Let f(x) be a function defined by `f(x)=int_1^xt(t^2-3t+2)dt,1<=x<=3` 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 find the range of the function \( f(x) = \int_1^x t(t^2 - 3t + 2) \, dt \) for \( 1 \leq x \leq 3 \), we will follow these steps:

### Step 1: Simplify the Integrand
First, we simplify the integrand:
\[ t(t^2 - 3t + 2) = t^3 - 3t^2 + 2t \]
Thus, we can rewrite the function as:
\[ f(x) = \int_1^x (t^3 - 3t^2 + 2t) \, dt \]

### Step 2: Compute the Integral
Now, we compute the integral:
\[ f(x) = \int_1^x t^3 \, dt - 3 \int_1^x t^2 \, dt + 2 \int_1^x t \, dt \]
Calculating each integral separately:
1. \(\int t^3 \, dt = \frac{t^4}{4}\)
2. \(\int t^2 \, dt = \frac{t^3}{3}\)
3. \(\int t \, dt = \frac{t^2}{2}\)

Now, substituting the limits:
\[ f(x) = \left[ \frac{t^4}{4} \right]_1^x - 3 \left[ \frac{t^3}{3} \right]_1^x + 2 \left[ \frac{t^2}{2} \right]_1^x \]
Calculating each term:
1. \(\frac{x^4}{4} - \frac{1^4}{4} = \frac{x^4}{4} - \frac{1}{4}\)
2. \(-3\left(\frac{x^3}{3} - \frac{1^3}{3}\right) = - (x^3 - 1) = -x^3 + 1\)
3. \(2\left(\frac{x^2}{2} - \frac{1^2}{2}\right) = (x^2 - 1)\)

Combining these results gives:
\[ f(x) = \frac{x^4}{4} - \frac{1}{4} - x^3 + 1 + x^2 - 1 \]
Simplifying further:
\[ f(x) = \frac{x^4}{4} - x^3 + x^2 - \frac{1}{4} \]

### Step 3: Find the Derivative
Next, we find the derivative \( f'(x) \):
\[ f'(x) = x^3 - 3x^2 + 2x \]

### Step 4: Analyze the Critical Points
To find the critical points, we 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:
\[ (x-1)(x-2) = 0 \implies x = 1, 2 \]

### Step 5: Evaluate \( f(x) \) at Critical Points and Endpoints
We evaluate \( f(x) \) at \( x = 1, 2, 3 \):
1. \( f(1) = 0 \)
2. \( f(2) = \int_1^2 t(t^2 - 3t + 2) dt \)
- Calculate \( f(2) \):
\[ f(2) = \int_1^2 (t^3 - 3t^2 + 2t) dt = \left[ \frac{t^4}{4} - t^3 + t^2 \right]_1^2 \]
- Evaluating gives:
\[ f(2) = \left( \frac{16}{4} - 8 + 4 \right) - \left( \frac{1}{4} - 1 + 1 \right) = 4 - 8 + 4 - \left( \frac{1}{4} \right) = 0 - \frac{1}{4} = -\frac{1}{4} \]
3. \( f(3) = \int_1^3 t(t^2 - 3t + 2) dt \)
- Calculate \( f(3) \):
\[ f(3) = \left[ \frac{t^4}{4} - t^3 + t^2 \right]_1^3 \]
- Evaluating gives:
\[ f(3) = \left( \frac{81}{4} - 27 + 9 \right) - \left( \frac{1}{4} - 1 + 1 \right) = \frac{81}{4} - 27 + 9 - \left( \frac{1}{4} \right) \]
- Simplifying:
\[ = \frac{81}{4} - \frac{108}{4} + \frac{36}{4} - \frac{1}{4} = \frac{81 - 108 + 36 - 1}{4} = \frac{8}{4} = 2 \]

### Step 6: Determine the Range
Now we have:
- \( f(1) = 0 \)
- \( f(2) = -\frac{1}{4} \)
- \( f(3) = 2 \)

The minimum value is \( -\frac{1}{4} \) and the maximum value is \( 2 \). Therefore, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = \left[-\frac{1}{4}, 2\right] \]