Step-by-step Derivation

1. Analyze the Implicit Function The curve is defined by the equation $y^3 - 3y + x = 0$. We can express $x$ as a function of $y$: $x = 3y - y^3$. Let's call this $x = h(y)$.

2. Determine the properties of f(x) The problem asks for the area in the region where $-\infty < a < b < -2$. So we need to understand the function $y=f(x)$ for $x \in (-\infty, -2)$. To analyze the relation between $x$ and $y$, let's find the critical points of $x = h(y)$. ${dx \over dy} = 3 - 3y^2 = 3(1-y^2)$ Setting ${dx \over dy} = 0$ gives $y = \pm 1$. These are the points where the curve has vertical tangents.

   • At $y=1$, $x = 3(1) - 1^3 = 2$. This is a local maximum for $x$.
   • At $y=-1$, $x = 3(-1) - (-1)^3 = -3 + 1 = -2$. This is a local minimum for $x$.

   Let's examine the intervals for $y$ based on the value of $x$:

   • When $y > 1$, ${dx \over dy} < 0$, so $x$ is a decreasing function of $y$. As $y$ goes from $1$ to $\infty$, $x$ goes from $h(1)=2$ to $-\infty$. Thus, for $x \in (-\infty, 2)$, there is a unique root with $y > 1$.
   • When $-1 < y < 1$, ${dx \over dy} > 0$, so $x$ is an increasing function of $y$. As $y$ goes from $-1$ to $1$, $x$ goes from $h(-1)=-2$ to $h(1)=2$. Thus, for $x \in (-2, 2)$, there is a unique root with $y \in (-1, 1)$.
   • When $y < -1$, ${dx \over dy} < 0$, so $x$ is a decreasing function of $y$. As $y$ goes from $-1$ to $-\infty$, $x$ goes from $h(-1)=-2$ to $\infty$. Thus, for $x \in (-2, \infty)$, there is a unique root with $y < -1$.

   The problem states that for $x \in (-\infty, -2) \cup (2, \infty)$, the function $y=f(x)$ is unique. Based on our analysis:

   • For $x \in (-\infty, -2)$, the unique real root for $y$ is in the interval $(1, \infty)$. So, $f(x) > 1$.
   • For $x \in (2, \infty)$, the unique real root for $y$ is in the interval $(-\infty, -1)$. So, $f(x) < -1$.

3. Set up the Area Integral We need to find the area of the region bounded by $y=f(x)$, the x-axis, and the lines $x=a$ and $x=b$, where $-\infty < a < b < -2$. In this interval, we have established that $f(x) > 1$. Since $f(x)$ is positive, the area $A$ is given by the definite integral: $A = \int_a^b f(x) dx$

4. Evaluate the Integral using Integration by Parts The integral $\int f(x) dx$ is difficult to compute directly. We can use integration by parts, where $\int u \, dv = uv - \int v \, du$. Let $u = f(x)$ and $dv = dx$. Then $du = f'(x) dx$ and $v = x$. Applying this to our area integral: $A = \int_a^b f(x) dx = [x f(x)]_a^b - \int_a^b x f'(x) dx$ $A = (b f(b) - a f(a)) - \int_a^b x f'(x) dx$

5. Find f'(x) using Implicit Differentiation We differentiate the original equation $y^3 - 3y + x = 0$ with respect to $x$, treating $y$ as $f(x)$: $3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 1 = 0$ $(3y^2 - 3) \frac{dy}{dx} = -1$ $\frac{dy}{dx} = f'(x) = \frac{-1}{3(y^2 - 1)} = \frac{-1}{3((f(x))^2 - 1)}$ Note that since $f(x) > 1$ for $x < -2$, the denominator is never zero, so the function is differentiable.

6. Substitute f'(x) into the Area Formula Now we substitute the expression for $f'(x)$ back into our formula for area $A$: $A = b f(b) - a f(a) - \int_a^b x \left( \frac{-1}{3((f(x))^2 - 1)} \right) dx$ $A = b f(b) - a f(a) + \int_a^b \frac{x}{3((f(x))^2 - 1)} dx$

7. Compare with the Options Rearranging the terms to match the format of the options: $A = \int_a^b \frac{x}{3((f(x))^2 - 1)} dx + b f(b) - a f(a)$ This expression matches option A.

Conclusion

The area of the specified region is given by the formula derived, which corresponds to option A.