Let `f(x)=x-x^(2)` and `g(x)=ax`. If the area bounded by f(x) and g(x) is equal to the area bounded by the curves `x=3y-y^(2)` and `x+y=3`, then find the value of `|[a]|`:
[Note: denotes the greatest integer less than or equal to k.]

Answer

To solve the problem, we need to find the value of \(|a|\) given the functions \(f(x) = x - x^2\) and \(g(x) = ax\). We also need to equate the area bounded by \(f(x)\) and \(g(x)\) with the area bounded by the curves \(x = 3y - y^2\) and \(x + y = 3\).

### Step 1: Find the area bounded by \(f(x)\) and \(g(x)\)

1. **Find the points of intersection** of \(f(x)\) and \(g(x)\):
\[ x - x^2 = ax \]
Rearranging gives:
\[ x^2 + (a - 1)x = 0 \]
Factoring out \(x\):
\[ x(x + (a - 1)) = 0 \]
This gives us the solutions:
\[ x = 0 \quad \text{or} \quad x = 1 - a \]

2. **Determine the area** between the curves from \(x = 0\) to \(x = 1 - a\):
\[ \text{Area} = \int_0^{1-a} (f(x) - g(x)) \, dx = \int_0^{1-a} ((x - x^2) - ax) \, dx \]
Simplifying the integrand:
\[ = \int_0^{1-a} (x - x^2 - ax) \, dx = \int_0^{1-a} ((1-a)x - x^2) \, dx \]

3. **Calculate the integral**:
\[ = \left[ \frac{(1-a)x^2}{2} - \frac{x^3}{3} \right]_0^{1-a} \]
Evaluating at the bounds:
\[ = \frac{(1-a)(1-a)^2}{2} - \frac{(1-a)^3}{3} \]
\[ = \frac{(1-a)^3}{2} - \frac{(1-a)^3}{3} \]
Finding a common denominator (6):
\[ = \frac{3(1-a)^3}{6} - \frac{2(1-a)^3}{6} = \frac{(1-a)^3}{6} \]

### Step 2: Find the area bounded by the curves \(x = 3y - y^2\) and \(x + y = 3\)

1. **Find the points of intersection** of \(x = 3y - y^2\) and \(x + y = 3\):
Substitute \(x = 3 - y\) into \(3y - y^2\):
\[ 3 - y = 3y - y^2 \]
Rearranging gives:
\[ y^2 - 4y + 3 = 0 \]
Factoring:
\[ (y - 1)(y - 3) = 0 \quad \Rightarrow \quad y = 1, 3 \]
Finding corresponding \(x\) values:
- For \(y = 1\): \(x = 2\)
- For \(y = 3\): \(x = 0\)

2. **Calculate the area**:
The area is given by:
\[ \text{Area} = \int_0^2 ((3 - y) - (3y - y^2)) \, dy = \int_0^2 (3 - y - 3y + y^2) \, dy \]
Simplifying:
\[ = \int_0^2 (y^2 - 4y + 3) \, dy \]
Calculating the integral:
\[ = \left[ \frac{y^3}{3} - 2y^2 + 3y \right]_0^2 \]
Evaluating at the bounds:
\[ = \left( \frac{8}{3} - 8 + 6 \right) - 0 = \frac{8}{3} - 8 + 6 = \frac{8}{3} - \frac{24}{3} + \frac{18}{3} = \frac{2}{3} \]

### Step 3: Equate the areas

Set the areas equal:
\[ \frac{(1-a)^3}{6} = \frac{2}{3} \]
Cross-multiplying gives:
\[ (1-a)^3 = 4 \]
Taking the cube root:
\[ 1 - a = \sqrt[3]{4} \quad \Rightarrow \quad a = 1 - \sqrt[3]{4} \]

### Step 4: Find \(|a|\)

Calculating \(|a|\):
\[ |a| = |1 - \sqrt[3]{4}| \]
Since \(\sqrt[3]{4} \approx 1.5874\), we have:
\[ |a| = |1 - 1.5874| \approx 0.5874 \]

### Step 5: Find the greatest integer less than or equal to \(|a|\)

The greatest integer less than or equal to \(0.5874\) is \(0\).

### Final Answer
\[ \lfloor |a| \rfloor = 0 \]