The set of values of `alpha(alphagt0)` for which the inequality
`int_(-alpha)^(alpha)e^(x)dxgt3/2` holds true is

Options

A. `(0,oo)`
B. `(2,oo)`
C. `("log"2,oo)`
D. none of these

Explanation

To solve the inequality

\[ \int_{-\alpha}^{\alpha} e^x \, dx > \frac{3}{2} \]

we will follow these steps:

### Step 1: Evaluate the Integral
The integral of \( e^x \) is \( e^x \). Thus, we can evaluate the definite integral:

\[ \int_{-\alpha}^{\alpha} e^x \, dx = e^{\alpha} - e^{-\alpha} \]

### Step 2: Set Up the Inequality
We need to set up the inequality based on our integral evaluation:

\[ e^{\alpha} - e^{-\alpha} > \frac{3}{2} \]

### Step 3: Rewrite the Inequality
We can rewrite the left-hand side:

\[ e^{\alpha} - \frac{1}{e^{\alpha}} > \frac{3}{2} \]

### Step 4: Multiply by \( e^{\alpha} \)
To eliminate the fraction, multiply both sides by \( e^{\alpha} \) (since \( e^{\alpha} > 0 \) for \( \alpha > 0 \)):

\[ e^{2\alpha} - 1 > \frac{3}{2} e^{\alpha} \]

### Step 5: Rearrange the Inequality
Rearranging gives us:

\[ e^{2\alpha} - \frac{3}{2} e^{\alpha} - 1 > 0 \]

### Step 6: Substitute \( t = e^{\alpha} \)
Let \( t = e^{\alpha} \), then the inequality becomes:

\[ t^2 - \frac{3}{2} t - 1 > 0 \]

### Step 7: Solve the Quadratic Inequality
We can solve the quadratic equation \( t^2 - \frac{3}{2} t - 1 = 0 \) using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{\frac{3}{2} \pm \sqrt{\left(\frac{3}{2}\right)^2 + 4}}{2} \]

Calculating the discriminant:

\[ \left(\frac{3}{2}\right)^2 + 4 = \frac{9}{4} + \frac{16}{4} = \frac{25}{4} \]

So we have:

\[ t = \frac{\frac{3}{2} \pm \frac{5}{2}}{2} \]

This gives us the roots:

\[ t_1 = \frac{8/2}{2} = 2 \quad \text{and} \quad t_2 = \frac{-2/2}{2} = -\frac{1}{2} \]

### Step 8: Analyze the Sign of the Quadratic
The quadratic \( t^2 - \frac{3}{2} t - 1 \) opens upwards (since the coefficient of \( t^2 \) is positive). The intervals to test are:

1. \( (-\infty, -\frac{1}{2}) \)
2. \( (-\frac{1}{2}, 2) \)
3. \( (2, \infty) \)

The quadratic is positive in the intervals \( (-\infty, -\frac{1}{2}) \) and \( (2, \infty) \).

### Step 9: Consider the Domain of \( t \)
Since \( t = e^{\alpha} > 0 \), we discard the interval \( (-\infty, -\frac{1}{2}) \). Therefore, we only consider \( t > 2 \).

### Step 10: Convert Back to \( \alpha \)
Since \( t = e^{\alpha} \), we have:

\[ e^{\alpha} > 2 \]

Taking the natural logarithm of both sides:

\[ \alpha > \ln(2) \]

### Conclusion
The set of values of \( \alpha \) for which the inequality holds true is:

\[ \alpha \in (\ln(2), \infty) \]