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 step by step, we will first analyze the functions and the areas they define.

### Step 1: Identify the functions
We have two functions:
- \( f(x) = x - x^2 \) (a downward-opening parabola)
- \( g(x) = ax \) (a straight line through the origin)

### Step 2: Find the intersection points of \( f(x) \) and \( g(x) \)
To find the area between these two curves, we first need to determine where they intersect. Set \( f(x) = g(x) \):
\[ x - x^2 = ax \]
Rearranging gives:
\[ x^2 + ax - x = 0 \implies x^2 + (a - 1)x = 0 \]
Factoring out \( x \):
\[ x(x + (a - 1)) = 0 \]
This gives us two solutions:
1. \( x = 0 \)
2. \( x = 1 - a \)

### Step 3: Determine the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 - a \) is given by:
\[ A = \int_0^{1-a} (f(x) - g(x)) \, dx = \int_0^{1-a} ((x - x^2) - ax) \, dx \]
This simplifies to:
\[ A = \int_0^{1-a} (x - x^2 - ax) \, dx = \int_0^{1-a} ((1-a)x - x^2) \, dx \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_0^{1-a} ((1-a)x - x^2) \, dx \]
Calculating the integral:
\[ = \left[ \frac{(1-a)x^2}{2} - \frac{x^3}{3} \right]_0^{1-a} \]
Substituting \( x = 1-a \):
\[ = \frac{(1-a)(1-a)^2}{2} - \frac{(1-a)^3}{3} \]
\[ = \frac{(1-a)^3}{2} - \frac{(1-a)^3}{3} \]
Finding a common denominator (which is 6):
\[ = \frac{3(1-a)^3}{6} - \frac{2(1-a)^3}{6} = \frac{(1-a)^3}{6} \]

### Step 5: Area bounded by the curves \( x = 3y - y^2 \) and \( x + y = 3 \)
Next, we need to find the area bounded by the curves \( x = 3y - y^2 \) and \( x + y = 3 \).

1. **Find the intersection points**:
Set \( 3y - y^2 = 3 - y \):
\[ y^2 - 4y + 3 = 0 \]
Factoring gives:
\[ (y-1)(y-3) = 0 \implies y = 1 \text{ or } y = 3 \]
For \( y = 1 \), \( x = 2 \) and for \( y = 3 \), \( x = 0 \).

2. **Calculate the area**:
The area \( A' \) can be calculated as:
\[ A' = \int_1^3 ((3y - y^2) - (3 - y)) \, dy \]
Simplifying the integrand:
\[ = \int_1^3 (4y - y^2 - 3) \, dy = \int_1^3 (-y^2 + 4y - 3) \, dy \]

3. **Evaluate the integral**:
\[ = \left[ -\frac{y^3}{3} + 2y^2 - 3y \right]_1^3 \]
Evaluating at the bounds:
\[ = \left( -\frac{27}{3} + 18 - 9 \right) - \left( -\frac{1}{3} + 2 - 3 \right) \]
\[ = (-9 + 18 - 9) - \left(-\frac{1}{3} - 1\right) = 0 + \frac{4}{3} = \frac{4}{3} \]

### Step 6: Set the areas equal
Now we set the areas equal:
\[ \frac{(1-a)^3}{6} = \frac{4}{3} \]
Cross-multiplying gives:
\[ (1-a)^3 = 8 \implies 1-a = 2 \implies a = -1 \]

### Step 7: Find the value of \(|[a]|\)
The greatest integer less than or equal to \(-1\) is \(-1\), and taking the absolute value gives:
\[ |[a]| = 1 \]

### Final Answer
Thus, the value of \(|[a]|\) is \( \boxed{1} \).