The mean of exp of Brownian motion

Idea. $W_1$ is standard normal, so this is the mean of a lognormal variable. Completing the square turns the integral into a normal density times a constant.

Step 1. $W_1\sim N(0,1)$, so $$\mathbb{E}\left[e^{W_1}\right]={\int }_{-\infty }^{\infty }{e}^x\frac{e^{-x^2/2}}{\sqrt{2\pi }}dx.$$

Step 2: complete the square. $x-\frac{x^2}{2}=\frac{1}{2}-\frac{(x-1{)}^2}{2}$, so $$\mathbb{E}\left[e^{W_1}\right]=e^{1/2}{\int }_{-\infty }^{\infty }\frac{e^{-(x-1{)}^2/2}}{\sqrt{2\pi }}dx=e^{1/2}.$$ The last integral is $1$: it is the density of $N(1,1)$.

So the answer is $\sqrt{e}\approx 1.649$.

The trap. Answering $e^{\mathbb{E}[W_1]}=e^0=1$. The exponential is convex, so by Jensen's inequality the mean of $e^{W_1}$ is larger than $e$ raised to the mean.

In general. For $Z\sim N(\mu ,{\sigma }^2)$, $\mathbb{E}[e^Z]=e^{\mu +{\sigma }^2/2}$, so $\mathbb{E}[e^{W_t}]=e^{t/2}$. This is why $e^{W_t-t/2}$ has constant mean 1: it is a martingale, and it is the reason for the $-\frac{1}{2}{\sigma }^2$ term in the Black-Scholes model.