The point of extremum of `f(x)=int_0^x(t-2)^2(t-1)dt` is a

Options

A. maximum at x=1
B. maximum at x=2
C. minimum at x=1
D. minimum at x=2

Answer

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

### Step 1: Differentiate the Function
To find the points of extremum, we first need to differentiate the function \( f(x) \) with respect to \( x \). According to the Fundamental Theorem of Calculus, we have:

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

### Step 2: Set the Derivative to Zero
To find the critical points, we set the derivative equal to zero:

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

### Step 3: Solve for \( x \)
Now, we solve the equation \( (x-2)^2 (x-1) = 0 \). This gives us two factors to consider:

1. \( (x-2)^2 = 0 \) which implies \( x = 2 \)
2. \( (x-1) = 0 \) which implies \( x = 1 \)

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

### Step 4: Determine the Nature of the Extremum
To determine whether these points are maxima or minima, we can use the second derivative test. We first find the second derivative \( f''(x) \).

Differentiating \( f'(x) = (x-2)^2 (x-1) \):

Using the product rule:

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

Simplifying this gives:

\[ f''(x) = 2(x-2)(x-1) + (x-2)^2 = (x-2) \left( 2(x-1) + (x-2) \right) \]

\[ = (x-2)(3x - 4) \]

Now we evaluate \( f''(x) \) at the critical points:

1. For \( x = 1 \):
\[ f''(1) = (1-2)(3(1) - 4) = (-1)(-1) = 1 > 0 \quad \text{(local minimum)} \]

2. For \( x = 2 \):
\[ f''(2) = (2-2)(3(2) - 4) = 0 \quad \text{(inconclusive)} \]

### Conclusion
Thus, we have determined that \( x = 1 \) is a local minimum. The behavior at \( x = 2 \) requires further investigation, but it is not a local extremum based on the second derivative test.

### Final Answer
The point of extremum of \( f(x) \) is at \( x = 1 \) (local minimum).

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