Key Concepts and Formulas

• Area between curves: The area between two curves $x = f(y)$ and $x = g(y)$ between $y = a$ and $y = b$, where $f(y) \ge g(y)$ on $[a, b]$, is given by $\int_a^b [f(y) - g(y)] \, dy$.
• Solving quadratic equations: For a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Integration of Polynomials: $\int y^n dy = \frac{y^{n+1}}{n+1} + C$, where $n \neq -1$.

Step-by-Step Solution

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

The given equations are:

1. $y^2 + 4x = 4$
2. $y - 2x = 2$

We rewrite these equations as $x = f(y)$ and $x = g(y)$:

From equation (1): $4x = 4 - y^2$, so $x = 1 - \frac{y^2}{4}$. Let $f(y) = 1 - \frac{y^2}{4}$.

From equation (2): $2x = y - 2$, so $x = \frac{y}{2} - 1$. Let $g(y) = \frac{y}{2} - 1$.

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

To find the points of intersection, we set $f(y) = g(y)$: $1 - \frac{y^2}{4} = \frac{y}{2} - 1$ Multiplying by 4, we get: $4 - y^2 = 2y - 4$ $y^2 + 2y - 8 = 0$ $(y + 4)(y - 2) = 0$ So, $y = -4$ or $y = 2$.

Step 3: Determine which function is greater on the interval $[-4, 2]$.

We need to determine if $f(y) \ge g(y)$ or $g(y) \ge f(y)$ on the interval $[-4, 2]$. Let's test a value within the interval, say $y = 0$: $f(0) = 1 - \frac{0^2}{4} = 1$ $g(0) = \frac{0}{2} - 1 = -1$ Since $f(0) > g(0)$, we have $f(y) \ge g(y)$ on the interval $[-4, 2]$.

Step 4: Set up the integral to find the area.

The area enclosed by the curves is given by: $A = \int_{-4}^{2} [f(y) - g(y)] \, dy = \int_{-4}^{2} \left[ \left(1 - \frac{y^2}{4}\right) - \left(\frac{y}{2} - 1\right) \right] \, dy$ $A = \int_{-4}^{2} \left(2 - \frac{y^2}{4} - \frac{y}{2}\right) \, dy$

Step 5: Evaluate the integral.

$A = \left[ 2y - \frac{y^3}{12} - \frac{y^2}{4} \right]_{-4}^{2}$ $A = \left[ 2(2) - \frac{(2)^3}{12} - \frac{(2)^2}{4} \right] - \left[ 2(-4) - \frac{(-4)^3}{12} - \frac{(-4)^2}{4} \right]$ $A = \left[ 4 - \frac{8}{12} - \frac{4}{4} \right] - \left[ -8 - \frac{-64}{12} - \frac{16}{4} \right]$ $A = \left[ 4 - \frac{2}{3} - 1 \right] - \left[ -8 + \frac{16}{3} - 4 \right]$ $A = \left[ 3 - \frac{2}{3} \right] - \left[ -12 + \frac{16}{3} \right]$ $A = 3 - \frac{2}{3} + 12 - \frac{16}{3}$ $A = 15 - \frac{18}{3}$ $A = 15 - 6$ $A = 9$

Step 6: Re-evaluate to find the correct answer.

There was an error in Step 5. Let's recalculate: $A = \left[ 2y - \frac{y^3}{12} - \frac{y^2}{4} \right]_{-4}^{2}$ $A = \left[ 2(2) - \frac{(2)^3}{12} - \frac{(2)^2}{4} \right] - \left[ 2(-4) - \frac{(-4)^3}{12} - \frac{(-4)^2}{4} \right]$ $A = \left[ 4 - \frac{8}{12} - 1 \right] - \left[ -8 + \frac{64}{12} - 4 \right]$ $A = \left[ 3 - \frac{2}{3} \right] - \left[ -12 + \frac{16}{3} \right]$ $A = 3 - \frac{2}{3} + 12 - \frac{16}{3}$ $A = 15 - \frac{18}{3}$ $A = 15 - 6 = 9$

It seems we are still getting 9, which is not the correct answer. The error lies in assuming $f(y) \ge g(y)$ throughout the interval. Let's re-evaluate the integral carefully.

$A = \int_{-4}^2 (2 - \frac{y^2}{4} - \frac{y}{2}) dy = [2y - \frac{y^3}{12} - \frac{y^2}{4}]_{-4}^2$ $A = (4 - \frac{8}{12} - 1) - (-8 - \frac{-64}{12} - 4) = (3 - \frac{2}{3}) - (-12 + \frac{16}{3}) = 3 - \frac{2}{3} + 12 - \frac{16}{3} = 15 - \frac{18}{3} = 15 - 6 = 9$.

Let's go back to Step 5 and check the calculation again.

$A = \left[ 2y - \frac{y^3}{12} - \frac{y^2}{4} \right]_{-4}^{2} = \left(4 - \frac{8}{12} - 1\right) - \left(-8 + \frac{64}{12} - 4\right) = 3 - \frac{2}{3} - (-12 + \frac{16}{3}) = 3 - \frac{2}{3} + 12 - \frac{16}{3} = 15 - \frac{18}{3} = 15 - 6 = 9$.

Still getting 9. Let's try another approach to check. The definite integral is: $\int_{-4}^2 (2 - \frac{y^2}{4} - \frac{y}{2}) dy = [2y - \frac{y^3}{12} - \frac{y^2}{4}]_{-4}^2 = (4 - \frac{8}{12} - 1) - (-8 + \frac{64}{12} - 4) = 3 - \frac{2}{3} + 12 - \frac{16}{3} = 15 - \frac{18}{3} = 15 - 6 = 9$

After reviewing the problem and the calculations multiple times, it appears there might be an error in the provided correct answer. The calculated area is consistently 9. However, we are given that the correct answer is $\frac{22}{3}$. Let's examine the integration one more time.

$f(y) = 1 - \frac{y^2}{4}$ and $g(y) = \frac{y}{2} - 1$. $A = \int_{-4}^2 (f(y) - g(y)) dy = \int_{-4}^2 (1 - \frac{y^2}{4} - (\frac{y}{2} - 1)) dy = \int_{-4}^2 (2 - \frac{y^2}{4} - \frac{y}{2}) dy = [2y - \frac{y^3}{12} - \frac{y^2}{4}]_{-4}^2 = (4 - \frac{8}{12} - 1) - (-8 + \frac{64}{12} - 4) = (3 - \frac{2}{3}) - (-12 + \frac{16}{3}) = 3 - \frac{2}{3} + 12 - \frac{16}{3} = 15 - 6 = 9$

It seems the correct answer provided may be incorrect.

Common Mistakes & Tips

• Incorrectly identifying the upper and lower functions: Always verify which function is greater than the other in the interval of integration.
• Sign Errors: Be careful with signs when substituting the limits of integration.
• Computational Errors: Double-check all arithmetic calculations, especially when dealing with fractions.

Summary

We found the points of intersection of the parabola $y^2 + 4x = 4$ and the line $y - 2x = 2$. We expressed both equations in the form $x = f(y)$. We then set up the integral to find the area between the curves and evaluated the integral. However, the calculated area is 9, which does not match the provided correct answer of $\frac{22}{3}$. After reviewing the calculations thoroughly, we conclude that there may be an error in the provided answer key.

Final Answer

The final answer is \boxed{9}, which does not correspond to any of the provided options.