Find the area of the region which is inside the parabola `y = - x^(2) + 6x - 5`, outside the parabola `y = - x^(2) + 4x - 3` and left of the straight line `y = 3x-15`.

Answer

To find the area of the region that is inside the parabola \( y = -x^2 + 6x - 5 \), outside the parabola \( y = -x^2 + 4x - 3 \), and left of the straight line \( y = 3x - 15 \), we will follow these steps:

### Step 1: Identify the equations of the curves
We have the following equations:
1. Parabola 1: \( y = -x^2 + 6x - 5 \)
2. Parabola 2: \( y = -x^2 + 4x - 3 \)
3. Line: \( y = 3x - 15 \)

### Step 2: Find the vertices of the parabolas
To find the vertices of the parabolas, we can rewrite them in vertex form.

- For the first parabola:
\[ y = -x^2 + 6x - 5 \implies y = -(x^2 - 6x + 9) + 4 \implies y = -(x - 3)^2 + 4 \]
The vertex is at \( (3, 4) \).

- For the second parabola:
\[ y = -x^2 + 4x - 3 \implies y = -(x^2 - 4x + 4) + 1 \implies y = -(x - 2)^2 + 1 \]
The vertex is at \( (2, 1) \).

### Step 3: Find the intersection points of the curves
To find the intersection points of the first parabola and the line, we set them equal:
\[ 3x - 15 = -x^2 + 6x - 5 \]
Rearranging gives:
\[ x^2 - x - 12 = 0 \]
Factoring:
\[ (x - 4)(x + 3) = 0 \implies x = 4 \text{ or } x = -3 \]

### Step 4: Determine the area between the curves
The area we want is bounded by \( x = 1 \) and \( x = 4 \) for the first parabola and the second parabola, and then from \( x = 4 \) to \( x = 5 \) for the area between the first parabola and the line.

#### Area from \( x = 1 \) to \( x = 4 \):
\[ \text{Area}_1 = \int_{1}^{4} \left( (-x^2 + 6x - 5) - (-x^2 + 4x - 3) \right) \, dx \]
This simplifies to:
\[ \text{Area}_1 = \int_{1}^{4} (2x - 2) \, dx \]

#### Area from \( x = 4 \) to \( x = 5 \):
\[ \text{Area}_2 = \int_{4}^{5} \left( (-x^2 + 6x - 5) - (3x - 15) \right) \, dx \]
This simplifies to:
\[ \text{Area}_2 = \int_{4}^{5} (-x^2 + 3x + 10) \, dx \]

### Step 5: Calculate the integrals
1. **Calculating Area 1**:
\[ \text{Area}_1 = \int_{1}^{4} (2x - 2) \, dx = [x^2 - 2x]_{1}^{4} = (16 - 8) - (1 - 2) = 8 + 1 = 9 \]

2. **Calculating Area 2**:
\[ \text{Area}_2 = \int_{4}^{5} (-x^2 + 3x + 10) \, dx = \left[-\frac{x^3}{3} + \frac{3x^2}{2} + 10x\right]_{4}^{5} \]
Evaluating at the limits:
\[ = \left[-\frac{125}{3} + \frac{75}{2} + 50\right] - \left[-\frac{64}{3} + 24 + 40\right] \]

### Step 6: Combine the areas
The total area is:
\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 \]

### Final Result
After evaluating the integrals and simplifying, we find that the area of the region is \( \frac{73}{6} \) square units.