Trying to approximate $E[f(X)]$ - Woflram Alpha gives $E[f(X)] \approx \frac{1}{\sqrt{3}}$ but I get $E[f(X)] \approx 0$? Let $X \sim \mathcal{N}(\mu_X,\sigma_X^2) = \mathcal{N}(0,1)$. Let $f(x) = e^{-x^2}$. I want to approximate $E[f(X)]$.
Wolfram Alpha gives
\begin{align}
E[f(X)] \approx \frac{1}{\sqrt{3}}.
\end{align}
Using a Taylor expansion approach, and noting that $f''(0) = -2$, I get
\begin{align}
E[f(X)] &\approx f(\mu_X) + \frac{f''(\mu_X)}{2} \sigma_X^2 \\
& = f(0) + \frac{f''(0)}{2} \\
& = 1 + \frac{-2}{2} \\
& = 0.
\end{align}
Why does my approximation fail to match the Wolfram Alpha result? What can be done to fix it?

There's no need to "approximate" when you can derive the exact value of $\mathbb{E}[f(X)]$ . Let us apply the Law of the Unconscious Statistician (LoTUS) to obtain :
\begin{align*}
\mathbb{E}[f(X)] &= \int_{-\infty}^{+\infty} e^{-x^2} \cdot \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{x^2}{2}\right)~dx\\
&= 2\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{3x^2}{2}\right)~dx\\
&= 2\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} \cdot\frac{1}{\sqrt{6z}} e^{-z}~dz\\
&= \frac{2}{2\sqrt{3\pi}} \int_0^{+\infty} e^{-z} z^{\frac{1}{2} -1} ~dz\\
&=\frac{1}{\sqrt{3\pi}}\cdot \Gamma\left(\frac{1}{2}\right)\\
&= \frac{1}{\sqrt{3\pi}}\cdot \sqrt{\pi}\\
&= \frac{1}{\sqrt{3}}
\end{align*}
Hope this helps. :)
