Let `f(x)` be a function defined by `f(x)=int_(1)^(x)t(t^(2)-3t+2)dt,1ltxlt3` then the maximum value of `f(x)` is

Explanation

To find the maximum value of the function \( f(x) \) defined by the integral

\[ f(x) = \int_{1}^{x} t(t^2 - 3t + 2) \, dt \]

for \( 1 < x < 3 \), we will follow these steps:

### Step 1: Simplify the integrand
First, simplify the expression inside the integral:

\[ t(t^2 - 3t + 2) = t^3 - 3t^2 + 2t \]

### Step 2: Compute the integral
Now we can compute the integral:

\[ f(x) = \int_{1}^{x} (t^3 - 3t^2 + 2t) \, dt \]

We can integrate term by term:

\[ \int t^3 \, dt = \frac{t^4}{4}, \quad \int t^2 \, dt = \frac{t^3}{3}, \quad \int t \, dt = \frac{t^2}{2} \]

Thus,

\[ f(x) = \left[ \frac{t^4}{4} - t^3 + t^2 \right]_{1}^{x} \]

### Step 3: Evaluate the definite integral
Now, evaluate the integral from 1 to \( x \):

\[ f(x) = \left( \frac{x^4}{4} - x^3 + x^2 \right) - \left( \frac{1^4}{4} - 1^3 + 1^2 \right) \]

Calculating the constant part:

\[ \frac{1}{4} - 1 + 1 = \frac{1}{4} \]

So,

\[ f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4} \]

### Step 4: Find the derivative
Next, we find the derivative \( f'(x) \):

\[ f'(x) = x^3 - 3x^2 + 2x \]

### Step 5: Set the derivative to zero
To find the critical points, set \( f'(x) = 0 \):

\[ x^3 - 3x^2 + 2x = 0 \]

Factoring out \( x \):

\[ x(x^2 - 3x + 2) = 0 \]

This gives us:

\[ x = 0 \quad \text{or} \quad x^2 - 3x + 2 = 0 \]

Factoring the quadratic:

\[ (x - 1)(x - 2) = 0 \]

Thus, the critical points are \( x = 1, 2 \).

### Step 6: Analyze the intervals
We need to check the intervals \( (1, 2) \) and \( (2, 3) \) to determine where \( f(x) \) is increasing or decreasing.

- For \( x < 1 \), \( f'(x) < 0 \) (not in our interval).
- For \( 1 < x < 2 \), \( f'(x) < 0 \) (decreasing).
- For \( 2 < x < 3 \), \( f'(x) > 0 \) (increasing).

### Step 7: Evaluate \( f(x) \) at the endpoints
Now we evaluate \( f(x) \) at the endpoints \( x = 1 \) and \( x = 3 \):

1. **At \( x = 1 \)**:
\[ f(1) = 0 \]

2. **At \( x = 3 \)**:
\[ f(3) = \frac{3^4}{4} - 3^3 + 3^2 - \frac{1}{4} \]
\[ = \frac{81}{4} - 27 + 9 - \frac{1}{4} \]
\[ = \frac{81}{4} - \frac{108}{4} + \frac{36}{4} - \frac{1}{4} \]
\[ = \frac{81 - 108 + 36 - 1}{4} = \frac{8}{4} = 2 \]

### Conclusion
The maximum value of \( f(x) \) occurs at \( x = 3 \):

\[ \text{Maximum value of } f(x) = 2 \]