If `int_(0)^(1) x e^(x^(2) ) dx=alpha int_(0)^(1) e^(x^(2)) dx`, then

Options

A. `alpha in (0,1)`
B. `(1,2)`
C. `(-oo,0)`
D. `(0,oo)`

Explanation

To solve the problem, we need to find the range of \(\alpha\) given the equation:

\[ \int_{0}^{1} x e^{x^2} \, dx = \alpha \int_{0}^{1} e^{x^2} \, dx \]

### Step 1: Analyze the given integrals

We have two integrals to consider:

1. \(I_1 = \int_{0}^{1} x e^{x^2} \, dx\)
2. \(I_2 = \int_{0}^{1} e^{x^2} \, dx\)

### Step 2: Establish inequalities

Since \(0 < x < 1\) in the interval of integration, we can multiply \(x\) by \(e^{x^2}\):

\[ 0 < x e^{x^2} < e^{x^2} \quad \text{for } 0 < x < 1 \]

### Step 3: Integrate the inequalities

Integrating the inequalities from \(0\) to \(1\):

\[ \int_{0}^{1} 0 \, dx < \int_{0}^{1} x e^{x^2} \, dx < \int_{0}^{1} e^{x^2} \, dx \]

This simplifies to:

\[ 0 < I_1 < I_2 \]

### Step 4: Substitute the original equation

From the original equation, we have:

\[ I_1 = \alpha I_2 \]

### Step 5: Substitute into the inequality

Substituting \(I_1\) into the inequality gives:

\[ 0 < \alpha I_2 < I_2 \]

### Step 6: Divide by \(I_2\)

Assuming \(I_2 > 0\) (which is true since \(e^{x^2}\) is positive for all \(x\)), we can divide through by \(I_2\):

\[ 0 < \alpha < 1 \]

### Conclusion

Thus, we find that the range of \(\alpha\) is:

\[ \alpha \in (0, 1) \]