The poins of extremum of `phi(x)=underset(1)overset(x)inte^(-t^(2)//2) (1-t^(2))dt` are

Options

A. `x=1,-1`
B. `x=-1,2`
C. `x=2,1`
D. `x=-2,1`

Answer

To find the points of extremum of the function

\[ \phi(x) = \int_{1}^{x} e^{-\frac{t^2}{2}} (1 - t^2) dt, \]

we will follow these steps:

### Step 1: Differentiate the function
To find the points of extremum, we first need to differentiate the function \(\phi(x)\). We can use the Leibniz rule for differentiation under the integral sign. According to this rule:

\[ \frac{d}{dx} \left( \int_{g(x)}^{h(x)} f(t) dt \right) = f(h(x)) h'(x) - f(g(x)) g'(x). \]

In our case, \(g(x) = 1\) and \(h(x) = x\), and \(f(t) = e^{-\frac{t^2}{2}} (1 - t^2)\).

Thus, we have:

\[ \phi'(x) = f(x) \cdot 1 - f(1) \cdot 0 = e^{-\frac{x^2}{2}} (1 - x^2). \]

### Step 2: Set the derivative to zero
To find the points of extremum, we set the derivative equal to zero:

\[ e^{-\frac{x^2}{2}} (1 - x^2) = 0. \]

### Step 3: Analyze the equation
Since \(e^{-\frac{x^2}{2}}\) is never zero (as the exponential function is always positive), we focus on the term \(1 - x^2 = 0\):

\[ 1 - x^2 = 0 \implies x^2 = 1. \]

### Step 4: Solve for x
Taking the square root of both sides, we find:

\[ x = 1 \quad \text{or} \quad x = -1. \]

### Conclusion
The points of extremum of the function \(\phi(x)\) are:

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