Let f(x) be a function defined by
`f(x)=int_(1)^(x)t(t^2-3t+2)dt,x in [1,3]`
Then the range of f(x), is

Options

A. [0,2]
B. `[-1/4,4]`
C. `[-1/4,2]`
D. none of these

Explanation

To find the range of the function \( f(x) = \int_{1}^{x} t(t^2 - 3t + 2) \, dt \) for \( x \in [1, 3] \), we will follow these steps:

### Step 1: Simplify the integrand
First, we simplify the expression inside the integral:
\[ t(t^2 - 3t + 2) = t^3 - 3t^2 + 2t \]

### Step 2: Set up the integral
Now we can express \( f(x) \) as:
\[ f(x) = \int_{1}^{x} (t^3 - 3t^2 + 2t) \, dt \]

### Step 3: Integrate term by term
We will integrate each term separately:
\[ \int t^3 \, dt = \frac{t^4}{4}, \quad \int t^2 \, dt = \frac{t^3}{3}, \quad \int t \, dt = \frac{t^2}{2} \]
Thus,
\[ f(x) = \left[ \frac{t^4}{4} - 3 \cdot \frac{t^3}{3} + 2 \cdot \frac{t^2}{2} \right]_{1}^{x} \]
This simplifies to:
\[ f(x) = \left[ \frac{t^4}{4} - t^3 + t^2 \right]_{1}^{x} \]

### Step 4: Evaluate the definite integral
Now we evaluate the integral at the limits:
\[ f(x) = \left( \frac{x^4}{4} - x^3 + x^2 \right) - \left( \frac{1^4}{4} - 1^3 + 1^2 \right) \]
Calculating the value at \( t = 1 \):
\[ \frac{1}{4} - 1 + 1 = \frac{1}{4} \]
Thus,
\[ f(x) = \left( \frac{x^4}{4} - x^3 + x^2 \right) - \frac{1}{4} \]
This gives:
\[ f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4} \]

### Step 5: Find the critical points
To find the range, we need to find the critical points by differentiating \( f(x) \):
\[ f'(x) = x^3 - 3x^2 + 2x \]
Setting \( f'(x) = 0 \):
\[ x^3 - 3x^2 + 2x = 0 \]
Factoring out \( x \):
\[ x(x^2 - 3x + 2) = 0 \]
This gives:
\[ x(x-1)(x-2) = 0 \]
Thus, \( x = 0, 1, 2 \).

### Step 6: Evaluate \( f(x) \) at the endpoints and critical points
We evaluate \( f(x) \) at \( x = 1, 2, 3 \):
1. \( f(1) = 0 \)
2. \( f(2) = \frac{2^4}{4} - 2^3 + 2^2 - \frac{1}{4} = \frac{16}{4} - 8 + 4 - \frac{1}{4} = 4 - 8 + 4 - \frac{1}{4} = 0 - \frac{1}{4} = -\frac{1}{4} \)
3. \( f(3) = \frac{3^4}{4} - 3^3 + 3^2 - \frac{1}{4} = \frac{81}{4} - 27 + 9 - \frac{1}{4} = \frac{81 - 108 + 36 - 1}{4} = \frac{8}{4} = 2 \)

### Step 7: Determine the range
From the evaluations:
- \( f(1) = 0 \)
- \( f(2) = -\frac{1}{4} \)
- \( f(3) = 2 \)

Thus, the range of \( f(x) \) as \( x \) varies from 1 to 3 is:
\[ [-\frac{1}{4}, 2] \]

### Final Answer
The range of \( f(x) \) is \([-1/4, 2]\).