Consider the function defined implicitly by the equation `y^3-3y+x=0` on various intervals in the real line. If `x in (-oo,-2) uu (2,oo)`, the equation implicitly defines a unique real-valued defferentiable function `y=f(x)`. If `x in (-2,2)`, the equation implicitly defines a unique real-valud diferentiable function `y-g(x)` satisfying `g_(0)=0`.
The area of the region bounded by the curve `y=f(x)`, the X-axis and the line `x=a` and `x=b`, where `-oo lt a lt b lt -2` is

Options

A. `int_(a)^(b)(x)/(3[{f(x)}^(2)-1])dx+by(b)-af(a)`
B. `-int_(a)^(b)(x)/(3[{f(x)}^(2)-1])dx-by(b)+af(a)`
C. `int_(a)^(b)(x)/(3[{f(x)}^(2)-1])dx-by(b)+af(a)`
D. `-int_(a)^(b)(x)/(3[{f(x)}^(2)-1])dx+by(b)=af(a)`

Answer

To find the area of the region bounded by the curve \( y = f(x) \), the x-axis, and the lines \( x = a \) and \( x = b \) where \( a, b \in (-\infty, -2) \), we will follow these steps:

### Step 1: Understand the Implicit Function
The function is defined implicitly by the equation:
\[ y^3 - 3y + x = 0 \]
This means that for a given \( x \), we can find \( y \) such that the equation holds true.

### Step 2: Identify the Area to be Calculated
We need to calculate the area under the curve \( y = f(x) \) from \( x = a \) to \( x = b \):
\[ \text{Area} = \int_{a}^{b} f(x) \, dx \]

### Step 3: Use Integration by Parts
To compute the integral, we can use integration by parts. We will let:
- \( u = f(x) \)
- \( dv = dx \)

Then, we have:
- \( du = f'(x) \, dx \)
- \( v = x \)

Using integration by parts, we have:
\[ \int u \, dv = uv - \int v \, du \]
Thus,
\[ \int f(x) \, dx = x f(x) \bigg|_a^b - \int_a^b x f'(x) \, dx \]

### Step 4: Evaluate the Boundary Terms
Now we evaluate the boundary terms:
\[ x f(x) \bigg|_a^b = b f(b) - a f(a) \]

### Step 5: Find the Derivative \( f'(x) \)
To find \( f'(x) \), we differentiate the implicit equation:
\[ y^3 - 3y + x = 0 \]
Differentiating both sides with respect to \( x \):
\[ 3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 1 = 0 \]
This simplifies to:
\[ (3y^2 - 3) \frac{dy}{dx} + 1 = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{3(y^2 - 1)} \]
So,
\[ f'(x) = -\frac{1}{3(f(x)^2 - 1)} \]

### Step 6: Substitute Back into the Area Formula
Now we substitute \( f'(x) \) back into our area formula:
\[ \text{Area} = b f(b) - a f(a) - \int_a^b x \left(-\frac{1}{3(f(x)^2 - 1)}\right) \, dx \]
This simplifies to:
\[ \text{Area} = b f(b) - a f(a) + \frac{1}{3} \int_a^b \frac{x}{f(x)^2 - 1} \, dx \]

### Final Area Expression
Thus, the area of the region bounded by the curve \( y = f(x) \), the x-axis, and the lines \( x = a \) and \( x = b \) is given by:
\[ \text{Area} = b f(b) - a f(a) + \frac{1}{3} \int_a^b \frac{x}{f(x)^2 - 1} \, dx \]