The area of the region bounded by the `x`-axis , the function `y =-x^(2) + 4x - 8`, `x = -1` and `x = 4` is equal to:

Options

A. (a) `31""(2)/(3)` sq. units
B. (b) `31` sq. units
C. (c) `32""(2)/(3)` sq. units
D. (d) `32` sq. units

Answer

To find the area of the region bounded by the x-axis, the function \( y = -x^2 + 4x - 8 \), and the vertical lines \( x = -1 \) and \( x = 4 \), we will follow these steps:

### Step 1: Identify the function and the bounds
The function we are working with is:
\[ f(x) = -x^2 + 4x - 8 \]
We need to calculate the area between this curve and the x-axis from \( x = -1 \) to \( x = 4 \).

### Step 2: Determine if the function is above or below the x-axis
To find where the function intersects the x-axis, we set \( f(x) = 0 \):
\[ -x^2 + 4x - 8 = 0 \]
This can be rearranged to:
\[ x^2 - 4x + 8 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 8 = 16 - 32 = -16 \]
Since the discriminant is negative, the function does not intersect the x-axis, meaning it is always below the x-axis.

### Step 3: Set up the integral for the area
Since the function is below the x-axis, the area \( A \) can be calculated as:
\[ A = -\int_{-1}^{4} f(x) \, dx = -\int_{-1}^{4} (-x^2 + 4x - 8) \, dx \]
This simplifies to:
\[ A = \int_{-1}^{4} (x^2 - 4x + 8) \, dx \]

### Step 4: Compute the integral
Now we compute the integral:
\[ A = \int_{-1}^{4} (x^2 - 4x + 8) \, dx \]
Calculating the integral:
\[ A = \left[ \frac{x^3}{3} - 2x^2 + 8x \right]_{-1}^{4} \]

### Step 5: Evaluate the integral at the bounds
Now we evaluate at the upper and lower limits:
1. Evaluate at \( x = 4 \):
\[ \frac{4^3}{3} - 2(4^2) + 8(4) = \frac{64}{3} - 32 + 32 = \frac{64}{3} \]
2. Evaluate at \( x = -1 \):
\[ \frac{(-1)^3}{3} - 2(-1)^2 + 8(-1) = -\frac{1}{3} - 2 - 8 = -\frac{1}{3} - 10 = -\frac{31}{3} \]

### Step 6: Calculate the area
Now we combine the results:
\[ A = \left( \frac{64}{3} - \left( -\frac{31}{3} \right) \right) = \frac{64}{3} + \frac{31}{3} = \frac{95}{3} \]

### Step 7: Final area
Since the area is positive, we conclude:
\[ A = \frac{95}{3} = 31 \frac{2}{3} \text{ square units} \]

### Conclusion
Thus, the area of the region bounded by the x-axis, the function, and the vertical lines is:
\[ \boxed{31 \frac{2}{3}} \text{ square units} \]