Problem 55 - Calculus

Question

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 \(3 + \ln(2)\).

Step 1: Identify the Boundaries

The problem describes a region bounded by the lines \(x = 2\) and \(y = 2\), above the curve \(y = \frac{1}{x}\). We need to focus on the area formed between these boundaries and the curve.

Step 2: Determine the Intersection Points

We need to find where the curve \(y = \frac{1}{x}\) intersects with both the line \(y = 2\) and the line \(x = 2\). Solving these equations gives the intersections: \((0.5, 2)\) for \(y = 2\) and \((2, 0.5)\) for \(x = 2\). However, relevant intersections for our area are within the first quadrant, only \((2, 0.5)\).

Step 3: Define the Limits of Integration

The limits of integration for \(x\) are from the intersection with the y-axis at \(x = 0.5\) to \(x = 2\). So, we will integrate from \(x = 0.5\) to \(x = 2\).

Step 4: Set Up the Integral for Area

The area of the region above \(y = \frac{1}{x}\) is given by the integral of \(2 - \frac{1}{x}\) over the interval from \(x = 0.5\) to \(x = 2\). The integral to be solved is: \[\int_{0.5}^{2} \left(2 - \frac{1}{x}\right) \, dx.\]

Step 5: Calculate the Integral

Let's calculate the integral: \[\int_{0.5}^{2} \left(2 - \frac{1}{x}\right) \, dx = \left[2x - \ln|x|\right]_{0.5}^{2}.\]Evaluating this, we find: \[(4 - \ln(2)) - (1 - \ln(0.5)) = 3 + \ln(2).\]

Step 6: Simplify the Result

The area of the hyperbolic quarter-circle above \(y = \frac{1}{x}\) and enclosed by the lines \(x = 2\) and \(y = 2\) is \(3 + \ln(2)\).

Key Concepts

Integration Limits

When working with integrals, one of the first steps is to set the integration limits correctly. Integration limits define the start and end points of the region we want to examine on the x-axis.
In our exercise, we identified these limits by finding the intersection points of the curve with the boundary lines of the region. The curve given is \(y = \frac{1}{x}\), and the boundaries are \(x = 2\) and \(y = 2\).
To determine the integration limits for our region, we found the relevant intersection points, which are (2, 0.5) for \(x=2\) and the projection of the curve on the x-axis \(x=0.5\) when \(y=2\). Thus, the integral is evaluated from \(x = 0.5\) to \(x = 2\).

• Integration limits are crucial for defining the scope of an integral.
• Finding intersection points helps in setting these limits accurately.
• Always double-check that the limits reflect the actual region being analyzed.

Definite Integral

A definite integral calculates the area under a curve and between boundary limits. It is quantified by evaluating the integral between specified limits.
In our scenario, the definite integral is used to find the area above the curve \(y = \frac{1}{x}\) and between \(x=0.5\) and \(x=2\). The function to integrate is \(2 - \frac{1}{x}\), representing the distance from the line \(y=2\) to the curve.
This generates the integral \(\int_{0.5}^{2} (2 - \frac{1}{x}) \, dx\). Solving this integral involves anti-differentiation and applying the fundamental theorem of calculus.

• Definite integrals give a numerical value representing an area or accumulated quantity.
• The process involves finding an antiderivative for the function within the limits.
• These integrals are different from indefinite integrals, which include a constant of integration.

Area Under Curve

Finding the area under a curve is a frequent application of integration in calculus. This concept helps in determining the space beneath the curve and above the x-axis, or other reference lines.
Our task was to find the area above \(y = \frac{1}{x}\) and below the line \(y=2\), bounded by \(x=0.5\) and \(x=2\). By setting up the integral \(\int_{0.5}^{2} (2 - \frac{1}{x}) \, dx\), we have described the area in question.
The result of this evaluated integral, \(3 + \ln(2)\), gives us the precise measure of this "hyperbolic quarter-circle" area.

• Integration is used to find the area under curves, capturing the total space below a curve.
• When calculating these areas, be sure to include all boundaries and apply the correct limits.
• This concept is foundational in physical sciences for studying various phenomena.