Question
\( ( \int _ { - \infty } ^ { + \infty } e ^ { - x ^ { 2 } } d x ) ^ { 2 \cdot \frac { 1 } { 2 } } \)
Ask by Potter Jimenez. in France
Mar 08,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\[
\sqrt{\pi}
\]
Solution
1. Рассмотрим данное выражение:
\[
\left( \int_{-\infty}^{+\infty} e^{-x^2} dx \right)^{2\cdot\frac{1}{2}}
\]
2. Вычислим показатель степени:
\[
2\cdot\frac{1}{2} = 1
\]
Таким образом, выражение упрощается до:
\[
\left( \int_{-\infty}^{+\infty} e^{-x^2} dx \right)^1 = \int_{-\infty}^{+\infty} e^{-x^2} dx
\]
3. Известно, что интеграл Гаусса имеет значение:
\[
\int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt{\pi}
\]
4. Подставляя найденное значение, получаем окончательный ответ:
\[
\sqrt{\pi}
\]
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution The Deep Dive
Did you know that the integral \( \int _{ - \infty }^{ + \infty } e^{-x^2} \, dx \) equals \( \sqrt{\pi} \)? This integral is a classic example in probability and statistics and is often used to derive properties of the normal distribution. The beauty of this integral lies in its symmetry and connection to Gaussian functions. Now, if we square the result, \( ( \sqrt{\pi} )^2 \), we end up with \( \pi \). This transformation is useful in many areas, including quantum mechanics and statistical mechanics, where the Gaussian distribution plays a pivotal role in model behaviors and phenomena. How cool is that?
