Discussion
Given \(\int_{-\infty}^{\infty} e^{-x^{2}}\,dx = \sqrt{\pi}\).
If \(a\) and \(b\) are positive integers, the value of \(\int_{-\infty}^{\infty} e^{-a(x+b)^{2}}\,dx\) is ________.
Answer: B
The value is \(\sqrt{\pi/a}\).
Substitute \(u = \sqrt{a}\,(x + b)\), so \(du = \sqrt{a}\,dx\). The limits stay \(-\infty\) to \(\infty\), because shifting by \(b\) and scaling by a positive \(\sqrt{a}\) does not change an infinite range:
\[ \begin{aligned} \int_{-\infty}^{\infty} e^{-a(x+b)^{2}}\,dx &= \frac{1}{\sqrt{a}}\int_{-\infty}^{\infty} e^{-u^{2}}\,du \\ &= \frac{\sqrt{\pi}}{\sqrt{a}} = \sqrt{\frac{\pi}{a}} \end{aligned} \]
The shift \(b\) only slides the bell curve along the axis, so it cannot change the area: both options with a factor \(b\) are wrong. \(\sqrt{\pi a}\) multiplies by \(\sqrt{a}\) where it should divide; a larger \(a\) makes the curve narrower, so the area must shrink.
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