Find the points of minima for `f(x)=int_0^x t(t-1)(t-2)dt`

Explanation

To find the points of minima for the function \( f(x) = \int_0^x t(t-1)(t-2) \, dt \), we will follow these steps:

### Step 1: Compute the integral
First, we need to compute the integral \( f(x) \):
\[ f(x) = \int_0^x t(t-1)(t-2) \, dt \]

### Step 2: Differentiate \( f(x) \)
Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = x(x-1)(x-2) \]

### Step 3: Find critical points
To find the critical points, we set \( f'(x) = 0 \):
\[ x(x-1)(x-2) = 0 \]
This gives us the critical points:
\[ x = 0, \quad x = 1, \quad x = 2 \]

### Step 4: Determine the nature of critical points
We will analyze the sign of \( f'(x) \) around the critical points to determine whether they are points of minima or maxima.

1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \):
\[ f'(-1) = (-1)(-2)(-3) = -6 \quad (\text{negative}) \]

2. **Interval \( (0, 1) \)**: Choose \( x = 0.5 \):
\[ f'(0.5) = (0.5)(-0.5)(-1.5) = 0.375 \quad (\text{positive}) \]

3. **Interval \( (1, 2) \)**: Choose \( x = 1.5 \):
\[ f'(1.5) = (1.5)(0.5)(-0.5) = -0.375 \quad (\text{negative}) \]

4. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[ f'(3) = (3)(2)(1) = 6 \quad (\text{positive}) \]

### Step 5: Conclusion about critical points
From the sign analysis:
- At \( x = 0 \): \( f' \) changes from negative to positive, indicating a local minimum.
- At \( x = 1 \): \( f' \) changes from positive to negative, indicating a local maximum.
- At \( x = 2 \): \( f' \) changes from negative to positive, indicating a local minimum.

### Final Answer
The points of minima for \( f(x) \) are \( x = 0 \) and \( x = 2 \).