Vtyjkyuk

$f(x) = int_1^infty frac2sqrt2pie^-frac12x^2 dx$

find $P(X > 1)$

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

$P(X > 1) = 1 - P(X leq 1) = 1 - 2 phi(1) = 1-2(1-phi(-1)) = 1 - 2(1-0.1587) = -0.6826$.

Yikes. Negative number. What am I doing wrong?

Let´s say the pdf is $$f(x)=frac2sqrt2pie^-frac12x^2$$

(without the integral)

Now you want to calculate $P(X> 1)$ which is

$int_1^infty frac2sqrt2pie^-frac12x^2 dx$

First we can factor out $2$.

$2cdot int_1^infty frac1sqrtpie^-frac12x^2 dx=2cdot P(Y>1)$, where $Y$ is standard normal distributed as $Ysimmathcal N(0,1)$

Since $Y$ is symmetric distributed around $0$ we can say that $2cdot P(Y> 1)= P(|Y|>1)$.

From the 68–95–99.7 rule you probably know that $1-P(|Y|>1)=P(|Y|<1)=0.6827$

Consequently $P(|Y|>1)=1-0.6827=0.3173=P(X>1)$

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:
$$P(X > 1) = 1 - P(X leq 1) = 1 - colorred2phi(1) = dots$$

thus $$P(Xle1)=phi(1)$$

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