If `f(x) = a + bx + cx^2` where `a, b, c in R` then `int_o ^1 f(x)dx`

Options

A. `1/3{f(0) + f(1) +2f(1/2)}`
B. `1/6{f(0)+f(1)+4f(1/2)}`
C. `1/6{f(0)+f(1)-4f(1/2)}`
D. `1/6{f(0)-f(1)-4f(1/2)}`

Answer

To solve the problem, we need to evaluate the integral of the function \( f(x) = a + bx + cx^2 \) from 0 to 1. Let's go through the steps systematically.

### Step 1: Define the function
We are given the function:
\[ f(x) = a + bx + cx^2 \]

### Step 2: Set up the integral
We need to evaluate the integral:
\[ \int_0^1 f(x) \, dx = \int_0^1 (a + bx + cx^2) \, dx \]

### Step 3: Break down the integral
We can break this integral into three parts:
\[ \int_0^1 f(x) \, dx = \int_0^1 a \, dx + \int_0^1 bx \, dx + \int_0^1 cx^2 \, dx \]

### Step 4: Evaluate each integral
1. **First Integral**:
\[ \int_0^1 a \, dx = a \cdot [x]_0^1 = a \cdot (1 - 0) = a \]

2. **Second Integral**:
\[ \int_0^1 bx \, dx = b \cdot \left[\frac{x^2}{2}\right]_0^1 = b \cdot \left(\frac{1^2}{2} - \frac{0^2}{2}\right) = b \cdot \frac{1}{2} = \frac{b}{2} \]

3. **Third Integral**:
\[ \int_0^1 cx^2 \, dx = c \cdot \left[\frac{x^3}{3}\right]_0^1 = c \cdot \left(\frac{1^3}{3} - \frac{0^3}{3}\right) = c \cdot \frac{1}{3} = \frac{c}{3} \]

### Step 5: Combine the results
Now, we can combine the results from the three integrals:
\[ \int_0^1 f(x) \, dx = a + \frac{b}{2} + \frac{c}{3} \]

### Step 6: Final result
Thus, the value of the integral is:
\[ \int_0^1 f(x) \, dx = a + \frac{b}{2} + \frac{c}{3} \]