If `I=int_(0)^(2)e^(x^(4))(x-alpha)dx=0`, then `alpha` lies in the interval

Options

A. `(0, 2)`
B. `(-1, 0)`
C. `(2, 3)`
D. `(-2, -1)`

Answer

To solve the problem where \( I = \int_{0}^{2} e^{x^4}(x - \alpha) \, dx = 0 \), we need to find the interval in which \( \alpha \) lies.

### Step-by-Step Solution:

1. **Set Up the Integral**:
Given:
\[ I = \int_{0}^{2} e^{x^4}(x - \alpha) \, dx = 0 \]
We can separate the integral:
\[ I = \int_{0}^{2} e^{x^4} x \, dx - \alpha \int_{0}^{2} e^{x^4} \, dx = 0 \]

2. **Rearranging the Equation**:
Rearranging gives:
\[ \alpha \int_{0}^{2} e^{x^4} \, dx = \int_{0}^{2} e^{x^4} x \, dx \]
Thus, we can express \( \alpha \) as:
\[ \alpha = \frac{\int_{0}^{2} e^{x^4} x \, dx}{\int_{0}^{2} e^{x^4} \, dx} \]

3. **Evaluate the Integrals**:
- Let \( A = \int_{0}^{2} e^{x^4} x \, dx \)
- Let \( B = \int_{0}^{2} e^{x^4} \, dx \)

Both integrals \( A \) and \( B \) are positive since \( e^{x^4} \) is always positive for \( x \) in the interval [0, 2].

4. **Analyze the Expression for \( \alpha \)**:
Since both \( A \) and \( B \) are positive, \( \alpha \) will also be positive. Therefore:
\[ \alpha = \frac{A}{B} > 0 \]

5. **Determine the Interval for \( \alpha \)**:
- As \( x \) varies from 0 to 2, \( e^{x^4} \) increases rapidly, thus \( B \) is significantly larger than \( A \) because \( e^{x^4} \) grows faster than \( x \) in the interval.
- Thus, \( \alpha \) will be less than 2.

6. **Conclusion**:
Therefore, \( \alpha \) lies in the interval:
\[ \alpha \in (0, 2) \]

### Final Answer:
The value of \( \alpha \) lies in the interval \( (0, 2) \).