Question

The value of $\int_0^\infty e^{-x^2} \, dx$ is:

• A. $\sqrt{\pi}$
• B. $\frac{\sqrt{\pi}}{2}$
• C. $\frac{\sqrt{2}}{\pi}$
• D. $\frac{\sqrt{\pi}}{2}$

Hint:

Remember that the Gaussian integral over the entire real line gives $\sqrt{\pi}$, and the integral from 0 to infinity is simply half of that.

Answer 1

The Correct Option is B

We are asked to evaluate the integral:

\[ I = \int_0^\infty e^{-x^2} \, dx \]

Step 1: Recognize the Gaussian integral
The Gaussian integral is a well-known result. For the entire range from \(-\infty\) to \(\infty\), we know that:

\[ \int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi} \]

Step 2: Use symmetry of the function
The function \(e^{-x^2}\) is symmetric around \(x = 0\), which means:

\[ \int_{-\infty}^\infty e^{-x^2} \, dx = 2 \int_0^\infty e^{-x^2} \, dx \]

Thus, we can express the integral over \(0\) to \(\infty\) as half of the integral over the entire real line:

\[ I = \frac{1}{2} \int_{-\infty}^\infty e^{-x^2} \, dx \]

Step 3: Substitution from the known result
From Step 1, we know the value of the full Gaussian integral:

\[ \int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi} \]

Therefore, substituting this result into the equation for \(I\):

\[ I = \frac{1}{2} \times \sqrt{\pi} = \frac{\sqrt{\pi}}{2} \]