The probability that the maximum value of the three random variables is some value between 0 and $x$ is given by $$ P(\max \leq x) = \prod_{k=1}^{3} P(x_k \leq x) $$ and because $x_k$ is a uniformly distributed random variable on $[0,1]$, then $$ P(x_k \leq x) = x. $$ Therefore, the cumulative distribution of the maximum of the three random variables is $$ P(\max \leq x) = F_{\max}(x) = x^3. $$ Recall that the probability density function is the derivative of the cumulative distribution function, $$ f_{\max}(x) = \frac{\text{d} F_{\max}(x)}{\text{d}x} = 3x^2. $$ Now all we have to do is multiply the probability density function by $x$ and integrate to calculate the expected value $$ \text{E}[\max] = \int_0^1 x \cdot f_{\max}(x) \text{d}x $$ $$ \text{E}[\max] = 3 \int_0^1 x^3 \text{d}x = 3 \cdot \left[ \frac{1}{4} x^4 \right]_0^1 $$ which evaluates to $$ \text{E}[\max] = \frac{3}{4}. $$