The problem is evaluate $\int_{2}^{3} e^{-2(x-3)^2} dx$. I'm absolutely lost and couldn't find help in the textbook or any section in the textbook reading that had keywords about the steps taken to arrive at the solution. The steps are as follows:
$$\int_{2}^{3} e^{-2(x-3)^2} dx =\int_2^3 e^{\frac{-(x-3)^2}{2(\frac{1}{4})}}dx $$
$$= \sqrt{2\pi} \sqrt{1} \int_{2}^{3} e^{-2(x-3)^2} dx$$
$$=\sqrt{\frac{\pi}{2}}[\Phi (\frac{3-3}{1/2})-\Phi (\frac{2-3}{1/2})]$$
$$=\sqrt{\frac{\pi}{2}}[\frac{1}{2}-\Phi (-2)]$$
I'm not sure how the first step works. I see that the normal pdf is used between steps 2 and 3 and that usage roughly makes sense. I just don't understand the first and second lines. Any help/references for where to read are appreciated. Thanks in advance!
$\frac{1}{\sqrt{2\pi\sigma^2}} e^{-(x-\mu)^2/(2\sigma^2)}$ is the density of a Gaussian distributed random variable with mean $\mu$ and variance $\sigma^2$
so $\frac{1}{\sqrt{2\pi/4}} e^{-2(x-3)^2}$ is the density of a Gaussian distributed random variable with mean $3$ and variance $\frac14$
and thus $\int\limits_2^3 \frac{1}{\sqrt{\pi/2}} e^{-2(x-3)^2}\, dx$ is the probability that a Gaussian distributed random variable with mean $3$ and variance $\frac14$ takes values between $2$ and $3$. It is also what you want divided by $\sqrt{\pi/2}$.
If the variance is $\frac14$ then the standard deviation is $\frac12$. So $3$ is the mean and $2$ is two standard deviations below the mean. So what you want divided by $\sqrt{\pi/2}$ is also (from a standard normal distribution) $\int\limits_{-2}^0 \phi(x)\, dx = \Phi(0)-\Phi(-2) = \frac12-\Phi(-2)$
so, multiplying by $\sqrt{\pi/2}$, what you want is $\sqrt{\pi/2}\left(\frac12-\Phi(-2)\right)$
(which is also $\sqrt{\pi/2}\left(\Phi(2)-\frac12\right)$ if your tables do not stretch to negative values)