Asymptotes:** These represent a constant value that the function approaches as \(x\) tends to infinity or negative infinity.

Understanding asymptotes helps to predict whether an integral might converge or diverge, particularly with improper integrals that involve these asymptotic behaviors at their boundaries.

Limits of Integration

Limits of integration are the values that define the interval over which the integration is carried out. They are represented as the lower and upper bounds of a definite integral, such as \( [a, b]\).

• For the integral in this exercise, the limits are \(x = -4\) and \(x = 4\).
• Considering the behavior of the function at these limits is crucial, especially when dealing with improper integrals. If a function is undefined at one or both of the limits of integration, as \( f(x) = \frac{1}{16-x^2} \) is at \(x=\pm 4\), special techniques involving limits are needed.

These limits are adjusted by breaking the integral into parts and calculating limits as the interval approaches the points of discontinuity. In cases where both parts do not result in a finite value upon evaluation, the integral is deemed divergent, indicating no finite area under the curve between the given limits.