Question

Area versus net area Graph the following functions. Then use geometry (not Riemann sums) to find the area and the net area of the region described. The region between the graph of \(y=4 x-8\) and the \(x\) -axis, for \(-4 \leq x \leq 8\)

Short answer

Answer: The area of the region is \(90\), and the net area is \(42\).

Step-by-step solution

Graph the given function

First, we will graph the linear function, \(y = 4x - 8\), within the interval \(-4 \leq x \leq 8\). We can also identify the slope and the y-intercept of this equation. The slope is \(4\), and the y-intercept is \(-8\). To find the points where the graph intersects the x-axis, we need to find the x-values where \(y=0\). In other words, we want to solve the equation, \(0 = 4x - 8\).

Find the points where the graph intersects the x-axis

To find the points where the graph intersects the x-axis, we need to solve the equation \(0 = 4x - 8\). Adding \(8\) to both sides, we get \(4x = 8\), and dividing by \(4\), we have \(x = 2\). Therefore, the graph intersects the x-axis at the point \((2,0)\).

Find the area of the region above the x-axis

Looking at the graph, we can see that the line is above the x-axis for an interval where \(2\leq x\leq8\). The shape of the region between the line and the x-axis is a trapezoid with base lengths of \(6\) and \(16\), and a height of \(6\). To find the area of this trapezoid, we can apply the formula for the area of a trapezoid: \(A=\frac{1}{2}(b_1+b_2)h\), where \(b_1\) and \(b_2\) are the lengths of the two bases, and \(h\) is the height. We have: \(A=\frac{1}{2}(6+16)6\) \(A=\frac{1}{2}(22)6\) \(A=66\) So the area of the region above the x-axis is \(66\).

Find the area of the region below the x-axis

The line is below the x-axis when the interval is \(-4\leq x\leq2\). This forms a triangle with a base length of \(6\) and a height of \(8\). To find the area of this triangle, we can use the formula for the area of a triangle: \(A=\frac{1}{2}bh\), where \(b\) is the base and \(h\) is the height. We have: \(A=\frac{1}{2}(6)(8)\) \(A=\frac{1}{2}(48)\) \(A=24\) So the area of the region below the x-axis is \(24\).

Calculate the area and net area of the region

Now we have the areas of both the region above and below the x-axis. To find the total area of the region, we simply add the two areas together: Area = \(66+24=90\) To find the net area, we need to account for the area below the x-axis being negative, so we subtract the area below the x-axis from the area above the x-axis: Net Area = \(66 - 24 = 42\) Therefore, the area of the region is \(90\), and the net area is \(42\).

Key concepts

Linear Function Graphing

Graphing a linear function like \(y=4x-8\) is an essential skill in calculus and algebra. Each linear function can be illustrated as a straight line graph. The line's slope and y-intercept help determine its exact placement on the graph. The slope indicates how steep the line is, while the y-intercept tells us where the line crosses the y-axis. In this case, the slope is \(4\), meaning the line rises four units vertically for every one unit it moves horizontally. The y-intercept is \(-8\), so the line crosses the y-axis at the point \((0, -8)\). Identifying where the function intersects the x-axis is another step. Here, set \(y=0\) to solve for \(x\). This gives \(x=2\). Graphing the line on an interval from \(-4\) to \(8\), you'll find the line changes its position at \(x=2\), which will be crucial in identifying areas related to the x-axis.

Trapezoid and Triangle Area Formulas

In the realm of geometry, knowing how to calculate areas of trapezoids and triangles is valuable. When graphing \(y=4x-8\), you can form these shapes which help in calculating the area between the line and the x-axis.

For the segment where the graph is above the x-axis \((2\leq x\leq 8)\), the figure forms a trapezoid. The trapezoid formula is \(A=\frac{1}{2}(b_1+b_2)h\), where \(b_1\) and \(b_2\) are base lengths, and \(h\) is the height. In this example, the height is from \(y=0\) to \(y=8\) over the x-range of \(2\) to \(8\), and the total area is \(66\).
Below the x-axis \((-4\leq x\leq2)\), you get a triangular region. The triangle's area is calculated using \(A=\frac{1}{2}bh\). In this graph, the base length is \(6\) and height is \(8\), resulting in an area of \(24\). Understanding these formulas aids in calculating the geometry-based area efficiently.

Calculus Concepts of Area and Net Area

In calculus, distinguishing between 'area' and 'net area' is fundamental. These concepts become crucial when assessing regions under and over the x-axis on a graph. The area accounts for the combined size of regions, regardless of their position concerning the x-axis. Contrast this with the net area, which considers direction; regions below the x-axis contribute negative values.
In practice, the total area consists of summing all individual areas without regard for their signs. In the case of \(y=4x-8\), the combined areas are \(66\) (above the x-axis) and \(24\) (below the x-axis), providing a total area of \(90\). Net area, on the other hand, requires subtracting the negative area from the positive one, resulting in a net area of \(42\).
These calculations demonstrate key calculus principles related to graphical analysis and provide a solid footing for further studies in integral calculus.