Key Concepts and Formulas

• Area Between Curves: The area between two curves $y = f(x)$ and $y = g(x)$ from $x = a$ to $x = b$, where $f(x) \ge g(x)$ on the interval $[a, b]$, is given by $\int_a^b [f(x) - g(x)] \, dx$.
• Intersection Points: Finding the intersection points of curves $y = f(x)$ and $y = g(x)$ involves solving the equation $f(x) = g(x)$.
• Integration: Basic integration techniques are required.

Step-by-Step Solution

Step 1: Visualize the Region and Find Intersection Points

We are given the region defined by $y^2 \le 4x$, $x + y \le 1$, $x \ge 0$, and $y \ge 0$. The inequalities represent:

• $y^2 \le 4x$ : The region to the right of the parabola $x = \frac{y^2}{4}$
• $x + y \le 1$ : The region below the line $x = 1 - y$
• $x \ge 0$ and $y \ge 0$ : The first quadrant

To find the area of this region, we need to determine the intersection points of the curves $x = \frac{y^2}{4}$ and $x = 1 - y$. Setting them equal gives us: $\frac{y^2}{4} = 1 - y$ $y^2 = 4 - 4y$ $y^2 + 4y - 4 = 0$ Using the quadratic formula, we find: $y = \frac{-4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{-4 \pm \sqrt{32}}{2} = \frac{-4 \pm 4\sqrt{2}}{2} = -2 \pm 2\sqrt{2}$ Since $y \ge 0$, we only consider the positive root: $y = -2 + 2\sqrt{2}$. Then $x = 1 - y = 1 - (-2 + 2\sqrt{2}) = 3 - 2\sqrt{2}$. Thus, the intersection point is $(3 - 2\sqrt{2}, -2 + 2\sqrt{2})$.

Step 2: Set up the Integral

Since we are integrating with respect to $y$, we need to express the area as an integral with respect to $y$. The area is bounded by the curves $x = \frac{y^2}{4}$ and $x = 1 - y$, with $0 \le y \le -2 + 2\sqrt{2}$. In this region, $1 - y \ge \frac{y^2}{4}$. Therefore, the area is given by: $A = \int_0^{-2 + 2\sqrt{2}} \left[ (1 - y) - \frac{y^2}{4} \right] dy$

Step 3: Evaluate the Integral

$A = \int_0^{-2 + 2\sqrt{2}} \left( 1 - y - \frac{y^2}{4} \right) dy = \left[ y - \frac{y^2}{2} - \frac{y^3}{12} \right]_0^{-2 + 2\sqrt{2}}$ Let $y_0 = -2 + 2\sqrt{2}$. Then $A = y_0 - \frac{y_0^2}{2} - \frac{y_0^3}{12}$ We have $y_0 = -2 + 2\sqrt{2}$, so $y_0^2 = (-2 + 2\sqrt{2})^2 = 4 - 8\sqrt{2} + 8 = 12 - 8\sqrt{2}$ $y_0^3 = (-2 + 2\sqrt{2})^3 = (-2)^3 + 3(-2)^2(2\sqrt{2}) + 3(-2)(2\sqrt{2})^2 + (2\sqrt{2})^3 = -8 + 24\sqrt{2} - 48 + 16\sqrt{2} = -56 + 40\sqrt{2}$ Then \begin{align*} A &= (-2 + 2\sqrt{2}) - \frac{12 - 8\sqrt{2}}{2} - \frac{-56 + 40\sqrt{2}}{12} \ &= -2 + 2\sqrt{2} - (6 - 4\sqrt{2}) - \left( -\frac{14}{3} + \frac{10\sqrt{2}}{3} \right) \ &= -2 + 2\sqrt{2} - 6 + 4\sqrt{2} + \frac{14}{3} - \frac{10\sqrt{2}}{3} \ &= -8 + \frac{14}{3} + 6\sqrt{2} - \frac{10\sqrt{2}}{3} \ &= \frac{-24 + 14}{3} + \frac{18\sqrt{2} - 10\sqrt{2}}{3} \ &= \frac{-10}{3} + \frac{8\sqrt{2}}{3} = \frac{8}{3}\sqrt{2} - \frac{10}{3} \end{align*} The area is $\frac{8}{3}\sqrt{2} - \frac{10}{3}$.

Step 4: Find a - b

We are given that the area is $a\sqrt{2} + b$, so $a = \frac{8}{3}$ and $b = -\frac{10}{3}$. We need to find $a - b$: $a - b = \frac{8}{3} - \left( -\frac{10}{3} \right) = \frac{8}{3} + \frac{10}{3} = \frac{18}{3} = 6$ There seems to be a mistake in the problem statement or the given answer. Let's re-evaluate the intersection point. $y^2 + 4y - 4 = 0$ $y = \frac{-4 + \sqrt{16 + 16}}{2} = \frac{-4 + \sqrt{32}}{2} = -2 + 2\sqrt{2}$ $x = 1 - y = 1 - (-2 + 2\sqrt{2}) = 3 - 2\sqrt{2}$ The area calculation looks correct. The problem states the area is $a \sqrt{2} + b$, and we found it to be $\frac{8}{3}\sqrt{2} - \frac{10}{3}$. So, $a = \frac{8}{3}$ and $b = -\frac{10}{3}$. Then $a - b = \frac{8}{3} - (-\frac{10}{3}) = \frac{18}{3} = 6$.

Step 5: Re-examine the region

The region is bounded by $y^2 \le 4x$, $x+y \le 1$, $x \ge 0$, $y \ge 0$. Rewrite as $x \ge y^2/4$ and $x \le 1-y$. Area = $\int_0^{2\sqrt{2}-2} (1-y - y^2/4) dy = [y - y^2/2 - y^3/12]_0^{2\sqrt{2}-2}$. $= (2\sqrt{2}-2) - (12-8\sqrt{2})/2 - (-56+40\sqrt{2})/12 = 2\sqrt{2}-2 - 6 + 4\sqrt{2} + 14/3 - 10\sqrt{2}/3 = 6\sqrt{2} - 8 + 14/3 - 10\sqrt{2}/3 = 8\sqrt{2}/3 - 10/3$. $a = 8/3$ and $b = -10/3$. $a-b = 8/3 - (-10/3) = 18/3 = 6$.

Common Mistakes & Tips

• Incorrect Integration Limits: Carefully determine the intersection points to find the correct limits of integration.
• Incorrect Subtraction Order: Ensure you subtract the lower curve from the upper curve in the integral.
• Sign Errors: Pay close attention to signs when evaluating the integral and simplifying the expression.

Summary

We first found the intersection points of the parabola and the line, and then set up a definite integral to represent the area of the region. We evaluated the integral and simplified the result to the form $a\sqrt{2} + b$. Finally, we calculated $a - b$. The area is $\frac{8}{3}\sqrt{2} - \frac{10}{3}$, so $a = \frac{8}{3}$ and $b = -\frac{10}{3}$. Thus, $a - b = \frac{8}{3} - (-\frac{10}{3}) = 6$.

Final Answer

The final answer is \boxed{6}, which corresponds to option (C).