The area of the region bounded by y – x = 2 and `x^(2) = y` is equal to

Options

A. `(16)/(3)`
B. `(2)/(3)`
C. `(9)/(2)`
D. `(4)/(3)`

Answer

To find the area of the region bounded by the curves \( y - x = 2 \) and \( x^2 = y \), we will follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equations of the curves in a more usable form:
1. The line \( y - x = 2 \) can be rewritten as:
\[ y = x + 2 \]
2. The parabola \( x^2 = y \) is already in a usable form.

### Step 2: Find the points of intersection
Next, we need to find the points where the two curves intersect. We set the equations equal to each other:
\[ x + 2 = x^2 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Factoring the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]
Now we can find the corresponding \( y \) values:
- For \( x = 2 \):
\[ y = 2 + 2 = 4 \]
- For \( x = -1 \):
\[ y = -1 + 2 = 1 \]
So, the points of intersection are \( (2, 4) \) and \( (-1, 1) \).

### Step 3: 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:
\[ A = \int_{-1}^{2} \left( (x + 2) - (x^2) \right) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:
\[ A = \int_{-1}^{2} (x + 2 - x^2) \, dx \]
This simplifies to:
\[ A = \int_{-1}^{2} (-x^2 + x + 2) \, dx \]

Calculating the integral:
\[ A = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} \]

### Step 5: Evaluate the definite integral
Now we evaluate it at the limits:
1. Upper limit \( x = 2 \):
\[ -\frac{2^3}{3} + \frac{2^2}{2} + 2 \cdot 2 = -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + \frac{6}{3} + \frac{12}{3} = \frac{10}{3} \]
2. Lower limit \( x = -1 \):
\[ -\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2 \cdot (-1) = \frac{1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{3}{6} - \frac{12}{6} = \frac{1}{3} - \frac{9}{6} = \frac{1}{3} - \frac{3}{2} = \frac{1 - 4.5}{3} = -\frac{3.5}{3} = -\frac{7}{6} \]

### Step 6: Final calculation
Now we find the area:
\[ A = \left( \frac{10}{3} \right) - \left( -\frac{7}{6} \right) = \frac{10}{3} + \frac{7}{6} \]
To add these fractions, we need a common denominator:
\[ \frac{10}{3} = \frac{20}{6} \]
Thus,
\[ A = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \]

### Final Answer
The area of the region bounded by the curves is:
\[ \boxed{\frac{9}{2}} \]