If `f(x)= ax^(2) +bx +c` such that f(0)=2 `f'(0)= -3, f''(0) =4`, then `int_(-1)^(1) f(x) dx`=

Options

A. `-3`
B. `16//3`
C. 0
D. none

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Define the function
We are given the function \( f(x) = ax^2 + bx + c \).

### Step 2: Use the given conditions
We have the following conditions:
1. \( f(0) = 2 \)
2. \( f'(0) = -3 \)
3. \( f''(0) = 4 \)

### Step 3: Find the value of \( c \)
From the condition \( f(0) = 2 \):
\[ f(0) = a(0)^2 + b(0) + c = c = 2 \]
Thus, \( c = 2 \).

### Step 4: Differentiate the function to find \( b \)
The first derivative of \( f(x) \) is:
\[ f'(x) = 2ax + b \]
Using the condition \( f'(0) = -3 \):
\[ f'(0) = 2a(0) + b = b = -3 \]
Thus, \( b = -3 \).

### Step 5: Differentiate again to find \( a \)
The second derivative of \( f(x) \) is:
\[ f''(x) = 2a \]
Using the condition \( f''(0) = 4 \):
\[ f''(0) = 2a = 4 \implies a = 2 \]

### Step 6: Write the complete function
Now we have:
- \( a = 2 \)
- \( b = -3 \)
- \( c = 2 \)

Thus, the function is:
\[ f(x) = 2x^2 - 3x + 2 \]

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

### Step 8: Compute the integral
We can split the integral:
\[ \int_{-1}^{1} (2x^2 - 3x + 2) \, dx = \int_{-1}^{1} 2x^2 \, dx - \int_{-1}^{1} 3x \, dx + \int_{-1}^{1} 2 \, dx \]

Calculating each part:
1. **For \( \int_{-1}^{1} 2x^2 \, dx \)**:
\[ \int 2x^2 \, dx = \frac{2x^3}{3} \Big|_{-1}^{1} = \left(\frac{2(1)^3}{3} - \frac{2(-1)^3}{3}\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} \]

2. **For \( \int_{-1}^{1} 3x \, dx \)**:
\[ \int 3x \, dx = \frac{3x^2}{2} \Big|_{-1}^{1} = \left(\frac{3(1)^2}{2} - \frac{3(-1)^2}{2}\right) = \frac{3}{2} - \frac{3}{2} = 0 \]

3. **For \( \int_{-1}^{1} 2 \, dx \)**:
\[ \int 2 \, dx = 2x \Big|_{-1}^{1} = 2(1) - 2(-1) = 2 + 2 = 4 \]

### Step 9: Combine results
Now, we combine the results:
\[ \int_{-1}^{1} f(x) \, dx = \frac{4}{3} - 0 + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-1}^{1} f(x) \, dx = \frac{16}{3} \]