To evaluate the integral $\int_{-1}^{5} |x - 3| - 1 \, dx$ using geometry, let's analyze the graph of the function $f(x) = |x - 3| - 1$ over the interval $[-1, 5]$.

Step 1: Understand the Function and its Key Points

1. The function $f(x) = |x - 3| - 1$ is an absolute value function, shifted right by 3 and down by 1.
2. This function will have a "V" shape centered at $x = 3$.
3. At $x = 3$, $f(x) = -1$.
4. For $x < 3$, $f(x) = -(x - 3) - 1 = -x + 2$.
5. For $x > 3$, $f(x) = x - 3 - 1 = x - 4$.

Step 2: Identify Regions of the Graph

The interval $[-1, 5]$ covers two linear sections of the function:

• From $x = -1$ to $x = 3$, where $f(x) = -x + 2$.
• From $x = 3$ to $x = 5$, where $f(x) = x - 4$.

Step 3: Calculate the Areas of Each Region

1. From $x = -1$ to $x = 3$:

   • The line $f(x) = -x + 2$ goes from $f(-1) = 3$ to $f(3) = -1$.
   • This forms a triangle with base 4 (from $-1$ to $3$) and height 4 (from 3 to -1).

   The area of this triangle is: $\text{Area}_1 = \frac{1}{2} \times 4 \times 4 = 8$

2. From $x = 3$ to $x = 5$:

   • The line $f(x) = x - 4$ goes from $f(3) = -1$ to $f(5) = 1$.
   • This forms a triangle with base 2 (from 3 to 5) and height 2 (from -1 to 1).

   The area of this triangle is: $\text{Area}_2 = \frac{1}{2} \times 2 \times 2 = 2$

Step 4: Sum the Areas

The total area under the curve from $x = -1$ to $x = 5$ is the sum of these two areas: $\int_{-1}^{5} |x - 3| - 1 \, dx = 8 + 2 = 10$

Final Answer

$\int_{-1}^{5} |x - 3| - 1 \, dx = 10$

Would you like a detailed explanation of any specific part?

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1. How do you interpret absolute value functions in integrals geometrically?
2. How can you confirm the limits of integration using points of intersection?
3. Why does this function form triangles within the given interval?
4. How would the answer change if the function were shifted differently?
5. How do you calculate areas under non-linear functions?

Tip: For absolute value functions, always split the integral at points where the expression inside the absolute value changes signs.