The difference between the greatest and least values of the function `phi(x)=int_(0)^(x) (t+1)`dt on [2,3], is\

Options

A. 3
B. 2
C. `7//2`
D. `11//2`

Answer

To solve the problem, we need to find the difference between the greatest and least values of the function \(\phi(x) = \int_{0}^{x} (t + 1) \, dt\) on the interval \([2, 3]\).

### Step-by-Step Solution:

1. **Define the function**:
\[ \phi(x) = \int_{0}^{x} (t + 1) \, dt \]

2. **Evaluate the integral**:
To evaluate the integral, we find:
\[ \phi(x) = \int_{0}^{x} (t + 1) \, dt = \left[ \frac{t^2}{2} + t \right]_{0}^{x} \]
This simplifies to:
\[ \phi(x) = \left( \frac{x^2}{2} + x \right) - \left( \frac{0^2}{2} + 0 \right) = \frac{x^2}{2} + x \]

3. **Find the derivative**:
To determine the behavior of \(\phi(x)\), we compute its derivative:
\[ \phi'(x) = \frac{d}{dx} \left( \frac{x^2}{2} + x \right) = x + 1 \]

4. **Analyze the derivative**:
Since \(\phi'(x) = x + 1\) is positive for all \(x \geq 2\), we conclude that \(\phi(x)\) is an increasing function on the interval \([2, 3]\).

5. **Determine the greatest and least values**:
- The least value occurs at \(x = 2\):
\[ \phi(2) = \frac{2^2}{2} + 2 = \frac{4}{2} + 2 = 2 + 2 = 4 \]
- The greatest value occurs at \(x = 3\):
\[ \phi(3) = \frac{3^2}{2} + 3 = \frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2} \]

6. **Calculate the difference**:
Now, we find the difference between the greatest and least values:
\[ \text{Difference} = \phi(3) - \phi(2) = \frac{15}{2} - 4 = \frac{15}{2} - \frac{8}{2} = \frac{7}{2} \]

### Final Answer:
The difference between the greatest and least values of the function \(\phi(x)\) on the interval \([2, 3]\) is \(\frac{7}{2}\).