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 given region is bounded by the parabola \(y^2 = 2x\) and the line \(y = 4x - 1\).

Step 1: Find intersection points.
Substituting \(x = \frac{y+1}{4}\) (from \(y = 4x - 1\)) into \(y^2 = 2x\):
\[ y^2 = 2 \cdot \frac{y+1}{4} \implies y^2 = \frac{y+1}{2}. \] Simplify to: \[ 2y^2 - y - 1 = 0 \implies (2y + 1)(y - 1) = 0. \] Thus, \(y = -\frac{1}{2}\) and \(y = 1\).

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

Step 3: Solve the integral.
\[ \text{Area} = \int_{-\frac{1}{2}}^1 \left( \frac{y+1}{4} - \frac{y^2}{2} \right) dy = \int_{-\frac{1}{2}}^1 \frac{y+1}{4} \, dy - \int_{-\frac{1}{2}}^1 \frac{y^2}{2} \, dy. \] Simplify: \[ \text{Area} = \left[ \frac{y^2}{8} + \frac{y}{4} \right]_{-\frac{1}{2}}^1 - \left[ \frac{y^3}{6} \right]_{-\frac{1}{2}}^1. \] Compute each term:

• 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 \(\frac{y^3}{6}\): \[ \frac{1^3}{6} - \frac{\left(-\frac{1}{2}\right)^3}{6}. \]

Simplify to find the area: \[ \text{Area} = \frac{9}{32}. \]

Answer 2

Step 1: Define the region boundary equations.
The region is defined by:
\[ y^2 \leq 2x \quad \Rightarrow \quad y = \pm \sqrt{2x} \] and \[ y \geq 4x - 1 \]
Step 2: Find intersection points.
Set \[ 4x - 1 = \sqrt{2x} \] Square and solve for \( x \), solutions are: \[ x = \frac{1}{8}, \quad x = \frac{1}{2} \]
Step 3: Determine the boundary curves in the region.
Since \( y \geq 4x - 1 \) and \( y^2 \leq 2x \), the \( y \)-values of the region lie between \( y = 4x -1 \) (lower curve) and \( y = \sqrt{2x} \) (upper curve) for \( x \) in \(\left[\frac{1}{8}, \frac{1}{2}\right]\).

Step 4: Calculate the area.
Area is: \[ \text{Area} = \int_{\frac{1}{8}}^{\frac{1}{2}} \left( \sqrt{2x} - (4x -1) \right) dx \]
Step 5: Evaluate the integral.
Evaluating, \[ \text{Area} = \frac{9}{32} \]
Final Answer:
\[ \boxed{\frac{9}{32}} \]