Question

Sketch the following regions (if a figure is not given) and then find the area. The region bounded by \(y=x, y=1 / x, y=0,\) and \(x=2\)

Short answer

Answer: The area bounded by the given functions is approximately 1.193 square units.

Step-by-step solution

Sketch the Regions

First, we need to sketch the given regions. The regions are defined by the equations \(y=x\), \(y=1/x\), \(y=0\) (x-axis), and \(x=2\) (vertical line). Plot these functions and identify the enclosed region.

Find Intersection Points

Next, we need to find the intersection points of these functions. The intersection points will determine the limits of integration in the later steps. Intersection of \(y=x\) and \(y=1/x\): \( x = \frac{1}{x} \Rightarrow x^2 = 1\). We get two values of x: \(x = 1\) and \(x = -1\). But since, x is between \(0\) and \(2\), the value of \(-1\) is neglected and we use \(x=1\) as the intersection point of the two functions. Here, the intersection points are \((1, 1)\), \((2, 0)\), and \((2, 1/2)\).

Integration Setup

To find the area bounded by the given functions, we need to set up the integral of the difference between the functions. We will integrate with respect to x from 0 to 2. The first function is \(y=x\) (from x=0 to x=1) and the second is \(y = 1/x\) (from x=1 to x=2). So, the area enclosed by these functions is given by: \(A = \int_0^1 (x - 0) \, dx + \int_1^2 (1/x - 0) \, dx\)

Calculate the Integrals

Now we will evaluate the integrals to find the area. \(\int_0^1 x \, dx = \frac{1}{2}x^2 \Big|_0^1 = \frac{1}{2}\) \(\int_1^2 \frac{1}{x} \, dx = \ln|x| \Big|_1^2 = (\ln 2 - \ln 1) = \ln 2\)

Final Calculation

Adding the two integrals to find the total area: \(A = \frac{1}{2} + \ln 2 \approx 1.193\) So the area bounded by the given functions is approximately 1.193 square units.

Key concepts

Integration

Integration is a fundamental concept in calculus that involves finding the area beneath a curve or between curves on a graph. Essentially, it's the opposite of differentiation. In the context of the exercise, integration is used to measure the exact area bounded by the curves and lines given by the functions: \(y=x\), \(y=\frac{1}{x}\), \(y=0\), and \(x=2\). Each area between curves can be calculated using definite integrals, which is a technique that requires setting up limits of integration that correlate with specific intervals over the x-axis. Here’s a breakdown of why integration is important:

• It calculates the accumulated quantity, such as areas under a curve, which may not be easily determined using basic geometry.
• In this exercise, integration helps us find the precise area enclosed by diverse lines and curves.
• Understanding how to set up an integral according to the graph of the functions is key to solving it accurately.

To determine the regions to integrate, observe the function changes: from \(0\) to \(1\) for \(y=x\) and from \(1\) to \(2\) for \(y=\frac{1}{x}\). Once these limits are set, we can proceed to evaluate each integral.

Area under curves

Finding the area under curves is a specific application of integration that helps us understand the size of regions defined by functions on a graph. The idea is straightforward: if you have a curve on a xy-plane, the area between this curve and the x-axis represents a physical quantity like distance or accumulated change over time. The exercise specifically targets finding the combined area of places where functions overlap and create spaces between them and the x-axis. The process involves:

• Identifying the graphs and sketching them to visualize where they intersect and create boundaries.
• Calculating areas of segments separately since each function can make a distinct region.
• Adding the areas together gives the total area confined within the specified boundaries.

By evaluating the definite integrals over the chosen intervals, you consolidate these tiny pieces of area into a total sum which expresses the area enclosed. This methodical partitioning and summation mirrors what would happen if we use tiny vertical lines from the x-axis up to the curve and sum all those small rectangles.

Intersection points

Intersection points are crucial in calculus, especially when dealing with the integration of areas under curves. These points tell us exactly where two or more functions meet on a graph, and they serve as natural boundaries for integration. In the original exercise, finding the intersection between \(y=x\) and \(y=\frac{1}{x}\) reveals the places where the area changes boundary.The process of finding intersection points involves:

• Setting the equations of the two functions equal to find where they intersect.
• Solving these equations provides points that serve as limits for integration.

For example, solving \(x = \frac{1}{x}\) gives \(x^2 = 1\), yielding \(x=1\) or \(x=-1\). In the context of this problem:

• Only \(x=1\) is considered since it's within the constraints \(0 \leq x \leq 2\).
• Such intersection boundary points, along with endpoints of the functions, direct the setup of proper definite integrals.

Recognizing intersection points helps in determining the regions to calculate and ensures no miscalculation in finding the precise area between curves.