The points of extremum of the function `F(x)= int_(1)^(x) e^(-t^(2)) (1-t^(2)) dt` are

Options

A. x=0
B. x=1
C. x=1/2
D. `x=-1`

Answer

To find the points of extremum of the function \( F(x) = \int_{1}^{x} e^{-t^2} (1 - t^2) \, dt \), we will follow these steps:

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

\[ F'(x) = \frac{d}{dx} \left( \int_{1}^{x} e^{-t^2} (1 - t^2) \, dt \right) = e^{-x^2} (1 - x^2) \]

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

\[ F'(x) = e^{-x^2} (1 - x^2) = 0 \]

### Step 3: Solve for \( x \)
The expression \( e^{-x^2} \) is never zero for any real number \( x \). Therefore, we only need to solve:

\[ 1 - x^2 = 0 \]

This simplifies to:

\[ x^2 = 1 \]

Taking the square root of both sides gives:

\[ x = \pm 1 \]

### Step 4: Identify the Points of Extremum
Thus, the points of extremum of the function \( F(x) \) are:

\[ x = 1 \quad \text{and} \quad x = -1 \]

### Summary
The points of extremum of the function \( F(x) = \int_{1}^{x} e^{-t^2} (1 - t^2) \, dt \) are \( x = 1 \) and \( x = -1 \).

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