Let $X$ and $Y$ be two jointly continuous random variables with joint PDF
$$f_{XY}(x, y) = \begin{cases} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2} & x \in \mathbb{R}, \quad x-1<y<x\\ 0 & \text{otherwise} \end{cases}$$
Find $EY$.
What I tried:
$$EY = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} y f_{XY}(x, y) dy dx = \int_{-\infty}^{\infty} \int_{x-1}^{x} y \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2} dy dx$$
but I don't know how to evaluate this integral.
Guide:
\begin{align}\int_{-\infty}^{\infty} \int_{x-1}^{x} y \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2} dy dx &=\int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-\frac12x^2} \frac{x^2-(x-1)^2}{2} \, dx \\ &= \int_{-\infty}^\infty \frac1{\sqrt{2\pi}} e^{-\frac12x^2} \frac{(2x-1)}{2} \, dx\\ &= \int_{-\infty}^\infty x \frac1{\sqrt{2\pi}} e^{-\frac12x^2}\, dx - \frac12 \int_{-\infty}^\infty\frac1{\sqrt{2\pi}} e^{-\frac12x^2} \, dx \end{align}
Using the property of normal pdf and definition of expectation, can you complete the rest?