I have the following question:
Consider the following probability Density function:
$$f_n(x) = \begin{cases}c\cdot(x^n-x^{2n}) & : 0\leq x\leq 1 \\[1ex] 0 & : \textsf{otherwise}\end{cases}$$
And I am tasked to determine the value of $c$ in terms of $n$. Would this mean I have to integrate this function from $0$ to $1$ with respect to $n$ and treat $x$ as a constant? I am a little confused as to the wording of the question and would greatly appreciate any guidance.
Not quite. Integrate $f_n(x)$ from $0$ to $1$ with respect to $\color{blue}x$, holding $n$ as constant, equat it to $1$ (since it is total probability), then solve for $c$.
$$\begin{align}1= & ~ \int_0^1 c(x^n-x^{2n})\operatorname d x \\[2ex] \vdots \\[2ex]\therefore c= & ~ \ldots \end{align}$$
Thus $c$ will be evaluated as some expression in terms of $n$.