Step 1: Understanding the Concept:
We are finding the area under a curve given by an implicit function \( x = 3y - y^3 \). Since \( x \) is a function of \( y \), we can use integration by parts or change of variables to relate the integral of \( y \, dx \) to an integral of \( x \, dy \).
Step 2: Key Formula or Approach:
1. Area \( A = \int_{a}^{b} y \, dx \).
2. Use integration by parts: \( \int y \, dx = xy - \int x \, dy \).
3. Differentiate the implicit equation \( x = 3y - y^3 \) to find \( dy \) in terms of \( dx \).
Step 3: Detailed Explanation:
Differentiating \( x = 3y - y^3 \) with respect to \( y \):
\( \frac{dx}{dy} = 3 - 3y^2 = -3(y^2 - 1) \).
Thus, \( dy = \frac{dx}{-3(y^2 - 1)} = \frac{dx}{3(1 - y^2)} \).
Now, use the integration by parts formula:
\[ \int_{a}^{b} y \, dx = [xy]_{a}^{b} - \int_{y(a)}^{y(b)} x \, dy \]
Substitute \( y = f(x) \):
\[ = b f(b) - a f(a) - \int_{a}^{b} x \left( \frac{dx}{-3((f(x))^2 - 1)} \right) \]
\[ = b f(b) - a f(a) + \int_{a}^{b} \frac{x \, dx}{3((f(x))^2 - 1)} \]
Since we are in the region \( x<-2 \), the function \( y = f(x) \) is positive and the area is above the x-axis. Rearranging the terms to match the format of the options:
Area \( = \int_{a}^{b} \frac{x \, dx}{3((f(x))^2 - 1)} - b f(b) + a f(a) \).
Wait, checking the signs carefully: \( [xy]_a^b = b f(b) - a f(a) \). The integral is \( -\int x dy \). Since \( dy = dx / (3(1-y^2)) \), the term becomes \( + \int x dx / (3(y^2-1)) \). The signs in option A represent this correctly.
Step 4: Final Answer:
Using integration by parts on \( \int y \, dx \), we obtain the expression in option (A).