Area enclosed between the curve `y=x^(2)-5x+15` and the line `y=3x+3` is

Options

A. `(32)/(3)`
B. `(16)/(3)`
C. `(8)/(3)`
D. `(4)/(3)`

Answer

To find the area enclosed between the curve \( y = x^2 - 5x + 15 \) and the line \( y = 3x + 3 \), we will follow these steps:

### Step 1: Set the equations equal to find the points of intersection
We start by setting the equations equal to each other:

\[ x^2 - 5x + 15 = 3x + 3 \]

### Step 2: Rearrange the equation
Rearranging the equation gives:

\[ x^2 - 5x - 3x + 15 - 3 = 0 \]

This simplifies to:

\[ x^2 - 8x + 12 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - 6)(x - 2) = 0 \]

### Step 4: Find the roots
Setting each factor to zero gives us the points of intersection:

\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 5: Set up the integral for the area
The area \( A \) between the curves from \( x = 2 \) to \( x = 6 \) can be found using the integral:

\[ A = \int_{2}^{6} \left( (x^2 - 5x + 15) - (3x + 3) \right) \, dx \]

### Step 6: Simplify the integrand
Simplifying the integrand:

\[ A = \int_{2}^{6} \left( x^2 - 5x + 15 - 3x - 3 \right) \, dx \]
\[ = \int_{2}^{6} \left( x^2 - 8x + 12 \right) \, dx \]

### Step 7: Integrate the function
Now we integrate:

\[ A = \left[ \frac{x^3}{3} - 4x^2 + 12x \right]_{2}^{6} \]

### Step 8: Evaluate the definite integral
Calculating the integral at the bounds:

1. Evaluate at \( x = 6 \):

\[ = \frac{6^3}{3} - 4(6^2) + 12(6) = \frac{216}{3} - 4(36) + 72 = 72 - 144 + 72 = 0 \]

2. Evaluate at \( x = 2 \):

\[ = \frac{2^3}{3} - 4(2^2) + 12(2) = \frac{8}{3} - 16 + 24 = \frac{8}{3} + 8 = \frac{8 + 24}{3} = \frac{32}{3} \]

### Step 9: Calculate the area
Now, substituting back into the integral:

\[ A = 0 - \frac{32}{3} = -\frac{32}{3} \]

Since area cannot be negative, we take the positive value:

\[ A = \frac{32}{3} \]

### Final Answer
Thus, the area enclosed between the curve and the line is:

\[ \boxed{\frac{32}{3}} \]