Problem 21 - Calculus Early Transcendentals

Question

Sketch the following regions (if a figure is not given) and then find the area. The region bounded by \(y=|x-3|\) and \(y=x / 2\)

Step-by-Step Solution

Answer

Question: Find the area of the bounded region between the equations \(y=|x-3|\) and \(y=x / 2\). Answer: The area of the bounded region is \(A = A_1 + A_2 = \int_{2}^{3} \left(\frac{x}{2} - (3-x)\right) dx + \int_{3}^{6} \left(\frac{x}{2} - (x-3)\right) dx\).

Step 1: Find the points of intersection

The points of intersection occur when the two equations are equal. So, let's first solve the equation: $$x/2 = |x - 3|$$ We must consider the two cases where expressions inside the absolute value can take both positive and negative values: Case 1: \(x-3 \geq 0\) $$x/2 = x - 3$$ $$x = 6$$ So, we have one point of intersection \((6, 3)\). Case 2: \(x-3 < 0\) $$x/2 = -(x - 3)$$ $$x/2 = 3 - x$$ $$x = 2$$ So, we have another point of intersection \((2, 1)\). Step 2: Sketch the regions

Step 2: Sketch the regions

Now, we need to sketch the two equations \(y=|x-3|\) and \(y=x / 2\) and locate the bounded region. Highlight the bounded region between the two functions. Step 3: Find the area of the bounded region

Step 3: Calculate the area

To find the area of the bounded region, we need to integrate the difference between the two functions. Since we have points of intersection at \(x=2\) and \(x=6\), we will integrate with respect to x over the interval \([2,6]\). For \(2\leq x \leq 6\), we will have the following difference between the functions: $$y = \frac{x}{2} - |x-3|$$ Now, we will consider the two sub-cases as we did earlier: Case 1: \(2 \leq x \leq 3\) $$y = \frac{x}{2} - (3 - x)$$ Integrate this equation: $$A_1 = \int_{2}^{3} \left(\frac{x}{2} - (3-x)\right) dx$$ Case 2: \(3 < x \leq 6\) $$y = \frac{x}{2} - (x - 3)$$ Integrate this equation: $$A_2 = \int_{3}^{6} \left(\frac{x}{2} - (x-3)\right) dx$$ Finally, add \(A_1\) and \(A_2\): $$A = A_1 + A_2$$ Calculate the value of the area A, and that will be the final answer.

Key Concepts

Area between curves

The area between two curves is a fundamental concept in calculus. It involves determining the region covered between two given functions over a specified interval. Imagine you have two curves on a graph. The space between them is what we're interested in finding. To calculate this area, you typically subtract one function's value from the other at every point along their defined interval. This difference in height is then integrated over the interval.
Using this approach helps to account for any differences in placement above or below the x-axis.

• Identify the two curves: In our case, these are \(y=|x-3|\) and \(y=x/2\).
• Locate the points where these curves intersect, which will set up the bounds of integration.
• Integrate the difference between the two functions across this interval, taking into account any changes in direction or crossing locations.

These steps ensure you're calculating an accurate representation of the area between two functions.

Integration

Integration is a key concept in calculus and is used to calculate the area under curves. It's crucial for determining the area between two functions, just as we've done in this problem. The core idea of integration is to sum an infinite number of infinitesimally small areas under a curve to find a total area.
When we find the area between two curves, we essentially perform an integration of the difference between the two functions, over a specific interval defined by their intersection points.

• Set up the integral: You'll evaluate the integral with the upper curve minus the lower curve across the defined boundaries.
• Evaluate the integral: Perform the arithmetic and manipulative steps to simplify and solve the integral expression.
• Combine results: If the function changes form, you may need to break the integral into parts and solve each over their respective domains.

This process reveals the exact area difference between the curves over the given segment.

Absolute value function

The absolute value function is distinct, typically written as \(|x|\). It essentially takes any input and returns its non-negative counterpart. This characteristic creates a unique challenge when solving equations because the expression within the absolute value can change sign, depending on the domain. When dealing with an absolute value function like \(y = |x - 3|\), you consider two cases:

• If \(x - 3 \geq 0\), then \(|x - 3| = x - 3\).
• If \(x - 3 < 0\), then \(|x - 3| = -(x - 3)\), or \(3 - x\).

By considering both cases, you can accurately solve equations and determine the behavior of the function across its domain. This split into positive and negative cases is what makes absolute value functions a bit tricky but interesting to work with.

Points of intersection

Points of intersection are crucial when dealing with multiple curves. These are the points where the graphs of two or more functions meet or cross each other. Calculating these intersection points is an important step because they often determine the bounds for the area you wish to find between the curves.
To find the points of intersection, equate the functions and solve the resulting equation. For example, to find where \(y = |x - 3|\) intersects with \(y = x / 2\), solve \(x / 2 = |x - 3|\). The solution involves dividing the problem based on the properties of the absolute value:

• Case 1: \(x - 3 \geq 0\), leading to an intersection at \((6, 3)\).
• Case 2: \(x - 3 < 0\), resulting in an intersection at \((2, 1)\).

These intersections define the range over which you integrate to calculate the area. It’s like finding the endpoints of a segment where you need to know the entire path or area.