Problem 21 - Calculus of a Single Variable

Question

Finding the Area of a Region In Exercises \(17-30,\) sketch the region bounded by the graphs of the equations and find the area of the region. $$ y=x, \quad y=2-x, \quad y=0 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the equations \(y = x\), \(y = 2 - x\), and \(y = 0\) is 1 square unit.

Step 1: Sketch the Region

Plot the three given equations on the same graph: \(y = x\), \(y = 2 - x\), and \(y = 0\) (which is the x-axis). The region bounded by these three equations is a triangular region.

Step 2: Identify the Intersection Points

Solve \(y = x\) and \(y = 2 - x\) simultaneously to find the points where these two lines intersect. Here, setting \(x\) for \(y\) in both equations and solving gives \(x = 1\). Therefore, these two lines intersect at the point \(x = 1\). So the region is bounded by \(x = 0\) and \(x = 1\).

Step 3: Develop the Area Expression

The area A of the region can be found by integrating the difference of the two functions from the left intersection point to the right intersection point: \(A = \int_{0}^{1}(2 - x - x) dx = \int_{0}^{1} (2 - 2x) dx \)

Step 4: Evaluate the Integral

Evaluate the integral through step by step integration. The antiderivative of \(2 - 2x\) is \(2x - x^2\). So, \(A = [2x - x^2]_{0}^{1} = (2*1 - (1)^2) - (2*0 - (0)^2) = 1\)

Key Concepts

Area Between Curves

Finding the area between curves is a fundamental concept in calculus and involves calculating the space that lies between two or more functions over a specified interval. This can help in understanding many real-world problems where differences between two conditions or states are important. To find the area between two curves, one typically follows these steps:

• Identify the functions that bound the region of interest.
• Determine the points of intersection of these functions – the limits for integration.
• The area is found by integrating the difference of the two functions over the interval determined by their intersection points.

For our specific problem, the area bounded by the curves \( y = x \), \( y = 2 - x \), and \( y = 0 \) (the x-axis), forms a triangular region. Here, we find the area by integrating the difference \( (2 - x - x) \) from \( x = 0 \) to \( x = 1 \), leading to the integral \( \int_0^1 (2 - 2x)\,dx \). Simplifying this result provides the area of the specified region.

Intersection Points

Finding intersection points is essential to accurately determine the boundaries of the region for which we want the area. When two functions intersect, they typically do so at one or more points, unless they are equivalent (coincident). These points help define the interval where we conduct our operations, like integration.

To find the intersection of \( y = x \) and \( y = 2 - x \), we set these equations equal to each other: \( x = 2 - x \). Solving for \( x \) gives us \( x = 1 \), meaning the two lines intersect at the point \( x = 1 \), as seen in the original solution. Additionally, the x-axis (\( y = 0 \)) intersects at \( x = 0 \) when \( y = x \) is solved for \( y = 0 \). Knowing these points (\( x = 0 \) and \( x = 1 \)) is crucial, as they provide the endpoints for our definite integral.

Graph Sketching

Sketching graphs of the functions involved is a valuable skill that helps visualize the problem, offering clear insights into the region enclosed. This step is essential for identifying the type of region (such as triangular, trapezoidal, etc.) formed by the intersection of the curves. Upon seeing the graph, we can better understand how the integration works step-by-step.

In our example, sketching the graphs of \( y = x \), \( y = 2 - x \), and \( y = 0 \) reveals a triangular shape. Each equation represents a straight line. The line \( y = x \) has a slope of 1, passing through the origin. Line \( y = 2 - x \) has a slope of -1, intersecting the y-axis at 2. Finally, the line \( y = 0 \) is the x-axis itself. Visualizing this makes it easy to see that the area we seek indeed fits between these intersecting lines, corresponding to a triangular region with its base along the x-axis from 0 to 1.