The points of extremum of `underset(0)overset(x^(2))int (t^(2)-5t+4)/(2+e^(t))dt` are

Answer

To find the points of extremum of the integral

\[ y = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} dt, \]

we will follow these steps:

### Step 1: Identify the function
We have the function defined as:

\[ y = \int_{0}^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} dt. \]

### Step 2: Differentiate using Leibniz's Rule
To find the points of extremum, we need to differentiate \(y\) with respect to \(x\). According to Leibniz's rule, if we have an integral of the form

\[ y = \int_{a}^{g(x)} f(t) dt, \]

the derivative is given by:

\[ \frac{dy}{dx} = f(g(x)) \cdot g'(x). \]

In our case, \(g(x) = x^2\) and \(f(t) = \frac{t^2 - 5t + 4}{2 + e^t}\). Thus, we have:

\[ \frac{dy}{dx} = f(x^2) \cdot \frac{d}{dx}(x^2) = f(x^2) \cdot 2x. \]

### Step 3: Substitute \(g(x)\) into \(f(t)\)
Now we need to evaluate \(f(x^2)\):

\[ f(x^2) = \frac{(x^2)^2 - 5(x^2) + 4}{2 + e^{x^2}} = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}}. \]

### Step 4: Write the derivative
Thus, we can write:

\[ \frac{dy}{dx} = \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x. \]

### Step 5: Set the derivative to zero
To find the points of extremum, we set the derivative equal to zero:

\[ \frac{x^4 - 5x^2 + 4}{2 + e^{x^2}} \cdot 2x = 0. \]

This gives us two cases to consider:

1. \(2x = 0\) which implies \(x = 0\).
2. \(x^4 - 5x^2 + 4 = 0\).

### Step 6: Solve the polynomial equation
Let \(u = x^2\). Then the equation becomes:

\[ u^2 - 5u + 4 = 0. \]

Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[ u = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2}. \]

This gives us:

\[ u_1 = \frac{8}{2} = 4 \quad \text{and} \quad u_2 = \frac{2}{2} = 1. \]

### Step 7: Find \(x\) values
Since \(u = x^2\):

1. \(x^2 = 4 \Rightarrow x = \pm 2\).
2. \(x^2 = 1 \Rightarrow x = \pm 1\).

### Step 8: Compile all points of extremum
Thus, the points of extremum are:

\[ x = 0, \quad x = 2, \quad x = -2, \quad x = 1, \quad x = -1. \]

### Final Answer
The points of extremum are \(x = 0, \pm 1, \pm 2\).

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