If `f(x)` is a polynomial of degree 2, such that `f(0)=3,f'(0)=-7,f''(0)=8 int_(1)^(2)f(x)dx=`

Options

A. `11/6`
B. `13/6`
C. `17/6`
D. `19/6`

Answer

To solve the problem, we need to find the definite integral of the polynomial function \( f(x) \) over the interval from 1 to 2. We start by determining the polynomial \( f(x) \) based on the given conditions.

### Step 1: Form the polynomial
Since \( f(x) \) is a polynomial of degree 2, we can express it in the standard form:
\[ f(x) = ax^2 + bx + c \]

### Step 2: Use the given conditions
We have the following conditions:
1. \( f(0) = 3 \)
2. \( f'(0) = -7 \)
3. \( f''(0) = 8 \)

#### Condition 1: \( f(0) = 3 \)
Substituting \( x = 0 \) into the polynomial:
\[ f(0) = a(0)^2 + b(0) + c = c = 3 \]
Thus, \( c = 3 \).

#### Condition 2: \( f'(0) = -7 \)
First, we find the derivative \( f'(x) \):
\[ f'(x) = 2ax + b \]
Now substituting \( x = 0 \):
\[ f'(0) = 2a(0) + b = b = -7 \]
Thus, \( b = -7 \).

#### Condition 3: \( f''(0) = 8 \)
Next, we find the second derivative \( f''(x) \):
\[ f''(x) = 2a \]
Now substituting \( x = 0 \):
\[ f''(0) = 2a = 8 \implies a = 4 \]

### Step 3: Write the polynomial
Now we can substitute the values of \( a \), \( b \), and \( c \) back into the polynomial:
\[ f(x) = 4x^2 - 7x + 3 \]

### Step 4: Calculate the definite integral
We need to compute the integral:
\[ \int_{1}^{2} f(x) \, dx = \int_{1}^{2} (4x^2 - 7x + 3) \, dx \]

### Step 5: Integrate the polynomial
We integrate term by term:
\[ \int (4x^2) \, dx = \frac{4x^3}{3}, \quad \int (-7x) \, dx = -\frac{7x^2}{2}, \quad \int (3) \, dx = 3x \]
Thus,
\[ \int (4x^2 - 7x + 3) \, dx = \frac{4x^3}{3} - \frac{7x^2}{2} + 3x \]

### Step 6: Evaluate the integral from 1 to 2
Now we evaluate the definite integral:
\[ \left[ \frac{4x^3}{3} - \frac{7x^2}{2} + 3x \right]_{1}^{2} \]

Calculating at \( x = 2 \):
\[ = \frac{4(2)^3}{3} - \frac{7(2)^2}{2} + 3(2) = \frac{32}{3} - 14 + 6 = \frac{32}{3} - \frac{42}{3} = -\frac{10}{3} \]

Calculating at \( x = 1 \):
\[ = \frac{4(1)^3}{3} - \frac{7(1)^2}{2} + 3(1) = \frac{4}{3} - \frac{7}{2} + 3 = \frac{4}{3} - \frac{21}{6} + \frac{18}{6} = \frac{4}{3} - \frac{3}{6} = \frac{4}{3} - \frac{1}{2} = \frac{8}{6} - \frac{3}{6} = \frac{5}{6} \]

### Step 7: Final calculation
Now we compute:
\[ \int_{1}^{2} f(x) \, dx = \left(-\frac{10}{3}\right) - \left(\frac{5}{6}\right) \]
Finding a common denominator (which is 6):
\[ = -\frac{20}{6} - \frac{5}{6} = -\frac{25}{6} \]

### Final Answer
Thus, the value of the definite integral \( \int_{1}^{2} f(x) \, dx \) is:
\[ \boxed{-\frac{25}{6}} \]