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

Options

A. `8`
B. `2`
C. `6`
D. `11//2`

Answer

To find the difference between the greatest and least values of the function

\[ F(x) = \int_{0}^{x} (t + 1) \, dt \]

on the interval \([1, 3]\), we will follow these steps:

### Step 1: Calculate the integral \(F(x)\)

We start by calculating \(F(x)\):

\[ F(x) = \int_{0}^{x} (t + 1) \, dt \]

To solve this integral, we can break it down:

\[ F(x) = \int_{0}^{x} t \, dt + \int_{0}^{x} 1 \, dt \]

Calculating each part:

1. \(\int_{0}^{x} t \, dt = \left[\frac{t^2}{2}\right]_{0}^{x} = \frac{x^2}{2}\)
2. \(\int_{0}^{x} 1 \, dt = [t]_{0}^{x} = x\)

Combining these results, we have:

\[ F(x) = \frac{x^2}{2} + x \]

### Step 2: Find the derivative \(F'(x)\)

Next, we find the derivative of \(F(x)\) to determine the critical points:

\[ F'(x) = \frac{d}{dx}\left(\frac{x^2}{2} + x\right) = x + 1 \]

### Step 3: Determine critical points

Set the derivative equal to zero to find critical points:

\[ F'(x) = 0 \implies x + 1 = 0 \implies x = -1 \]

Since \(-1\) is not in the interval \([1, 3]\), we will evaluate \(F(x)\) at the endpoints of the interval.

### Step 4: Evaluate \(F(x)\) at the endpoints

Calculate \(F(1)\):

\[ F(1) = \frac{1^2}{2} + 1 = \frac{1}{2} + 1 = \frac{3}{2} \]

Calculate \(F(3)\):

\[ F(3) = \frac{3^2}{2} + 3 = \frac{9}{2} + 3 = \frac{9}{2} + \frac{6}{2} = \frac{15}{2} \]

### Step 5: Find the greatest and least values

From the evaluations:

- The least value of \(F(x)\) on \([1, 3]\) is \(F(1) = \frac{3}{2}\).
- The greatest value of \(F(x)\) on \([1, 3]\) is \(F(3) = \frac{15}{2}\).

### Step 6: Calculate the difference

Now, we find the difference between the greatest and least values:

\[ \text{Difference} = F(3) - F(1) = \frac{15}{2} - \frac{3}{2} = \frac{15 - 3}{2} = \frac{12}{2} = 6 \]

### Final Answer

The difference between the greatest and least values of the function \(F(x)\) on the interval \([1, 3]\) is

\[ \boxed{6} \]

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