Understanding the Problem: Finding the Area of a Bounded Region

The question asks us to find the area of the region enclosed by two given equations: \(x - |y| = 0\) and \(x - 2 = 0\). To solve this, we first need to understand what these equations represent graphically and then find the points where they intersect to define the boundaries of the region.

Analyzing the Equations

Let's examine each equation:

1. Equation 1: \(x - |y| = 0\)
2. This equation can be rewritten as \(x = |y|\).
3. The absolute value means we have two cases:
   • If \(y \ge 0\), then \(|y| = y\), so the equation becomes \(x = y\). This is a straight line passing through the origin with a slope of 1 in the first quadrant.
   • If \(y < 0\), then \(|y| = -y\), so the equation becomes \(x = -y\). This is a straight line passing through the origin with a slope of -1 in the fourth quadrant.
4. Together, \(x = |y|\) forms a V-shaped graph symmetric about the x-axis, with its vertex at the origin (0,0).
5. Equation 2: \(x - 2 = 0\)
6. This equation can be rewritten as \(x = 2\).
7. This is a vertical line parallel to the y-axis, passing through the point (2,0) on the x-axis.

Identifying the Bounded Region

The region bounded by these two equations is the area enclosed by the V-shaped curve \(x = |y|\) and the vertical line \(x = 2\). Let's find the intersection points of these two graphs.

To find the intersection points, we set the x-values equal:

\(|y| = 2\)

This means \(y = 2\) or \(y = -2\).

• When \(y = 2\), \(x = |2| = 2\). Intersection point: (2, 2).
• When \(y = -2\), \(x = |-2| = 2\). Intersection point: (2, -2).

The graph of \(x = |y|\) passes through the origin (0,0). The vertical line \(x = 2\) intersects the two parts of \(x = |y|\) at (2, 2) and (2, -2). The bounded region is a triangle with vertices at (0,0), (2,2), and (2,-2).

Calculating the Area

We can calculate the area of this triangular region using either geometry or integration.

Method 1: Using Geometry (Area of a Triangle)

The bounded region is a triangle with vertices (0,0), (2,2), and (2,-2). We can consider the side along the line \(x = 2\) as the base of the triangle.

• The length of the base is the distance between (2,2) and (2,-2). This distance is \(|2 - (-2)| = |2 + 2| = 4\).
• The height of the triangle is the perpendicular distance from the third vertex (0,0) to the line \(x = 2\). The distance from the origin (0,0) to the vertical line \(x=2\) is simply the x-coordinate of the line, which is 2.

Using the formula for the area of a triangle, Area = \(\frac{1}{2} \times \text{base} \times \text{height}\):

\(\text{Area} = \frac{1}{2} \times 4 \times 2\)

\(\text{Area} = \frac{1}{2} \times 8\)

\(\text{Area} = 4\)

Method 2: Using Integration

We can integrate with respect to x or y. Integrating with respect to x is simpler here.

The region is bounded by \(x=|y|\) (which means \(y = x\) for \(y \ge 0\) and \(y = -x\) for \(y < 0\)) and the line \(x=2\).

Looking at the region from \(x=0\) to \(x=2\), for any given x, the region extends from the lower boundary \(y = -x\) to the upper boundary \(y = x\). We integrate the difference between the upper and lower boundary curves from the minimum x-value (0) to the maximum x-value (2).

\(\text{Area} = \int_{0}^{2} (\text{Upper boundary } - \text{ Lower boundary}) \, dx\)

\(\text{Area} = \int_{0}^{2} (x - (-x)) \, dx\)

\(\text{Area} = \int_{0}^{2} (x + x) \, dx\)

\(\text{Area} = \int_{0}^{2} 2x \, dx\)

Now, we perform the integration:

\(\text{Area} = \left[x^2\right]_{0}^{2}\)

\(\text{Area} = (2^2) - (0^2)\)

\(\text{Area} = 4 - 0\)

\(\text{Area} = 4\)

Both methods give the same result for the area of the bounded region.

Conclusion

The area of the region bounded by the equations \(x - |y| = 0\) and \(x - 2 = 0\) is 4 square units.

Revision Table: Key Concepts

Additional Information: Graphing Absolute Value Functions

Functions involving absolute values often create graphs with sharp corners or "vertices". For an equation like \(x = |y|\):

• Start with the graph of \(x = y\).
• For \(y < 0\), the original graph \(x=y\) would be in the third quadrant. However, because of \(x = |y|\), the x-value must be positive (since it equals an absolute value). The part of the graph where y is negative is reflected across the x-axis (or more accurately, the values are such that x is positive for negative y, following \(x = -y\)).
• This reflection creates the symmetric V-shape opening to the right along the positive x-axis. The vertex is always where the expression inside the absolute value is zero (here, \(y=0\), so \(x=|0|=0\), giving the vertex (0,0)).

Understanding how absolute values transform basic graphs is crucial for correctly identifying bounded regions in calculus problems.