Find $P(X > 1)$ if $X$ is standard normal $f(x) = \int_{1}^{\infty} \frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} dx$

find $P(X &gt; 1)$

This is $X$ ~ $Norm(0, 1)$.

$P(X &gt; 1)  = 1 - P(X \leq 1) = 1 - 2 \phi(1) = 1-2(1-\phi(-1)) = 1 - 2(1-0.1587) = -0.6826$. 

Yikes. Negative number. What am I doing wrong?
I assume that by $\phi(t)$ you mean the area in the left tail of a normal distribution. If so, then the two below is incorrect and should be removed:
$$P(X &gt; 1) = 1 - P(X \leq 1) = 1 - \color{red}{2}\phi(1) = \dots$$
thus $$P(X\le1)=\phi(1)$$

$$your answer =1-\phi(1)$$