General questions about Normal Distribution characteristics I am very weak in understanding what my lecturer says because of many gaps in what I know.
The Density Function of the Normal with parameters $(\mu,\sigma^2)$ is $f(x)=\frac{1}{\sqrt{2\pi}\sigma}exp\{-\frac{(x-\mu)^2}{2\sigma^2}\}$


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*My lecturer said that $\frac{1}{\sqrt{2\pi}\sigma}$ is a normalizing constant, which I understand to be sort of like a balance to ensure that $\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}exp\{-\frac{(x-\mu)^2}{2\sigma^2}\}=1$. Is that right?


*Next he mentioned that $(x-\mu)^2$ is to center the distribution at $0$. Is it because this is always positive, so $f(-x)=f(x)$, hence the distribution will be symmetrical about $0$.?


*Lastly, he said that $2\sigma^2$ is to center the variance by some scaling factor. I do not understand.


*Is the above are true, what does the negative in the power do?

My very last Question:
Let $\mu=0$. What can you say about $E(X^k)$, where $k$ is an odd positive integer? (EDIT: E(X) is the mean of the normal)

My attempt is to google up the definition of $E(X^k)$, which gives me $E(X^k)=\int_{-\infty}^{\infty}f(x)*x^k$. Then I am stuck.
EDIT: Maybe I should try to do by parts?

Also I would like to learn the motivations for coming up with this distribution. The formula seems as random to me as it is. Why was this invented?
 A: 1)$$c\cdot\int_{-\infty}^{\infty}e^{\frac{(x-\mu)^2}{2\sigma^2}}=1.  $$
Solve for c and you get $c=\frac{1}{\sqrt(2\pi)}$, so yes, it is a normalizing constant.
2)  If $\mu=0$, the distribution is centered.  If $\mu\neq0$, the distribution is centered at $\mu$.  The form you have with parameters $(\mu,\sigma^2)$ is standard normal if $\mu=0$ and $\sigma=1$.
3) To find probability using the standard normal distribution you have 
$$P(X\le{x})=\frac{1}{\sqrt(2\pi)}\int_{-\infty}^{x}{e^{\frac{-t^2}{2}}}dt$$
THe reason we have the above formula using $\mu$ and $\sigma$ is that if we set a random variable 
$$Z=\frac{x-\mu}{\sigma}\sim{N(0,1)}$$
4)  The reason for the negative in the exponent is for convergence reasons.  Look at
$$\lim_{x\to\infty}{e^x}\text{  and  }\lim_{x\to\infty}{e^{-x}}$$ and I think you will see why the negative is necessary.
Here's a little brief history into the function:http://onlinestatbook.com/2/normal_distribution/history_normal.html
For #5, see $X$ standard normal distribution, $E[X^k]=?$
