Let `f(x)=ax^(2)+bx + c`, where `a in R^(+)` and `b^(2)-4ac lt 0`. Area bounded by `y = f(x)`, x-axis and the lines x = 0, x = 1, is equal to :

Options

A. `(1)/(6)(3f(1)+f(-1)+2f(0))`
B. `(1)/(12)(5 f(1)+f(-1)+8f(0))`
C. `(1)/(6)(3f(1)-f(-1)+2f(0))`
D. `(1)/(12)(5f(1)-f(-1)+8f(0))`

Answer

To find the area bounded by the curve \( y = f(x) = ax^2 + bx + c \), the x-axis, and the lines \( x = 0 \) and \( x = 1 \), we follow these steps:

### Step 1: Set up the integral for the area
Since \( f(x) \) is a quadratic function and given that \( a > 0 \) and the discriminant \( b^2 - 4ac < 0 \), the parabola opens upwards and does not intersect the x-axis. Therefore, the area under the curve from \( x = 0 \) to \( x = 1 \) can be found by integrating \( f(x) \):

\[ \text{Area} = \int_0^1 f(x) \, dx = \int_0^1 (ax^2 + bx + c) \, dx \]

### Step 2: Compute the integral
Now, we compute the integral:

\[ \int_0^1 (ax^2 + bx + c) \, dx = \int_0^1 ax^2 \, dx + \int_0^1 bx \, dx + \int_0^1 c \, dx \]

Calculating each integral separately:

1. For \( \int_0^1 ax^2 \, dx \):
\[ \int_0^1 ax^2 \, dx = a \left[ \frac{x^3}{3} \right]_0^1 = a \cdot \frac{1^3}{3} = \frac{a}{3} \]

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

3. For \( \int_0^1 c \, dx \):
\[ \int_0^1 c \, dx = c \left[ x \right]_0^1 = c \cdot (1 - 0) = c \]

### Step 3: Combine the results
Now, we combine the results of the integrals:

\[ \text{Area} = \frac{a}{3} + \frac{b}{2} + c \]

### Conclusion
Thus, the area bounded by the curve \( y = f(x) \), the x-axis, and the lines \( x = 0 \) and \( x = 1 \) is:

\[ \text{Area} = \frac{a}{3} + \frac{b}{2} + c \]