Question

Find the area of the region bounded by the graphs of the equations. Use a graphing utility to verify your results. $$ y=4 / x \quad y=0, \quad x=1, \quad \text { and } x=3 $$

Short answer

The area of the region bounded by the graphs of the equations \(y = 4/x\), \(y = 0\), \(x = 1\), and \(x = 3\) is approximately 4.3944.

Step-by-step solution

Define the Bounded Region

Firstly, draw the graph of the equation \( y = 4/x \), then highlight the area bounded by the curve, the x-axis (\(y=0\)), and the lines \( x=1 \) and \( x=3 \). This will give a visual understanding of the bounded region we are trying to find.

Set up the Integral

The bounded region is within the curve above the x-axis; hence, it is appropriate to use the formula for the area \(A\) between a function and the x-axis \(A = \int_{a}^{b} f(x) dx\). In this case, \( f(x) = 4/x \) and the boundaries are \( a=1 \) and \( b=3 \). Hence, the area of the region is given by \[ A = \int_{1}^{3} (4/x) dx \].

Evaluate the Integral

The integral can be simplified as: \[ A = [4 \ln|x|]_{1}^{3} \]. Evaluating this at the limits of integration gives \[ A = 4 \ln|3| - 4 \ln|1| = 4 \ln 3 - 0 = 4 \ln 3 \].

Approximate the Area

Using common logarithm properties, \( \ln 3 \) is approximately 1.0986. Multiply this by 4 to get the final area approximately equal to 4.3944. This is the area of the region bounded by the given equations.

Key concepts

Definite Integral

The definite integral is a fundamental concept in calculus. It's used to calculate the area under a curve between two points on the x-axis—known as the interval. In this exercise, the goal is to find the area between the curve defined by the function \(y = \frac{4}{x}\) and the x-axis, from \(x = 1\) to \(x = 3\). To solve it, we use the notation \(\int_{a}^{b} f(x) \, dx\) to represent the area under the curve \(f(x)\) from \(x = a\) to \(x = b\). Here, the function \(f(x) = \frac{4}{x}\), and \(a = 1\), \(b = 3\). By setting up this integral and evaluating it, the integral gives a numerical value that represents the total area of interest. Integrals are powerful because they allow us to account for changing slopes and undulating curves.

Area Under Curve

Understanding the area under a curve is crucial because it often depicts actual physical quantities, such as distance, probability, or even work done. Here, the curve is the hyperbolic function \(y = \frac{4}{x}\). Imagine a curve draping over a two-dimensional space; the area under it between two vertical lines (our boundary values \(x=1\) and \(x=3\)) is what we want to calculate. This area provides a concrete value that translates into real-world applications. For instance, this might be used to understand how quickly a fluid is dispersing or how fast a process is slowing down over time. Calculating this involves, as outlined, finding the integral of the curve from the specified start and end points along the x-axis.

Graphing Utility

A graphing utility is a digital tool that helps visualize functions and their integrals effectively. Using this in our exercise helps confirm the shape and orientation of the curve defined by \(y = \frac{4}{x}\). Plotting helps ensure that the curve behaves as expected—meaning it matches the theoretical behavior of hyperbolas. By entering the function into a graphing utility, you get a real-time visual depiction. This confirms the bounded area's location, ensuring each step of the integral is calculated for the correct parameters. It also aids in verifying that the mathematical integration aligns with the visual representation by shading the area and cross-checking values with the integral's result.

Boundary Values

Boundary values are the points where the integration starts and stops. They define the limits of our integral. In the exercise, the boundaries \(x = 1\) and \(x = 3\) indicate where our area-of-interest lies. When dealing with curves, these values are often where the function intersects the axes or other lines, forming the 'corners' of the area we're looking for. These constraints are crucial because they determine the exact section of the infinite curve contributing to the defined area. Knowing where to start and stop integrations ensures precision and avoids overestimating or underestimating the area. By only integrating between these points, we ensure our calculations match the physical scenario we're examining.