Key Concepts and Formulas

• Area Between Curves: The area between two curves $x = f(y)$ and $x = g(y)$, where $f(y) \ge g(y)$ on the interval $[c, d]$, is given by $\int_c^d [f(y) - g(y)] \, dy$.
• Parabola: The equation $y^2 = 4ax$ represents a parabola opening to the right.
• Solving Inequalities: To find the region defined by inequalities, we need to consider the boundaries and test points within the region.

Step-by-Step Solution

Step 1: Express the inequalities in terms of x as a function of y.

The first inequality, $y^2 \le 2x$, can be rewritten as $x \ge \frac{y^2}{2}$. The second inequality, $y \ge 4x - 1$, can be rewritten as $4x \le y + 1$, or $x \le \frac{y + 1}{4}$.

Step 2: Find the points of intersection of the two curves.

We need to find the points where the curves $x = \frac{y^2}{2}$ and $x = \frac{y + 1}{4}$ intersect. Setting the expressions for $x$ equal to each other, we have: $\frac{y^2}{2} = \frac{y + 1}{4}$ Multiplying both sides by 4 gives: $2y^2 = y + 1$ $2y^2 - y - 1 = 0$ Factoring the quadratic equation, we get: $(2y + 1)(y - 1) = 0$ So, the solutions are $y = 1$ and $y = -\frac{1}{2}$.

Step 3: Determine which function is greater in the region of interest.

For $-\frac{1}{2} \le y \le 1$, we need to determine whether $\frac{y+1}{4}$ or $\frac{y^2}{2}$ is greater. We can test a value of $y$ in this range, say $y = 0$. When $y = 0$, $\frac{y + 1}{4} = \frac{1}{4}$ and $\frac{y^2}{2} = 0$. Since $\frac{1}{4} > 0$, we have $\frac{y + 1}{4} \ge \frac{y^2}{2}$ in the interval $[-\frac{1}{2}, 1]$.

Step 4: Set up the integral for the area.

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

Step 5: Evaluate the integral.

$A = \int_{-\frac{1}{2}}^{1} \left( \frac{y}{4} + \frac{1}{4} - \frac{y^2}{2} \right) dy$ $A = \left[ \frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6} \right]_{-\frac{1}{2}}^{1}$ $A = \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{(-\frac{1}{2})^2}{8} + \frac{-\frac{1}{2}}{4} - \frac{(-\frac{1}{2})^3}{6} \right)$ $A = \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \right)$ $A = \frac{1}{8} + \frac{1}{4} - \frac{1}{6} - \frac{1}{32} + \frac{1}{8} - \frac{1}{48}$ $A = \frac{2}{8} + \frac{1}{4} - \frac{1}{6} - \frac{1}{32} - \frac{1}{48}$ $A = \frac{1}{4} + \frac{1}{4} - \frac{1}{6} - \frac{1}{32} - \frac{1}{48}$ $A = \frac{1}{2} - \frac{1}{6} - \frac{1}{32} - \frac{1}{48}$ $A = \frac{24}{48} - \frac{8}{48} - \frac{3}{96} - \frac{2}{96}$ $A = \frac{16}{48} - \frac{5}{96}$ $A = \frac{1}{3} - \frac{5}{96}$ $A = \frac{32}{96} - \frac{5}{96}$ $A = \frac{27}{96} = \frac{9}{32}$ There seems to be an error in the calculation. Let's recalculate after the substitution: $A = \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \right)$ $A = \left( \frac{3+6-4}{24} \right) - \left( \frac{3-12+2}{96} \right)$ $A = \frac{5}{24} - \left( \frac{-7}{96} \right)$ $A = \frac{5}{24} + \frac{7}{96}$ $A = \frac{20}{96} + \frac{7}{96} = \frac{27}{96} = \frac{9}{32}$ Still incorrect. Let's simplify differently. $A = \left( \frac{1}{8} + \frac{1}{4} - \frac{1}{6} \right) - \left( \frac{1}{32} - \frac{1}{8} + \frac{1}{48} \right) = \frac{5}{24} + \frac{7}{96} = \frac{20 + 7}{96} = \frac{27}{96} = \frac{9}{32}$ The calculation is still leading to $\frac{9}{32}$. Need to re-examine the setup. The correct answer is $\frac{15}{64}$. Let us rework the integral:

$A = \int_{-1/2}^{1} (\frac{y+1}{4} - \frac{y^2}{2}) dy = [\frac{y^2}{8} + \frac{y}{4} - \frac{y^3}{6}]_{-1/2}^{1}$ $= (\frac{1}{8} + \frac{1}{4} - \frac{1}{6}) - (\frac{1/4}{8} - \frac{1}{8} + \frac{1/8}{6}) = (\frac{3+6-4}{24}) - (\frac{1}{32} - \frac{1}{8} + \frac{1}{48}) = \frac{5}{24} - (\frac{3-12+2}{96})$ $= \frac{5}{24} - \frac{-7}{96} = \frac{5}{24} + \frac{7}{96} = \frac{20+7}{96} = \frac{27}{96} = \frac{9}{32}$ Still getting $\frac{9}{32}$. The correct answer is $\frac{15}{64}$. So, there's an error in the problem statement or options.

After reviewing the question and carefully re-calculating the integral, I believe there might be a mistake in the provided correct answer. All the steps are correct, and the final area should be $\frac{9}{32}$. But, the provided correct answer is $\frac{15}{64}$.

Common Mistakes & Tips

• Incorrect Limits of Integration: Make sure to find the correct y-values where the curves intersect to use as the limits of integration.
• Incorrect Subtraction Order: Ensure you are subtracting the "lower" function from the "upper" function within the integral.
• Algebra Errors: Be careful with algebraic manipulations, especially when dealing with fractions and negative signs.

Summary

We found the area of the region bounded by the given inequalities by first rewriting the inequalities as $x$ in terms of $y$. Then, we found the intersection points of the curves and set up a definite integral to calculate the area. After evaluating the integral, we get the area as $\frac{9}{32}$.

Final Answer The final answer is \boxed{\frac{9}{32}}, which is closest to option (B), but the provided correct answer is (A). There is likely an error in the provided options or the given answer.