The area bounded by the parabola `y=x^(2)+x+1,` its tangent at P(1, 3), line `x=-1` and the x - axis is A sq units. Then, the value of 6A is equal to

Options

A. a prime number
B. a composite number
C. an irrational number
D. a non - integer

Answer

To solve the problem, we need to find the area bounded by the parabola \( y = x^2 + x + 1 \), its tangent at the point \( P(1, 3) \), the line \( x = -1 \), and the x-axis. Then we will calculate \( 6A \), where \( A \) is the area we found.

### Step 1: Find the equation of the tangent line at point \( P(1, 3) \)

1. **Calculate the derivative of the parabola**:
\[ \frac{dy}{dx} = 2x + 1 \]
2. **Evaluate the derivative at \( x = 1 \)**:
\[ \frac{dy}{dx} \bigg|_{x=1} = 2(1) + 1 = 3 \]
The slope of the tangent line at point \( P(1, 3) \) is 3.

3. **Use the point-slope form of the line**:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (1, 3) \) and \( m = 3 \):
\[ y - 3 = 3(x - 1) \implies y = 3x \]

### Step 2: Determine the area bounded by the parabola, tangent line, line \( x = -1 \), and the x-axis

1. **Find the intersection points**:
- The tangent line \( y = 3x \) intersects the parabola at:
\[ 3x = x^2 + x + 1 \implies x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1 \]
- The parabola intersects the x-axis when \( y = 0 \):
\[ x^2 + x + 1 = 0 \]
The discriminant \( D = 1^2 - 4 \cdot 1 \cdot 1 = -3 < 0 \) indicates that it does not intersect the x-axis.

2. **Set up the area calculation**:
- The area \( A \) can be calculated as:
\[ A = \int_{-1}^{0} (x^2 + x + 1) \, dx + \int_{0}^{1} (x^2 + x + 1 - 3x) \, dx \]

### Step 3: Calculate the first integral \( A_1 \)

1. **Evaluate \( A_1 = \int_{-1}^{0} (x^2 + x + 1) \, dx \)**:
\[ A_1 = \left[ \frac{x^3}{3} + \frac{x^2}{2} + x \right]_{-1}^{0} \]
- At \( x = 0 \):
\[ 0 \]
- At \( x = -1 \):
\[ \frac{(-1)^3}{3} + \frac{(-1)^2}{2} + (-1) = -\frac{1}{3} + \frac{1}{2} - 1 = -\frac{1}{3} + \frac{3}{6} - \frac{6}{6} = -\frac{1}{3} - \frac{3}{6} = -\frac{1}{3} - \frac{1}{2} = -\frac{5}{6} \]
- Thus:
\[ A_1 = 0 - \left(-\frac{5}{6}\right) = \frac{5}{6} \]

### Step 4: Calculate the second integral \( A_2 \)

1. **Evaluate \( A_2 = \int_{0}^{1} (x^2 + x + 1 - 3x) \, dx = \int_{0}^{1} (x^2 - 2x + 1) \, dx \)**:
\[ A_2 = \left[ \frac{x^3}{3} - x^2 + x \right]_{0}^{1} \]
- At \( x = 1 \):
\[ \frac{1}{3} - 1 + 1 = \frac{1}{3} \]
- At \( x = 0 \):
\[ 0 \]
- Thus:
\[ A_2 = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 5: Calculate total area \( A \)

1. **Sum the areas**:
\[ A = A_1 + A_2 = \frac{5}{6} + \frac{1}{3} = \frac{5}{6} + \frac{2}{6} = \frac{7}{6} \]

### Step 6: Calculate \( 6A \)

1. **Calculate \( 6A \)**:
\[ 6A = 6 \cdot \frac{7}{6} = 7 \]

### Final Answer

Thus, the value of \( 6A \) is \( \boxed{7} \).