Question

Sketch the following regions (if a figure is not given and find the area. The region bounded by \(y=8-2 x, y=x+8,\) and \(y=0\) (Use integration.)

Short answer

The area of the region bounded by the given lines \(y=8-2x, y=x+8,\) and \(y=0\) is 56 square units. To find the area, we first sketched the given regions and found the points of intersection at \((0,8), (4,0),\) and \((-8,0)\). Then, we used integration to calculate the area, with the result being -56, but we consider the absolute value, thus making it 56 square units.

Step-by-step solution

Sketch the given regions (if a figure is not given)

The graphs of the three given functions are: - A line \(y=8-2x\) with a slope of \(-2\) and a y-intercept at \((0,8)\). - A line \(y=x+8\) with a slope of \(1\) and a y-intercept at \((0,8)\). - A horizontal line \(y=0\). Now, sketch these three functions on a coordinate plane to have a figure of the given regions.

Find the points of intersection

To find the points of intersection between the lines, set up and solve the following systems: - Intersection of lines 𝑦=8−2𝑥 and 𝑦=𝑥+8: Set the equation equal to each other: \(8 - 2x = x + 8\) Solve for x: \(3x = 0\) \(x = 0\) Substitute the value of x into the equation to find the value of y: \(y = 8 - 2(0) = 8\) So, the point of intersection between these two lines is \((0,8)\). - Intersection of lines 𝑦=8−2𝑥 and 𝑦=0: Set the equation equal to each other: \(8 - 2x = 0\) Solve for x: \(2x = 8\) \(x = 4\) Substitute the value of x into the equation to find the value of y: \(y = 8 - 2(4) = 0\) So, the point of intersection between these two lines is \((4,0)\). - Intersection of lines 𝑦=𝑥+8 and 𝑦=0: Set the equation equal to each other: \(x + 8 = 0\) Solve for x: \(x = -8\) Substitute the value of x into the equation to find the value of y: \(y = (-8) + 8 = 0\) So, the point of intersection between these two lines is \((-8,0)\).

Set up the integral to find the area

To find the area of the region bounded by the three lines, subtract the integral of the lower function from the integral of the upper function over the interval [-8, 4]. In this case, the lower function is \(y=0\) and the upper function changes from \(y=8-2x\) to \(y=x+8\) at the point \((0,8)\) Area = \(\int_{-8}^{4} [(8-2x)-(x+8)]dx + \int_{0}^{4} [(x+8)-0]dx\)

Evaluate the integral

Evaluate the integral: Area = \(\int_{-8}^{0} [(8-2x)-(x+8)]dx + \int_{0}^{4} [(x+8)]dx\) Area = \(\int_{-8}^{0} [-3x]dx + \int_{0}^{4} [(x+8)]dx\) Now, integrate and evaluate on limits. Area = \(\left[-\dfrac{3}{2}x^2\right]_{-8}^0 + \left[\dfrac{1}{2}x^2 + 8x\right]_{0}^{4}\) Area = \(\left[-\dfrac{3}{2}(0)^2 - (-\dfrac{3}{2}(-8)^2)\right] + \left[\dfrac{1}{2}(4)^2 + 8(4) - \left(\dfrac{1}{2}(0)^2 + 8(0)\right)\right]\) Area = \(\left[0 - (96)\right] + \left[8 + 32 - 0\right]\) Area = \(-96 + 40\) Area = -56 square units (Note: the area is negative because of the orientation of the graph, but the area should be considered as 56 square units in absolute value).

Key concepts

Integration

Integration is a fundamental concept in calculus that involves finding the area under a curve within a certain interval. In essence, it is the reverse process of differentiation. By integrating a function, we can compute accumulated quantities, such as distances or areas.

• The integral symbol \( \int \) represents the integration process.
• The limits of integration, \( a \) to \( b \), indicate the interval over which you are finding the area.
• The expression within the integral, often denoted as \( f(x) \), is the function whose area under the curve is to be calculated.

When we integrate, we are essentially adding up an infinite number of infinitely small quantities, which builds up to the total area between a curve and the x-axis. This technique is crucial in calculating areas of more complex shapes that can't be easily decomposed into basic geometric figures.

Area Calculation

Calculating area using integration involves determining the region between the graph of a function and a reference line, usually the x-axis. The process is about computing the definite integral of the function over a specified interval.

• Identify the upper and lower functions that bound the area.
• Set up the integral by placing the upper function minus the lower function within the integral symbol.
• Resolve the integral to find the numerical value of the area.

For instance, in the given problem, multiple integrals are used: one from \(-8\) to \(0\) for part of the region and another from \(0\) to \(4\) for the remaining region. By solving these integrals separately and summing their absolute values, we determine the total area enclosed by the curves.

Functions and Graphs

Functions are mathematical expressions that relate input values to output values. Graphs visually represent these relationships and provide a way to see how changes in input affect output. In this exercise, we are dealing with linear functions.

• Line equations, such as \(y=8-2x\) and \(y=x+8\), describe straight lines in the coordinate plane.
• The slope of a line determines its steepness, and the y-intercept is the point where the line crosses the y-axis.
• Graphs of these lines help visualize intersection points and the enclosed regions necessary for integration.

Understanding how these equations translate into graphs is key to setting up the correct integrals and accurately identifying the areas under study.

Points of Intersection

Finding the points of intersection is crucial when determining the region enclosed by multiple functions. These points determine where integration limits change and are necessary for setting up accurate integrals.

• To find intersections, set the equations of two lines equal and solve for their variables.
• Plug the solved value back into either function to find the corresponding coordinate, giving you the exact point of intersection.

In the exercise, three pairs of equations were solved, giving points at \((0,8)\), \((4,0)\), and \((-8,0)\). These intersections define the vertices of the triangular region under study, ensuring that the correct limits are applied in each part of the area calculation.