Let `f(x)` be a function defined by `f (x)=int_1^x x(x^2-3x+2)dx, 1 leq x leq 4`. Then, the range of f(x) is

Options

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

Answer

To solve the problem, we need to find the range of the function defined by the integral:

\[ f(x) = \int_1^x x(x^2 - 3x + 2) \, dx \quad \text{for } 1 \leq x \leq 4 \]

### Step 1: Differentiate \( f(x) \)

Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \):

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

### Step 2: Factor the derivative

We can factor the expression:

\[ f'(x) = x(x - 1)(x - 2) \]

### Step 3: Determine critical points

To find the critical points, we set \( f'(x) = 0 \):

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

This gives us the critical points:

\[ x = 0, \, x = 1, \, x = 2 \]

However, since \( x \) is restricted to the interval \( [1, 4] \), we only consider \( x = 1 \) and \( x = 2 \).

### Step 4: Analyze the intervals

We analyze the sign of \( f'(x) \) in the intervals \( [1, 2] \) and \( [2, 4] \):

- For \( 1 < x < 2 \), \( f'(x) < 0 \) (decreasing)
- For \( 2 < x < 4 \), \( f'(x) > 0 \) (increasing)

### Step 5: Evaluate \( f(x) \) at the endpoints and critical points

Now, we need to evaluate \( f(x) \) at the critical points and the endpoints of the interval:

1. **At \( x = 1 \)**:
\[ f(1) = \int_1^1 x(x^2 - 3x + 2) \, dx = 0 \]

2. **At \( x = 2 \)**:
\[ f(2) = \int_1^2 x(x^2 - 3x + 2) \, dx \]
We compute this integral:
\[ = \int_1^2 (x^3 - 3x^2 + 2x) \, dx \]
\[ = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_1^2 \]
Evaluating this:
\[ = \left( \frac{16}{4} - 8 + 4 \right) - \left( \frac{1}{4} - 1 + 1 \right) \]
\[ = (4 - 8 + 4) - (0.25 - 1 + 1) = 0 - 0.25 = -0.25 = -\frac{1}{4} \]

3. **At \( x = 4 \)**:
\[ f(4) = \int_1^4 x(x^2 - 3x + 2) \, dx \]
Compute this integral:
\[ = \int_1^4 (x^3 - 3x^2 + 2x) \, dx \]
\[ = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_1^4 \]
Evaluating this:
\[ = \left( \frac{256}{4} - 64 + 16 \right) - \left( \frac{1}{4} - 1 + 1 \right) \]
\[ = (64 - 64 + 16) - 0.25 = 16 - 0.25 = 15.75 \]

### Step 6: Determine the range of \( f(x) \)

From our evaluations, we have:

- \( f(1) = 0 \)
- \( f(2) = -\frac{1}{4} \)
- \( f(4) = 15.75 \)

Since \( f(x) \) is decreasing from \( x = 1 \) to \( x = 2 \) and increasing from \( x = 2 \) to \( x = 4 \), the minimum value of \( f(x) \) is \( -\frac{1}{4} \) and the maximum value is \( 15.75 \).

### Conclusion

Thus, the range of \( f(x) \) is:

\[ \text{Range of } f(x) = \left[-\frac{1}{4}, 15.75\right] \]