The problem asks to find the area of the region bounded by the curves:

$f(x) = 5 - |x - 3| \quad \text{and} \quad f(x) = |x|$

Step 1: Understanding the Curves

1. First curve: $f(x) = 5 - |x - 3|$

   • This is a V-shaped curve with its vertex at $x = 3$ and opens downward with a slope of $-1$ on either side of the vertex.

2. Second curve: $f(x) = |x|$

   • This is the standard absolute value function, also a V-shaped curve, with the vertex at $x = 0$ and opens upward with slopes $\pm 1$.

Step 2: Intersection Points

To find the area between the curves, we need to determine their points of intersection by solving:

$5 - |x - 3| = |x|$

Case 1: $x \geq 3$

In this region, $|x - 3| = x - 3$, so the equation becomes:

$5 - (x - 3) = x$ $5 - x + 3 = x \quad \Rightarrow \quad 8 = 2x \quad \Rightarrow \quad x = 4$

Case 2: $x \leq 3$

In this region, $|x - 3| = 3 - x$, so the equation becomes:

$5 - (3 - x) = |x|$

This leads to two sub-cases:

• For $x \geq 0$, $|x| = x$, so:

$5 - (3 - x) = x \quad \Rightarrow \quad 5 - 3 + x = x \quad \Rightarrow \quad 2 = 0$

This is a contradiction, meaning there is no solution in this range.

• For $x \leq 0$, $|x| = -x$, so:

$5 - (3 - x) = -x$ $5 - 3 + x = -x \quad \Rightarrow \quad 2 + x = -x \quad \Rightarrow \quad 2 = -2x \quad \Rightarrow \quad x = -1$

Step 3: Setting up the Area

The curves intersect at $x = -1$ and $x = 4$. Now, we calculate the area between these curves from $x = -1$ to $x = 4$.

The area is given by:

$\text{Area} = \int_{-1}^{4} \left( (5 - |x - 3|) - |x| \right) dx$

We will split the integral into two regions: $x \in [-1, 0]$, $x \in [0, 3]$, and $x \in [3, 4]$, because the definitions of the absolute values change in these intervals.

Step 4: Solve the Integral

Let's compute the area step by step.The area of the region bounded by the curves $f(x) = 5 - |x - 3|$ and $f(x) = |x|$ is $8$ square units.

Would you like further details or have any questions?

Here are 5 related questions that build upon this concept:

1. How do we determine the points of intersection for two different absolute value functions?
2. What is the significance of splitting the integral based on different ranges for absolute value functions?
3. How does the shape of the graph influence the integral's limits and setup?
4. Can this method be applied to other types of piecewise functions?
5. How would the area change if the curves were shifted vertically or horizontally?

Tip: Always carefully consider the different cases for absolute value functions, especially when solving for intersections or setting up integrals.