Question

Given \[ \int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi}. \] If $a$ and $b$ are positive integers, the value of 

\(\int_{-\infty}^{\infty} e^{-a(x+b)^2}\, dx \text{ is} \)______

• A. \(\sqrt{\pi a}\)
• B. \(\frac{\sqrt{\pi}}{\sqrt{a}}\)
• C. \(b\sqrt{\pi a}\)
• D. \(b\sqrt{\frac{\pi}{a}} \)

Hint:

A shift inside a Gaussian integral never changes its value—the width (controlled by $a$) is what matters.

Answer 1

The Correct Option is B

We evaluate \[ \int_{-\infty}^{\infty} e^{-a(x+b)^2}\, dx. \] Step 1: Substitution.
Let \(u = x+b\). Then: \[ \int_{-\infty}^{\infty} e^{-a(x+b)^2} dx = \int_{-\infty}^{\infty} e^{-a u^2} du. \] Step 2: Gaussian formula.
\[ \int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}. \] Thus: \[ \int_{-\infty}^{\infty} e^{-a u^2} du = \frac{1}{\sqrt{a}} \sqrt{\pi}. \] Step 3: Conclusion.
Therefore: \[ \int_{-\infty}^{\infty} e^{-a(x+b)^2} dx = \frac{\sqrt{\pi}}{\sqrt{a}}. \] Final Answer: \(\frac{\sqrt{\pi}}{\sqrt{a}}\)