Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Finding Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$.
• Power Rule of Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, where $n \ne -1$.

Step-by-Step Solution

Step 1: Identify the curves and the region

We are given the region defined by $x^2 \le y \le 3 - 2x$. This means we are looking for the area bounded by the parabola $y = x^2$ and the line $y = 3 - 2x$. We need to determine which function is the upper function and which is the lower function.

Step 2: Find the points of intersection

To find the points of intersection, we set the two equations equal to each other: $x^2 = 3 - 2x$ Rearranging the terms, we get a quadratic equation: $x^2 + 2x - 3 = 0$ Factoring the quadratic, we have: $(x + 3)(x - 1) = 0$ This gives us two solutions for $x$: $x = -3 \quad \text{or} \quad x = 1$ So the limits of integration are $a = -3$ and $b = 1$.

Step 3: Determine the upper and lower functions

We know the curves intersect at $x=-3$ and $x=1$. To determine which function is the upper function on the interval $[-3, 1]$, we can test a point within the interval, say $x=0$. For $y = x^2$, when $x = 0$, $y = 0^2 = 0$. For $y = 3 - 2x$, when $x = 0$, $y = 3 - 2(0) = 3$. Since $3 > 0$, the line $y = 3 - 2x$ is above the parabola $y = x^2$ on the interval $[-3, 1]$. Thus, $f(x) = 3 - 2x$ and $g(x) = x^2$.

Step 4: Set up the definite integral

Now we can set up the definite integral to find the area: $A = \int_{-3}^1 [(3 - 2x) - x^2] \, dx$ $A = \int_{-3}^1 (3 - 2x - x^2) \, dx$

Step 5: Evaluate the definite integral

We find the antiderivative: $\int (3 - 2x - x^2) \, dx = 3x - x^2 - \frac{x^3}{3} + C$ Now we evaluate the definite integral using the Fundamental Theorem of Calculus: $A = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^1$ $A = \left( 3(1) - (1)^2 - \frac{(1)^3}{3} \right) - \left( 3(-3) - (-3)^2 - \frac{(-3)^3}{3} \right)$ $A = \left( 3 - 1 - \frac{1}{3} \right) - \left( -9 - 9 - \frac{-27}{3} \right)$ $A = \left( 2 - \frac{1}{3} \right) - \left( -18 + 9 \right)$ $A = \frac{5}{3} - (-9)$ $A = \frac{5}{3} + 9$ $A = \frac{5}{3} + \frac{27}{3}$ $A = \frac{32}{3}$

Common Mistakes & Tips

• Reversing the Functions: Make sure you subtract the lower function from the upper function. If you reverse them, you'll get the negative of the correct answer.
• Sign Errors: Be careful with negative signs, especially when substituting the limits of integration into the antiderivative.
• Sketching the Graph: Sketching the graph can help you visualize the region and ensure you're setting up the integral correctly.

Summary

We found the area of the region bounded by $y = x^2$ and $y = 3 - 2x$ by finding the intersection points, determining the upper and lower functions, setting up the definite integral, and evaluating it. The area is $\frac{32}{3}$ square units.

Final Answer The final answer is $\boxed{\frac{32}{3}}$, which corresponds to option (D).