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

Step 1: Understanding the Concept:
The problem asks for the area of a region bounded by a parabola \(y = 2x^2\), a straight line \(y = 4 - 2x\), and the condition \(x \ge 0\).
We first identify the intersection points of these two curves to determine the limits of the integration.
Step 2: Key Formula or Approach:
Area $= \int_a^b (y_{\text{upper}} - y_{\text{lower}}) dx$.
Step 3: Detailed Explanation:
Find the intersection points of \(y = 2x^2\) and \(y = 4 - 2x\):
\[ 2x^2 = 4 - 2x \Rightarrow 2x^2 + 2x - 4 = 0 \Rightarrow x^2 + x - 2 = 0 \]
Factoring: \((x + 2)(x - 1) = 0\).
The roots are \(x = -2\) and \(x = 1\).
Since the condition \(x \ge 0\) is given, we only consider the interval \(x \in [0, 1]\).
In this interval, the line \(y = 4 - 2x\) is above the parabola \(y = 2x^2\).
Area $= \int_0^1 [(4 - 2x) - 2x^2] dx$.
\[ \text{Area} = [4x - x^2 - \frac{2x^3}{3}]_0^1 \]
\[ \text{Area} = (4(1) - 1^2 - \frac{2(1)^3}{3}) - 0 = 3 - \frac{2}{3} = \frac{7}{3} \]
Step 4: Final Answer:
The area is \(\frac{7}{3}\) sq. units.