Let P be the set of points `(x, y)` such that `x^2 le y le – 2x + 3`. Then area of region bounded by points in set P is

Options

A. `29/3`
B. `34/3`
C. `32/3`
D. `31/3`

Answer

To find the area of the region bounded by the points in the set \( P \) defined by the inequalities \( x^2 \leq y \leq -2x + 3 \), we will follow these steps:

### Step 1: Identify the curves
We have two equations to consider:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = -2x + 3 \) (a straight line)

### Step 2: Find the points of intersection
To find the area between these two curves, we first need to determine where they intersect. We set the equations equal to each other:
\[ x^2 = -2x + 3 \]
Rearranging gives us:
\[ x^2 + 2x - 3 = 0 \]
Now, we can factor this quadratic equation:
\[ (x - 1)(x + 3) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{and} \quad x = -3 \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = -3 \) to \( x = 1 \) can be calculated using the integral:
\[ A = \int_{-3}^{1} \left((-2x + 3) - (x^2)\right) \, dx \]
This simplifies to:
\[ A = \int_{-3}^{1} (3 - 2x - x^2) \, dx \]

### Step 4: Compute the integral
Now we will compute the integral:
\[ A = \int_{-3}^{1} (3 - 2x - x^2) \, dx \]
Calculating the integral term by term:
1. The integral of \( 3 \) is \( 3x \).
2. The integral of \( -2x \) is \( -x^2 \).
3. The integral of \( -x^2 \) is \( -\frac{x^3}{3} \).

Thus, we have:
\[ A = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^{1} \]

### Step 5: Evaluate the definite integral
Now we will evaluate this from \( -3 \) to \( 1 \):
1. For \( x = 1 \):
\[ 3(1) - (1)^2 - \frac{(1)^3}{3} = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]
2. For \( x = -3 \):
\[ 3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 + 9 = -9 \]

Now we compute:
\[ A = \left( \frac{5}{3} \right) - (-9) = \frac{5}{3} + 9 = \frac{5}{3} + \frac{27}{3} = \frac{32}{3} \]

### Final Answer
Thus, the area of the region bounded by the points in set \( P \) is:
\[ \boxed{\frac{32}{3}} \]