Problem 340 - Calculus Volume 1

Question

If you are unable to find intersection points analytically in the following exercises, use a calculator. Find the area of the hyperbolic quarter-circle enclosed by \(x=2\) and \(y=2\) above \(y=1 / x\).

Step-by-Step Solution

Answer

The area is \(3 - \ln(4)\), which simplifies to \(3 - 2\ln(2)\).

Step 1: Understand the Problem

We need to find the area of the region enclosed by the hyperbolic quarter-circle with the lines \(x=2\) and \(y=2\), above the hyperbola \(y=\frac{1}{x}\). The region is to the first quadrant under these constraints.

Step 2: Setup the Relevant Graphical Region

The graph involves the lines \(x=2\), \(y=2\), and \(y=\frac{1}{x}\). The region we are considering is the part in the first quadrant that is above the hyperbola and bounded by the lines. It's the small quarter-circle shape visible on the graph.

Step 3: Find the Points of Intersection

To find where the hyperbola \(y=\frac{1}{x}\) and the boundaries intersect, we solve \(y=2\) for \(\frac{1}{x} = 2\) giving us \(x=0.5\), and \(x=2\) for \(y=\frac{1}{2}\). Hence, the intersections are \((0.5, 2)\) and \((2, 0.5)\).

Step 4: Set Up the Integral

We need to integrate with respect to \(x\) from 0.5 to 2, the expression \(2 - \frac{1}{x}\), which is the line \(y=2\) minus the hyperbola \(y=\frac{1}{x}\). This integral will give us the area of interest.

Step 5: Calculate the Integral

Evaluate the integral: \[\int_{0.5}^{2} \left(2 - \frac{1}{x}\right) \, dx = \int_{0.5}^{2} 2 \, dx - \int_{0.5}^{2} \frac{1}{x} \, dx\]The evaluations are:- \(\int_{0.5}^{2} 2 \, dx = 2[x]_{0.5}^{2} = 2(2) - 2(0.5) = 3\)- \(\int_{0.5}^{2} \frac{1}{x} \, dx = [\ln|x|]_{0.5}^{2} = \ln(2) - \ln(0.5) = \ln(4)\).Thus, the area is: \[3 - \ln(4)\]

Step 6: Simplify the Result

The final step is simplifying the expression obtained: \[3 - \ln(4)\].Given \(\ln(4) = 2\ln(2)\), the area can also be represented as \[3 - 2\ln(2)\].

Key Concepts

Integration Techniques

Integration techniques are crucial tools in calculus, helping us solve problems that involve finding areas, among other things. In our exercise, we need to integrate the function given by the region between the line and the hyperbola.

There are several methods to perform integrations, such as substitution, integration by parts, and partial fractions. However, when dealing with straightforward functions like those in our problem, the direct application of the fundamental theorem of calculus and basic integration rules often suffices.

• The fundamental theorem of calculus allows us to find the area under a curve by evaluating antiderivatives at given limits.
• In this problem, we break the integral into two parts: integrating a constant and integrating a hyperbolic-like function, specifically \(\frac{1}{x}\). This simplifies computations significantly.
• Remember, when integrating \(\frac{1}{x}\), the result is \ln|x|\, a key point that often comes up in similar problems.

Understanding these techniques enables a more seamless calculation, providing a clear pathway to solving for areas and other integrals.

Area Under a Curve

Finding the area under a curve is a fundamental application of integration. It involves calculating the integral between two points, which gives the net area between the curve and the x-axis over that interval.

In our hyperbolic quarter-circle problem, the area is computed based on the region bounded by specified lines and curves. We want the area between the line \y=2\ and \(y=\frac{1}{x}\) from the points of intersection found.

• First, identify the bounded region. The intersections are critical here, providing the limits for our integral (0.5 and 2).
• Set up the integral by subtracting the lower function from the upper function within those limits. Here, it means integrating \2 - \frac{1}{x}\ over \x=[0.5, 2]\.
• This method efficiently calculates not just the area but the exact region of interest as defined by the given boundaries.

By calculating this area, we understand how integrals can provide concrete numerical representations of complex regions.

Hyperbolic Functions

Hyperbolic functions have similarities to trigonometric functions but are distinct in their growth and applications. They are defined using exponential functions and exhibit unique characteristics important in many mathematical contexts.

In our specific problem, the function \(y=\frac{1}{x}\) somewhat resembles hyperbolic curves though not a hyperbolic sine or cosine directly.

• The curve \y=\frac{1}{x}\ forms one arm of what can be visualized as a hyperbolic shape, but it is technically a reciprocal function.
• These curves are helpful in illustrating growth rates and decay patterns, often appearing in calculus due to their interesting properties.
• When manipulated within integral setups, they showcase calculus's power in tackling non-linear shapes and identifying bounded areas.

Thus, while not a classic hyperbolic function, \(y=\frac{1}{x}\) provides a framework for understanding more elaborate curves and their properties. Understanding these helps in applying integration to varied mathematical forms and challenges.