If `int_0^ooe^(-x^2) dx=(sqrtpi)/2 `, then `int_0^ooe^(-ax^2) dx ` where ` a gt 0` is:

Options

A. (a) `(sqrt(2))/(2)`
B. (b) `(sqrt(pi))/(2a)`
C. (c) `2(sqrt(pi))/(a)`
D. (d) `1/2sqrt((pi)/(a))`

Answer

To solve the integral \(\int_0^\infty e^{-ax^2} \, dx\) where \(a > 0\), we can use the information given about the integral \(\int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}\).

### Step-by-step Solution:

1. **Identify the given integral**:
We know that:
\[ I = \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \]

2. **Change of variables**:
We will evaluate the integral \(\theta = \int_0^\infty e^{-ax^2} \, dx\) using a substitution. Let:
\[ u = \sqrt{a} x \implies x = \frac{u}{\sqrt{a}} \implies dx = \frac{du}{\sqrt{a}} \]

3. **Change the limits of integration**:
When \(x = 0\), \(u = 0\). When \(x \to \infty\), \(u \to \infty\). Therefore, the limits remain the same:
\[ \theta = \int_0^\infty e^{-a\left(\frac{u}{\sqrt{a}}\right)^2} \cdot \frac{du}{\sqrt{a}} = \int_0^\infty e^{-u^2} \cdot \frac{du}{\sqrt{a}} \]

4. **Factor out the constant**:
We can factor out \(\frac{1}{\sqrt{a}}\):
\[ \theta = \frac{1}{\sqrt{a}} \int_0^\infty e^{-u^2} \, du \]

5. **Substitute the known integral**:
We know from step 1 that:
\[ \int_0^\infty e^{-u^2} \, du = \frac{\sqrt{\pi}}{2} \]
Therefore, we can substitute this into our expression for \(\theta\):
\[ \theta = \frac{1}{\sqrt{a}} \cdot \frac{\sqrt{\pi}}{2} \]

6. **Final expression**:
This simplifies to:
\[ \theta = \frac{\sqrt{\pi}}{2\sqrt{a}} = \frac{1}{2} \sqrt{\frac{\pi}{a}} \]

Thus, the final result is:
\[ \int_0^\infty e^{-ax^2} \, dx = \frac{1}{2} \sqrt{\frac{\pi}{a}} \]