The problem involves finding the area of the region bounded by $y = |x - 1|$, the x-axis, and the vertical lines $x = -3$ and $x = 2$.

Step-by-Step Solution:

1. Understand the Function:

   • The function $y = |x - 1|$ represents two lines:
     • $y = x - 1$ for $x \geq 1$,
     • $y = -(x - 1) = -x + 1$ for $x < 1$.

2. Break Down the Region:

   • The region is split at $x = 1$ because the piecewise nature of $|x - 1|$ changes there.
   • Consider:
     • From $x = -3$ to $x = 1$: $y = -x + 1$,
     • From $x = 1$ to $x = 2$: $y = x - 1$.

3. Set Up the Integral:

   • The area is given by the definite integrals over the respective intervals: $\text{Area} = \int_{-3}^{1} (-(x - 1)) \, dx + \int_{1}^{2} ((x - 1)) \, dx$

4. Solve the Integrals:

   • First Integral ($x = -3$ to $x = 1$, $y = -x + 1$): $\int_{-3}^{1} (-x + 1) \, dx = \left[ -\frac{x^2}{2} + x \right]_{-3}^{1}$ Substituting the limits: = \left( -\frac{1}{2} + 1 \right) - \left( -\frac{9}{2} - 3 \right) = \frac{1}{2} - \left( -\frac{9}{2} - 3 \right) = \frac{1}{2} + \frac{9}{2} + 3 = 6.$$
   • Second Integral ($x = 1$ to $x = 2$, $y = x - 1$): $\int_{1}^{2} (x - 1) \, dx = \left[ \frac{x^2}{2} - x \right]_{1}^{2}$ Substituting the limits: = \left( \frac{4}{2} - 2 \right) - \left( \frac{1}{2} - 1 \right) = (2 - 2) - \left( \frac{1}{2} - 1 \right) = 0 - \left( \frac{1}{2} - 1 \right) = \frac{1}{2}.$$

5. Total Area:

   • Add the areas from the two intervals: $\text{Total Area} = 6 + \frac{1}{2} = \frac{12}{2} + \frac{1}{2} = \frac{13}{2}.$

Thus, the total area is $\frac{13}{2}$ square units.

If you have further questions or want clarification, let me know!

Questions for Further Understanding:

1. What is the significance of splitting the integral based on the piecewise nature of $|x - 1|$?
2. Can we visualize the bounded region on a graph to verify the solution?
3. Why do absolute value functions typically result in two different intervals?
4. How does changing the limits of integration affect the computed area?
5. Can you apply this technique to a different absolute value function?

Tip:

When solving integrals involving absolute values, always split the function at the points where the expression inside the absolute value equals zero. This ensures accurate computation of the area!