Expectation of normal distribution

[kind coral]

Can anyone explain why E[N^k} turns into that integral. I thought if the standard normal distribution is given by f(x), as in the question, then E[N^k] would equal E[f(x)^k]

[bold spruce]

Yeah sure, so the key here is there is a difference between the random variable X and the probability distribution f(x). X represents the value you read out when you measure the random variable, while f(X) spits out that probability that you would see/measure that value

For this problem it'll be more clear if we let Y=X^k stand in for X^k. Then the expected value of Y, E(Y), is given by the integral of Y*f(Y)

But f(Y), the probability distribution of Y=X^k is the exact same as the probability distribution of f(X) because the exponent k is just a constant

So E(Y)=E(X^k) is essentially the integral X^k*f(x)