Let `f(a,b) = int_(a)^(b)(x^(2)-4x+3)dx, (bgt 0)` then

Options

A. `f(a,3)` is least when `a = 1`
B. `f(a,b)` is an increasing function `AA b ge 4`
C. `f(0,b)` is least for `b = 2`
D. `min{f(a,b)} = -4/3`

Answer

To solve the problem, we need to evaluate the definite integral \( f(a, b) = \int_{a}^{b} (x^2 - 4x + 3) \, dx \) and analyze its properties based on the limits of integration. Here’s a step-by-step solution:

### Step 1: Set up the integral
We start with the function given:
\[ f(a, b) = \int_{a}^{b} (x^2 - 4x + 3) \, dx \]

### Step 2: Factor the quadratic expression
The quadratic \( x^2 - 4x + 3 \) can be factored:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]
This indicates that the function has roots at \( x = 1 \) and \( x = 3 \).

### Step 3: Find the antiderivative
Next, we find the antiderivative of \( x^2 - 4x + 3 \):
\[ \int (x^2 - 4x + 3) \, dx = \frac{x^3}{3} - 2x^2 + 3x + C \]

### Step 4: Evaluate the definite integral
Now, we evaluate the definite integral from \( a \) to \( b \):
\[ f(a, b) = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{a}^{b} \]
This gives us:
\[ f(a, b) = \left( \frac{b^3}{3} - 2b^2 + 3b \right) - \left( \frac{a^3}{3} - 2a^2 + 3a \right) \]

### Step 5: Analyze the function
To find the minimum value of \( f(a, 3) \), we can evaluate \( f(1, 3) \):
\[ f(1, 3) = \int_{1}^{3} (x^2 - 4x + 3) \, dx \]

### Step 6: Calculate \( f(1, 3) \)
Using the antiderivative:
\[ f(1, 3) = \left[ \frac{x^3}{3} - 2x^2 + 3x \right]_{1}^{3} \]
Calculating at the limits:
1. For \( x = 3 \):
\[ \frac{3^3}{3} - 2(3^2) + 3(3) = \frac{27}{3} - 18 + 9 = 9 - 18 + 9 = 0 \]
2. For \( x = 1 \):
\[ \frac{1^3}{3} - 2(1^2) + 3(1) = \frac{1}{3} - 2 + 3 = \frac{1}{3} + 1 = \frac{4}{3} \]
Thus,
\[ f(1, 3) = 0 - \frac{4}{3} = -\frac{4}{3} \]

### Step 7: Conclusion
The minimum value of \( f(a, 3) \) occurs when \( a = 1 \), and the minimum value is \( -\frac{4}{3} \).