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

Answer: (D)

$\frac{9}{32}$

Answer 2

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: $\text{Area} = \frac{9}{32}.$