If the regions A and B are given by `A={(x,y)// : y gt x}B={(x,y):y lt 2-x^(2)}` then the area of `A nn B` is

Options

A. a) `(1)/(3)`
B. b) `(9)/(2)`
C. c) `(2)/(5)`
D. d) `(3)/(4)`

Answer

To find the area of the intersection of the regions A and B given by:

- \( A = \{(x,y) : y > x\} \)
- \( B = \{(x,y) : y < 2 - x^2\} \)

we need to follow these steps:

### Step 1: Find the intersection points of the curves

We have the line \( y = x \) and the parabola \( y = 2 - x^2 \). To find the intersection points, we set these two equations equal to each other:

\[ x = 2 - x^2 \]

Rearranging gives:

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

### Step 2: Solve the quadratic equation

We can factor the quadratic equation:

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

Thus, the solutions are:

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

### Step 3: Find corresponding y-values

Using \( y = x \):

- For \( x = -2 \), \( y = -2 \) (point: \((-2, -2)\))
- For \( x = 1 \), \( y = 1 \) (point: \((1, 1)\))

### Step 4: Set up the integral for the area

The area of the intersection \( A \cap B \) can be found by integrating the difference between the upper curve (the parabola) and the lower curve (the line) from \( x = -2 \) to \( x = 1 \):

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

### Step 5: Simplify the integrand

Simplifying the integrand gives:

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

### Step 6: Integrate

Now we integrate:

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

This can be split into three separate integrals:

\[ \int_{-2}^{1} 2 \, dx - \int_{-2}^{1} x \, dx - \int_{-2}^{1} x^2 \, dx \]

Calculating each integral:

1. \(\int_{-2}^{1} 2 \, dx = 2[x]_{-2}^{1} = 2(1 - (-2)) = 2 \cdot 3 = 6\)
2. \(\int_{-2}^{1} x \, dx = \left[\frac{x^2}{2}\right]_{-2}^{1} = \frac{1^2}{2} - \frac{(-2)^2}{2} = \frac{1}{2} - 2 = -\frac{3}{2}\)
3. \(\int_{-2}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-2}^{1} = \frac{1^3}{3} - \frac{(-2)^3}{3} = \frac{1}{3} + \frac{8}{3} = 3\)

### Step 7: Combine the results

Now substituting back into the area calculation:

\[ \text{Area} = 6 - \left(-\frac{3}{2}\right) - 3 = 6 + \frac{3}{2} - 3 = 3 + \frac{3}{2} = \frac{6}{2} + \frac{3}{2} = \frac{9}{2} \]

### Final Answer

The area of the intersection \( A \cap B \) is:

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