Let `I=int_(a)^(b) (x^4-2x^2)dx`. If I is minimum, then the ordered pair (a, b) is

Options

A. `(-sqrt2, 0)`
B. `(-sqrt2, sqrt2)`
C. `(0,sqrt2)`
D. `(sqrt2,-sqrt2)`

Answer

To solve the problem, we need to find the ordered pair \((a, b)\) such that the integral \(I = \int_{a}^{b} (x^4 - 2x^2) \, dx\) is minimized.

### Step-by-Step Solution:

1. **Identify the Function**:
We start with the function \(f(x) = x^4 - 2x^2\).

2. **Find Critical Points**:
To find the minimum of the integral, we need to find the points where the function \(f(x)\) is zero. We set:
\[ f(x) = 0 \implies x^4 - 2x^2 = 0 \]
Factoring out \(x^2\):
\[ x^2(x^2 - 2) = 0 \]
This gives us:
\[ x^2 = 0 \quad \text{or} \quad x^2 - 2 = 0 \]
Thus, the solutions are:
\[ x = 0, \quad x = \sqrt{2}, \quad x = -\sqrt{2} \]

3. **Analyze the Critical Points**:
We have three critical points: \(-\sqrt{2}\), \(0\), and \(\sqrt{2}\). To determine where the function changes from increasing to decreasing (or vice versa), we can evaluate the sign of \(f(x)\) in the intervals defined by these points.

4. **Sign Analysis**:
- For \(x < -\sqrt{2}\): \(f(x) > 0\)
- For \(-\sqrt{2} < x < 0\): \(f(x) < 0\)
- For \(0 < x < \sqrt{2}\): \(f(x) < 0\)
- For \(x > \sqrt{2}\): \(f(x) > 0\)

From this analysis, we see that \(f(x)\) has a minimum at \(x = 0\) and \(x = \sqrt{2}\) is a local maximum.

5. **Determine the Interval for Minimum Integral**:
Since we want to minimize the integral \(I\), we should choose the interval that includes the points where \(f(x)\) changes sign. The interval that captures the minimum value of \(I\) is from \(-\sqrt{2}\) to \(\sqrt{2}\).

6. **Conclusion**:
Therefore, the ordered pair \((a, b)\) that minimizes the integral \(I\) is:
\[ (a, b) = (-\sqrt{2}, \sqrt{2}) \]

### Final Answer:
The ordered pair \((a, b)\) is \((- \sqrt{2}, \sqrt{2})\).