if `f(x) = a+bx +cx^(2)`, then what is `int_(0)^(1)f(x) dx` equal to ?

Options

A. `[f(0)+4f(1//2) +f(1)]//6`
B. `[f(0) +4f(1//2) + f(1)]//3`
C. `[f(0) + 4f(1//2) + f(1)]`
D. `[f(0) + 2f(1//2) + f(1)]//6`

Explanation

To solve the problem, we need to evaluate the definite integral of the function \( f(x) = a + bx + cx^2 \) from 0 to 1.

### Step-by-step Solution:

1. **Set up the integral**:
\[ \int_0^1 f(x) \, dx = \int_0^1 (a + bx + cx^2) \, dx \]

2. **Break down the integral**:
We can separate the integral into three parts:
\[ \int_0^1 (a + bx + cx^2) \, dx = \int_0^1 a \, dx + \int_0^1 bx \, dx + \int_0^1 cx^2 \, dx \]

3. **Evaluate each integral**:
- For the first integral:
\[ \int_0^1 a \, dx = a \cdot [x]_0^1 = a \cdot (1 - 0) = a \]
- For the second integral:
\[ \int_0^1 bx \, dx = b \cdot \left[\frac{x^2}{2}\right]_0^1 = b \cdot \left(\frac{1^2}{2} - \frac{0^2}{2}\right) = b \cdot \frac{1}{2} = \frac{b}{2} \]
- For the third integral:
\[ \int_0^1 cx^2 \, dx = c \cdot \left[\frac{x^3}{3}\right]_0^1 = c \cdot \left(\frac{1^3}{3} - \frac{0^3}{3}\right) = c \cdot \frac{1}{3} = \frac{c}{3} \]

4. **Combine the results**:
Now, we can combine all the results from the three integrals:
\[ \int_0^1 f(x) \, dx = a + \frac{b}{2} + \frac{c}{3} \]

Thus, the final result is:
\[ \int_0^1 f(x) \, dx = a + \frac{b}{2} + \frac{c}{3} \]