The area (in sq. units) of the region `{(x,y) in R^2: x^2 le y le 3-2x}`, is :

Options

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

Answer

To find the area of the region defined by the inequalities \( x^2 \leq y \leq 3 - 2x \), we can follow these steps:

### Step 1: Identify the curves
We have two curves:
1. \( y = x^2 \) (a parabola opening upwards)
2. \( y = 3 - 2x \) (a straight line with a negative slope)

### Step 2: Find the points of intersection
To find the area between these two curves, we need to determine where they intersect. We set the equations equal to each other:

\[ x^2 = 3 - 2x \]

Rearranging gives us:

\[ x^2 + 2x - 3 = 0 \]

### Step 3: Solve the quadratic equation
We can factor the quadratic:

\[ (x + 3)(x - 1) = 0 \]

Thus, the solutions are:

\[ x = -3 \quad \text{and} \quad x = 1 \]

### Step 4: 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:

\[ A = \int_{-3}^{1} ((3 - 2x) - (x^2)) \, dx \]

### Step 5: Evaluate the integral
Now we compute the integral:

\[ A = \int_{-3}^{1} (3 - 2x - x^2) \, dx \]

Calculating the integral:

\[ = \left[ 3x - x^2 - \frac{x^3}{3} \right]_{-3}^{1} \]

Calculating at the upper limit \( x = 1 \):

\[ = 3(1) - (1)^2 - \frac{(1)^3}{3} = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3} \]

Calculating at the lower limit \( x = -3 \):

\[ = 3(-3) - (-3)^2 - \frac{(-3)^3}{3} = -9 - 9 + 9 = -9 \]

### Step 6: Subtract the lower limit from the upper limit
Now we subtract the value at the lower limit from the value at the upper limit:

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

### Conclusion
Thus, the area of the region is:

\[ \boxed{\frac{32}{3}} \text{ square units} \]