The area of the region `A={(x,y), 0 le y le x|x|+1 and -1 le x le 1}` in sq. units, is

Options

A. 2
B. `(1)/(3)`
C. `(4)/(3)`
D. `(2)/(3)`

Answer

To find the area of the region \( A = \{(x,y) : 0 \leq y \leq x|x| + 1 \text{ and } -1 \leq x \leq 1\} \), we can follow these steps:

### Step 1: Understand the bounds of the region
The region is defined by the inequalities:
- \( 0 \leq y \leq x|x| + 1 \)
- \( -1 \leq x \leq 1 \)

### Step 2: Analyze the function \( y = x|x| + 1 \)
The function \( x|x| \) can be expressed as:
- \( x^2 \) when \( x \geq 0 \)
- \( -x^2 \) when \( x < 0 \)

Thus, we can rewrite the function:
- For \( x \geq 0 \): \( y = x^2 + 1 \)
- For \( x < 0 \): \( y = -x^2 + 1 \)

### Step 3: Determine the area under the curves
We will calculate the area in two parts: from \( -1 \) to \( 0 \) and from \( 0 \) to \( 1 \).

#### Part 1: Area from \( -1 \) to \( 0 \)
Here, the upper curve is \( y = -x^2 + 1 \).
The area \( A_1 \) can be calculated as:
\[ A_1 = \int_{-1}^{0} (-x^2 + 1) \, dx \]

Calculating the integral:
\[ A_1 = \int_{-1}^{0} (-x^2 + 1) \, dx = \left[-\frac{x^3}{3} + x\right]_{-1}^{0} \]
Evaluating the limits:
\[ = \left[-\frac{0^3}{3} + 0\right] - \left[-\frac{(-1)^3}{3} + (-1)\right] \]
\[ = 0 - \left[\frac{1}{3} - 1\right] = 0 - \left[\frac{1}{3} - \frac{3}{3}\right] = 0 - \left[-\frac{2}{3}\right] = \frac{2}{3} \]

#### Part 2: Area from \( 0 \) to \( 1 \)
Here, the upper curve is \( y = x^2 + 1 \).
The area \( A_2 \) can be calculated as:
\[ A_2 = \int_{0}^{1} (x^2 + 1) \, dx \]

Calculating the integral:
\[ A_2 = \int_{0}^{1} (x^2 + 1) \, dx = \left[\frac{x^3}{3} + x\right]_{0}^{1} \]
Evaluating the limits:
\[ = \left[\frac{1^3}{3} + 1\right] - \left[\frac{0^3}{3} + 0\right] \]
\[ = \left[\frac{1}{3} + 1\right] - 0 = \frac{1}{3} + 1 = \frac{1}{3} + \frac{3}{3} = \frac{4}{3} \]

### Step 4: Total Area
The total area \( A \) is the sum of the two areas:
\[ A = A_1 + A_2 = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 \]

### Final Answer
The area of the region \( A \) is \( 2 \) square units.

---