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Chapter 7: Problem 87

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Graph the function \(f(x)=\frac{\sqrt{x^{2}-9}}{x}\) and consider the region bounded by the curve and the \(x\) -axis on \([-6,-3] .\) Then evaluate \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} d x .\) Be sure the result is consistent with the graph.

Short Answer

Expert verified
**Answer:** The area enclosed between the curve and the x-axis on the interval \([-6, -3]\) is infinite.

Step by step solution

01

Define the function

Our given function is \(f(x) = \frac{\sqrt{x^2-9}}{x}\).
02

Identify the integral to solve

We need to find the integral \(\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} dx\) to determine the area enclosed between the curve and the x-axis on the interval \([-6, -3]\).
03

Use substitution

Let's find the correct substitution for our case. Since we are looking for the area in the interval \([-6,-3]\), we can say that \(x\leq -3\). Therefore, using \(x=-3\sec\theta\) (with \(a=3\)), as \(x \leq -a\) implies \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\).
04

Compute the differentials

Now, let's compute the differentials. Differentiating both sides of \(x=-3\sec\theta\), we have \(dx = -3\sec\theta\tan\theta d\theta\).
05

Replace \(x\) and \(dx\) in the integral

With the substitution and differentials calculated, we can now transform our integral as follows: $$\int_{-6}^{-3} \frac{\sqrt{x^{2}-9}}{x} dx =-\int_\frac{\pi}{2}^{\pi} \sqrt{(-3\sec\theta)^2-9} \cdot \frac{-3\sec\theta\tan\theta d\theta}{-3\sec\theta}$$ After some simplifications, the integral can be represented as: $$-\int_\frac{\pi}{2}^{\pi} \frac{\sqrt{9\sec^2\theta-9}}{\sec\theta}\sec\theta\tan\theta d\theta$$
06

Simplify and solve the integral

Now, let's simplify and solve the integral. $$-\int_\frac{\pi}{2}^{\pi} \frac{\sqrt{9(\sec^2\theta-1)}}{\sec\theta}\sec\theta\tan\theta d\theta = -\int_\frac{\pi}{2}^{\pi} 3\tan\theta d\theta$$ Integrate with respect to \(\theta\): $$-\left[3(\ln|\sec\theta+\tan\theta|)\right]_\frac{\pi}{2}^{\pi}$$ The result of this integral is: $$=-3([\ln|\sec\pi+\tan\pi|]-[\ln|\sec\frac{\pi}{2}+\tan\frac{\pi}{2}|])=-3(\ln1-\ln\infty)$$
07

Evaluate the area

Since we can't take \(\ln\infty\), it's clear that the area enclosed between the curve and the x-axis on the interval \([-6, -3]\) is infinite.
08

Graph the function

Finally, let's graph the function \(f(x)=\frac{\sqrt{x^2-9}}{x}\). The graph will verify that the function has asymptotic behavior as x approaches -3, which is consistent with the infinite area result. Note: To graph the function, you can use graphing software such as Desmos or GeoGebra. The graph will display a vertical asymptote at \(x=-3\), which is consistent with the infinite area result.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Substitution
Trigonometric substitution is a powerful tool for evaluating integrals, especially when dealing with square roots of quadratic expressions. In essence, this technique involves replacing an algebraic expression with a trigonometric identity to simplify the integration process.

For example, when encountering an expression like \(\sqrt{x^2 - a^2}\), you can use the substitution \(x = a\sec\theta\) for \(x \geq a\), or \(x = -a\sec\theta\) for \(x \leq -a\). This transforms the square root into a trigonometric form that is often easier to integrate.

By making such a substitution, we're able to exploit the Pythagorean identities of trigonometry, simplifying the integrand into a form that is more straightforward to integrate. Once the integral is found, we then inverse the substitution to return to the original variable.
Definite Integral
A definite integral represents the accumulation of quantities, and in the context of calculus, it provides the net area under a curve between two points. The definite integral is formulated as \(\int_{a}^{b} f(x)dx\), where \(f(x)\) is the integrand, and \(a\) and \(b\) are the lower and upper limits of integration, respectively.

Calculating a definite integral involves finding the antiderivative (or primitive) of the function and then applying the Fundamental Theorem of Calculus, which states that the definite integral of a function can be evaluated by taking the difference of its antiderivative at the upper and lower limits.
Asymptotic Behavior
Asymptotic behavior in mathematics describes how a function behaves as its input goes towards infinity or towards a particular point. An asymptote itself is a line that the graph of the function approaches but never actually touches.

In the given problem, the function \(f(x) = \frac{\sqrt{x^2-9}}{x}\) displays such behavior as the variable \(x\) approaches -3. The function becomes unbounded, and the values of \(f(x)\) grow without limit. This turnout is indicative of a vertical asymptote at \(x = -3\), which is confirmed by the infinite area result obtained through the definite integral performed with trigonometric substitution and clearly demonstrated in a graph of the function.
Area Under a Curve
The area under the curve of a given function over a certain interval refers to the space confined between the graph of the function and the horizontal axis (usually the \(x\)-axis). This area can be positive or negative, depending on whether the graph lies above or below the axis.

To compute this area, one typically uses the definite integral of the function over the specified interval. The problem presents an unconventional case where the area under the curve is infinite due to the asymptotic behavior of the function as it approaches a vertical asymptote. This conveys that integrals do not always yield finite numbers; they also convey meaningful insights into the nature of functions over their domains.

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