Question

The area of the region $A = [(x, y): 0 \leq y \leq x|x|+1$ and $-1 \leq x \leq 1]$ in s units is :

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. 2
• D. $\frac{4}{3}$

Answer 1

The Correct Option is C

To find the area of the region defined by \( A = \{(x, y): 0 \leq y \leq x|x| + 1 \} \) and \(-1 \leq x \leq 1\), we need to clearly understand the inequalities and integrate over the specified domain.

The inequality can be broken down into two parts because of the absolute value function \( x|x| \):

• For \( 0 \leq x \leq 1 \), \( x|x| = x^2 \). Thus, the inequality becomes \( 0 \leq y \leq x^2 + 1 \).
• For \(-1 \leq x < 0 \), \( x|x| = -x^2 \). Thus, the inequality becomes \( 0 \leq y \leq -x^2 + 1 \).

Now, we perform integration over these two intervals separately and sum the areas.

1. For \( x \) ranging from \( 0 \) to \( 1 \): 

\[\text{Area}_1 = \int_{0}^{1} (x^2 + 1) \, dx\]

1. Integrate \( \int x^2 \, dx = \frac{x^3}{3} \) and \( \int 1 \, dx = x \).
2. Evaluate: 

\[= \left[ \frac{x^3}{3} + x \right]_{0}^{1} = \left( \frac{1^3}{3} + 1 \right) - \left( 0 + 0 \right) = \frac{4}{3}\]

1. For \( x \) ranging from \(-1\) to \( 0 \): 

\[\text{Area}_2 = \int_{-1}^{0} (-x^2 + 1) \, dx\]

1. Integrate \( \int -x^2 \, dx = -\frac{x^3}{3} \) and \( \int 1 \, dx = x \).
2. Evaluate: 

\[= \left[ -\frac{x^3}{3} + x \right]_{-1}^{0} = \left( 0 + 0 \right) - \left( -\frac{(-1)^3}{3} - 1 \right) = \frac{2}{3}\]

Summing these areas gives the total region area:

\[\text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{4}{3} + \frac{2}{3} = 2\]

Therefore, the area of the region is \( \mathbf{2} \) square units, corresponding to the correct answer.