To find the area of the shaded region bounded by the curves $y = x - 1$ and $x = 3 - y^2$, we can set up an integral by determining the points of intersection and integrating the difference of the functions in terms of either $x$ or $y$.

Step-by-Step Solution:

1. Find the Points of Intersection:

   To find where $y = x - 1$ and $x = 3 - y^2$ intersect, substitute $y = x - 1$ into $x = 3 - y^2$.

   Substitute: $x = 3 - (x - 1)^2$

   Expand $(x - 1)^2$: $x = 3 - (x^2 - 2x + 1)$ $x = 3 - x^2 + 2x - 1$ $x^2 - 3x + 2 = 0$ Factor the equation: $(x - 1)(x - 2) = 0$ So, $x = 1$ and $x = 2$.

   For $x = 1$: $y = 1 - 1 = 0$ For $x = 2$: $y = 2 - 1 = 1$

   Therefore, the points of intersection are $(1, 0)$ and $(2, 1)$.

2. Set Up the Integral:

   Since we have $x$ as a function of $y$ for $x = 3 - y^2$, we can integrate with respect to $y$ from $y = 0$ to $y = 1$.

   The area $A$ is given by: $A = \int_{0}^{1} \left( (y + 1) - (3 - y^2) \right) \, dy$ Simplify the integrand: $A = \int_{0}^{1} \left( y + 1 - 3 + y^2 \right) \, dy$ $A = \int_{0}^{1} \left( y^2 + y - 2 \right) \, dy$

3. Integrate:

   Integrate each term separately: $A = \int_{0}^{1} y^2 \, dy + \int_{0}^{1} y \, dy - \int_{0}^{1} 2 \, dy$

   Calculate each integral: $\int_{0}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{0}^{1} = \frac{1}{3}$ $\int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2}$ $\int_{0}^{1} 2 \, dy = \left[ 2y \right]_{0}^{1} = 2$

   So, $A = \frac{1}{3} + \frac{1}{2} - 2$

   Combine terms: [ A = \frac{1}{3} + \frac{1}{2} - \frac{6}{3} = \frac{1 + 1.5 - 6}{3} = \frac{-3.