Question

The area (in sq. units) of the region, given by the set $\{(x, y) \in R \times R \mid x \ge 0, 2x^2 \le y \le 4 - 2x\}$ is :

• A. $\frac{7}{3}$
• B. $\frac{8}{3}$
• C. $\frac{13}{3}$
• D. $\frac{17}{3}$

Hint:

For area problems, always draw a rough sketch. It helps confirm the upper and lower curves and the correct quadrant constraint.

Answer 1

The Correct Option is A

To find the area of the region defined by the set \(\{(x, y) \in \mathbb{R} \times \mathbb{R} \mid x \ge 0, 2x^2 \le y \le 4 - 2x\}\), we need to understand the boundaries given by the inequalities.

1. The inequality \(2x^2 \le y\) defines the region above the parabola \(y = 2x^2\).
2. The inequality \(y \le 4 - 2x\) defines the region below the line \(y = 4 - 2x\).
3. We first find the points of intersection of the parabola and the line to determine the limits of integration.
4. Set \(2x^2 = 4 - 2x\) to find the intersection points.

Simplifying \(2x^2 = 4 - 2x\):

• Rearrange the equation: \(2x^2 + 2x - 4 = 0\)
• Divide the entire equation by 2: \(x^2 + x - 2 = 0\)
• Factoring gives: \((x-1)(x+2) = 0\)
• Hence, the solutions are \(x = 1\) and \(x = -2\).

Since we are considering \(x \ge 0\), the relevant intersection point is \(x = 1\). Thus, the region of interest is from \(x = 0\) to \(x = 1\).

The area A can be computed as:

• Find the integral of the difference between the top function and the bottom function:
• \(A = \int_{0}^{1} ((4 - 2x) - (2x^2)) \, dx\)

Calculate the integral:

• Simplify the function: \(4 - 2x - 2x^2\)
• Integrate: \(\int (4 - 2x - 2x^2) \, dx = [4x - x^2 - \frac{2}{3}x^3]_{0}^{1}\)
• Evaluate from 0 to 1:
  • At \(x = 1\): \(4(1) - (1)^2 - \frac{2}{3}(1)^3 = 4 - 1 - \frac{2}{3}\)
  • At \(x = 0\): \(0\)
  • Thus, \(A = (4 - 1 - \frac{2}{3}) - 0 = 3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3}\)

Therefore, the area of the region is \(\frac{7}{3}\) square units.