Key Concepts and Formulas

• Area Between Curves: The area of the region bounded by $x = f(y)$ and $x = g(y)$ between $y = a$ and $y = b$, where $f(y) \geq g(y)$ on $[a, b]$, is given by $\int_a^b (f(y) - g(y)) \, dy$.
• Intersection Points: To find the intersection points of two curves, we set their equations equal to each other and solve for the variable(s).
• Solving Quadratic Equations: For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

Step 1: Find the intersection points of the curves

We are given the curves $y^2 \leq 2x$ and $y \geq 4x - 1$. We want to find where $y^2 = 2x$ and $y = 4x - 1$ intersect. To do this, we solve the system of equations. From the second equation, we have $x = \frac{y + 1}{4}$. Substituting this into the first equation, we get $y^2 = 2\left(\frac{y + 1}{4}\right)$ $y^2 = \frac{y + 1}{2}$ $2y^2 = y + 1$ $2y^2 - y - 1 = 0$

Step 2: Solve the quadratic equation

We solve the quadratic equation $2y^2 - y - 1 = 0$ for $y$. Factoring gives us $(2y + 1)(y - 1) = 0$ So, $y = 1$ or $y = -\frac{1}{2}$.

Step 3: Find the corresponding x-values

Using $x = \frac{y + 1}{4}$, when $y = 1$, we have $x = \frac{1 + 1}{4} = \frac{2}{4} = \frac{1}{2}$. When $y = -\frac{1}{2}$, we have $x = \frac{-\frac{1}{2} + 1}{4} = \frac{\frac{1}{2}}{4} = \frac{1}{8}$. Thus, the intersection points are $\left(\frac{1}{2}, 1\right)$ and $\left(\frac{1}{8}, -\frac{1}{2}\right)$.

Step 4: Express the curves as functions of y

We have $y^2 \leq 2x$, so $x \geq \frac{y^2}{2}$. Also, $y \geq 4x - 1$, so $4x \leq y + 1$, which means $x \leq \frac{y + 1}{4}$. Therefore, the region is bounded by $x = \frac{y^2}{2}$ and $x = \frac{y + 1}{4}$. Since $\frac{y+1}{4} \geq \frac{y^2}{2}$ in the region of interest, we will integrate $\frac{y+1}{4} - \frac{y^2}{2}$ with respect to $y$.

Step 5: Set up the integral

The area of the region is given by $A = \int_{-1/2}^1 \left(\frac{y + 1}{4} - \frac{y^2}{2}\right) \, dy$

Step 6: Evaluate the integral

$A = \int_{-1/2}^1 \left(\frac{1}{4}y + \frac{1}{4} - \frac{1}{2}y^2\right) \, dy$ $A = \left[\frac{1}{8}y^2 + \frac{1}{4}y - \frac{1}{6}y^3\right]_{-1/2}^1$ $A = \left(\frac{1}{8}(1)^2 + \frac{1}{4}(1) - \frac{1}{6}(1)^3\right) - \left(\frac{1}{8}\left(-\frac{1}{2}\right)^2 + \frac{1}{4}\left(-\frac{1}{2}\right) - \frac{1}{6}\left(-\frac{1}{2}\right)^3\right)$ $A = \left(\frac{1}{8} + \frac{1}{4} - \frac{1}{6}\right) - \left(\frac{1}{32} - \frac{1}{8} + \frac{1}{48}\right)$ $A = \frac{3 + 6 - 4}{24} - \left(\frac{3 - 12 + 2}{96}\right)$ $A = \frac{5}{24} - \left(\frac{-7}{96}\right)$ $A = \frac{5}{24} + \frac{7}{96}$ $A = \frac{20 + 7}{96} = \frac{27}{96} = \frac{9}{32}$

Common Mistakes & Tips

• Always sketch the region to visualize the curves and their intersection points. This helps in determining which function is "above" the other.
• Be careful with signs when evaluating the definite integral, especially when substituting negative values.
• Double-check your algebra and arithmetic to avoid errors in the calculations.

Summary

We found the area of the region by first determining the intersection points of the curves $y^2 = 2x$ and $y = 4x - 1$. Then, we expressed both equations in the form $x = f(y)$ and integrated the difference between these functions with respect to $y$ from the lower intersection point's y-coordinate to the upper intersection point's y-coordinate. The final area is $\frac{9}{32}$.

Final Answer

The final answer is \boxed{\frac{9}{32}}, which corresponds to option (A).