The difference between the greateset ahnd least vlaue of the function `f(x)=int_(0)^(x) (6t^2-24)" dt on " [1,3] " dt on " [1,3]` is

Options

A. 14
B. 10
C. 5
D. 4

Explanation

To solve the problem, we need to find the difference between the greatest and least values of the function \( f(x) = \int_{0}^{x} (6t^2 - 24) \, dt \) on the interval \([1, 3]\).

### Step-by-step Solution:

1. **Evaluate the Integral**:
We first need to compute the integral \( f(x) \):
\[ f(x) = \int_{0}^{x} (6t^2 - 24) \, dt \]
To evaluate this integral, we find the antiderivative of \( 6t^2 - 24 \):
\[ \int (6t^2 - 24) \, dt = 2t^3 - 24t + C \]
Now, we apply the limits from \( 0 \) to \( x \):
\[ f(x) = \left[ 2t^3 - 24t \right]_{0}^{x} = (2x^3 - 24x) - (2(0)^3 - 24(0)) = 2x^3 - 24x \]

2. **Find the Critical Points**:
To find the maximum and minimum values of \( f(x) \) on the interval \([1, 3]\), we first find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 24x) = 6x^2 - 24 \]
Set the derivative equal to zero to find critical points:
\[ 6x^2 - 24 = 0 \implies 6x^2 = 24 \implies x^2 = 4 \implies x = 2 \quad (\text{since } x \text{ must be in } [1, 3]) \]

3. **Evaluate \( f(x) \) at Critical Points and Endpoints**:
Now we evaluate \( f(x) \) at the critical point \( x = 2 \) and the endpoints \( x = 1 \) and \( x = 3 \):
- For \( x = 1 \):
\[ f(1) = 2(1)^3 - 24(1) = 2 - 24 = -22 \]
- For \( x = 2 \):
\[ f(2) = 2(2)^3 - 24(2) = 2(8) - 48 = 16 - 48 = -32 \]
- For \( x = 3 \):
\[ f(3) = 2(3)^3 - 24(3) = 2(27) - 72 = 54 - 72 = -18 \]

4. **Determine the Greatest and Least Values**:
From the evaluations:
- \( f(1) = -22 \)
- \( f(2) = -32 \)
- \( f(3) = -18 \)

The greatest value is \( -18 \) (at \( x = 3 \)) and the least value is \( -32 \) (at \( x = 2 \)).

5. **Calculate the Difference**:
The difference between the greatest and least values is:
\[ \text{Difference} = (-18) - (-32) = -18 + 32 = 14 \]

### Final Answer:
The difference between the greatest and least value of the function \( f(x) \) on the interval \([1, 3]\) is \( \boxed{14} \).