Question

The area (in square units) of the region described by: $$ \{(x, y) : y^2 \leq 2x, \, \text{and} \, y \geq 4x - 1\} $$ is:

• A. \( \frac{11}{32} \)
• B. \( \frac{8}{9} \)
• C. \( \frac{11}{12} \)
• D. \( \frac{9}{32} \)

Answer 1

The Correct Option is D

The region is defined by the curves \(y^2 = 2x\) and \(y = 4x - 1\).

Step 1: Determine intersection points.
Substitute \(x = \frac{y+1}{4}\) into \(y^2 = 2x\):
\[ y^2 = 2 \cdot \frac{y+1}{4} \implies y^2 = \frac{y+1}{2}. \] This simplifies to: \[ 2y^2 - y - 1 = 0 \implies (2y + 1)(y - 1) = 0. \] The solutions for \(y\) are \(y = -\frac{1}{2}\) and \(y = 1\).

Step 2: Formulate the integral for the shaded area.
The area is given by: \[ \text{Area} = \int_{-\frac{1}{2}}^1 (x_{\text{right}} - x_{\text{left}}) \, dy, \] where \(x_{\text{right}} = \frac{y+1}{4}\) represents the line and \(x_{\text{left}} = \frac{y^2}{2}\) represents the parabola.

Step 3: Evaluate the integral.
\[ \text{Area} = \int_{-\frac{1}{2}}^1 \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy = \int_{-\frac{1}{2}}^1 \frac{y+1}{4} \, dy - \int_{-\frac{1}{2}}^1 \frac{y^2}{2} \, dy. \] This simplifies to: \[ \text{Area} = \left[ \frac{y^2}{8} + \frac{y}{4} \right]_{-\frac{1}{2}}^1 - \left[ \frac{y^3}{6} \right]_{-\frac{1}{2}}^1. \] Evaluate each part:

• For \(\frac{y^2}{8} + \frac{y}{4}\): \[ \left( \frac{1^2}{8} + \frac{1}{4} \right) - \left( \frac{\left(-\frac{1}{2}\right)^2}{8} + \frac{-\frac{1}{2}}{4} \right) = \frac{1}{8} + \frac{2}{8} - \frac{1}{32} - \frac{1}{16}. \]
• For \(\frac{y^3}{6}\): \[ \frac{1^3}{6} - \frac{\left(-\frac{1}{2}\right)^3}{6}. \]

The final area is: \[ \text{Area} = \frac{9}{32}. \]