If `int _0^1e^(x^2)(x-alpha)dx=0`, then

Options

A. `1 lt alpha lt 2`
B. `alpha lt 0`
C. `0 lt alpha lt 1`
D. `alpha=0`

Explanation

To solve the problem, we need to find the range of the parameter \( \alpha \) such that the integral

\[ \int_0^1 e^{x^2}(x - \alpha) \, dx = 0. \]

### Step-by-Step Solution:

1. **Set Up the Equation**:
We start with the equation given in the problem:

\[ \int_0^1 e^{x^2}(x - \alpha) \, dx = 0. \]

This can be rewritten as:

\[ \int_0^1 e^{x^2} x \, dx = \alpha \int_0^1 e^{x^2} \, dx. \]

2. **Evaluate the Integral**:
We need to evaluate both integrals on the left and right sides.

- Let \( I_1 = \int_0^1 e^{x^2} x \, dx \).
- Let \( I_2 = \int_0^1 e^{x^2} \, dx \).

To evaluate \( I_1 \), we can use the substitution \( t = x^2 \), which gives \( dt = 2x \, dx \) or \( dx = \frac{dt}{2\sqrt{t}} \).

Changing the limits:
- When \( x = 0 \), \( t = 0 \).
- When \( x = 1 \), \( t = 1 \).

Thus, we have:

\[ I_1 = \int_0^1 e^{t} \frac{1}{2} dt = \frac{1}{2} \left[ e^t \right]_0^1 = \frac{1}{2}(e - 1). \]

3. **Evaluate \( I_2 \)**:
For \( I_2 \):

\[ I_2 = \int_0^1 e^{x^2} \, dx. \]

This integral does not have a simple closed form, but we can denote it as \( I_2 \).

4. **Substitute Back**:
Now substituting back into our equation:

\[ \frac{1}{2}(e - 1) = \alpha I_2. \]

Therefore, we can express \( \alpha \) as:

\[ \alpha = \frac{1}{2} \frac{(e - 1)}{I_2}. \]

5. **Determine the Range of \( \alpha \)**:
To find the range of \( \alpha \), we need to analyze \( I_2 \). Since \( e^{x^2} \) is an increasing function for \( x \) in the interval [0, 1], we can establish bounds for \( I_2 \):

- At \( x = 0 \), \( e^{x^2} = 1 \).
- At \( x = 1 \), \( e^{x^2} = e \).

Therefore, we have:

\[ 1 < I_2 < e. \]

This means:

\[ \alpha = \frac{1}{2} \frac{(e - 1)}{I_2} \text{ will be bounded by } \frac{1}{2}(e - 1) \text{ and } 0. \]

Since \( I_2 \) is between 1 and \( e \), we can conclude:

\[ 0 < \alpha < \frac{1}{2}(e - 1). \]

### Conclusion:
Thus, the range of \( \alpha \) is:

\[ \alpha \in \left(0, \frac{1}{2}(e - 1)\right). \]