Expectation of maximum in uniform distribution

Question

Trying to solve this:

Let X be a random variable uniformly distributed over the interval [0,2]. Find E(max(X,X^3))

I'm not quite sure how to approach this. I looked up a few things online and found several approaches and they all gave me an answer of 4/3, which is apparently wrong.

Answer 1

For $X \in [0,2]$, when is $X > X^3$, and when is $X \le X^3$? Then:

$$\operatorname{E}[\max(X,X^3)] = \int_{x=0}^2 \max(x,x^3) f_X(x) \, dx = \int_{x=?}^? x f_X(x) \, dx + \int_{x=?}^? x^3 f_X(x) \, dx,$$ where $$f_X(x) = \frac{1}{2}, \quad 0 \le x \le 2$$ is the uniform density on $[0,2]$, and the $?$ symbols are the things you need to figure out.