Find the area bounded by the curves `x ^(2) le y le x + 2:`

Options

A. `11/2`
B. `7/2`
C. `9/2`
D. `5/2`

Answer

To find the area bounded by the curves \(x^2 \leq y \leq x + 2\), we will follow these steps:

### Step 1: Identify the curves
The curves given are:
1. \(y = x^2\) (a parabola opening upwards)
2. \(y = x + 2\) (a straight line)

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

\[ x^2 = x + 2 \]

Rearranging gives us:

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

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

\[ (x - 2)(x + 1) = 0 \]

Thus, the solutions are:

\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 4: Set up the integral for the area
The area \(A\) between the curves from \(x = -1\) to \(x = 2\) can be found using the integral of the upper curve minus the lower curve:

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

### Step 5: Simplify the integrand
Now, simplify the integrand:

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

### Step 6: Calculate the integral
We can now calculate the integral:

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

Calculating the integral term by term:

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

### Step 7: Evaluate the definite integral
Now, we evaluate at the limits \(x = 2\) and \(x = -1\):

1. At \(x = 2\):

\[ -\frac{2^3}{3} + \frac{2^2}{2} + 2(2) = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3} \]

2. At \(x = -1\):

\[ -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1}{3} - \frac{9}{6} = \frac{2 - 9}{6} = -\frac{7}{6} \]

### Step 8: Combine results
Now we combine the results:

\[ A = \left(\frac{10}{3}\right) - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} \]

To add these fractions, we need a common denominator:

\[ \frac{10}{3} = \frac{20}{6} \]

So,

\[ A = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
The area bounded by the curves is:

\[ \boxed{\frac{9}{2}} \text{ square units} \]