The function $f(x)=e^{-x^{2} / 2}$ is part of the normal probability density function (or bell-shaped curve).Find the Maclaurin series for $\int e^{-x^{2} / 2} d x$ by first finding the Maclaurin series for $f(x)$ and then integrating it term by term.
A) $\sum_{i=0}^{\infty} \frac{(-1)^{i} x^{2 i+1}}{(2 i+1) 2^{i}(i !)}+C$
B) $\sum_{i=0}^{\infty} \frac{(-1)^{i} x^{2 i}}{(2 i) 2^{i}(i !)}+C$
C) $\sum_{i=0}^{\infty} \frac{(-1)^{i} x^{2 i+1}}{(2 i+1) 2^{i}(2 i !)}+C$
D) $\sum_{i=0}^{\infty} \frac{(-1)^{i} x^{2 i}}{(2 i) 2^{i}(2 i !)}+C$

Question

Quizplus · Accepted Answer

The Maclaurin series for $f(x) = e^{-x^2/2}$ is found by expanding $e^x$ and substituting $-x^2/2$ for $x$, leading to $\sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{2^i i!}$. Integrating this term by term gives $\sum_{i=0}^{\infty} \frac{(-1)^i x^{2i+1}}{(2i+1)2^i i!} + C$, which matches choice A.

The Function f(x)=e−x2/2f(x)=e^{-x^{2} / 2}f(x)=e−x2/2 Is Part of the Normal Probability Density Function (Or Bell-Shaped

The Function $f(x)=e^{-x^{2} / 2}$ Is Part of the Normal Probability Density Function (Or Bell-Shaped