Many dimensions – 3

By linearity of the integral, we can integrate each $x_k^3$ separately to obtain $\frac{1}{4}$ . Thus the answer is $2018\times\frac{1}{4}=\boxed{504.5}$ .

$\int_0^1 (x_1^3 + x_2^3 + \cdots + x_n^3)\,\mathrm{d}x_i = x_1^3 + x_2^3 + \cdots + x_{i-1}^3 + \frac{1}{4} + x_{i+1}^3 + \cdots + x_n$ Thus, the $i$ th integral will replace $x_i^3$ with $\frac{1}{4}$ and we will be left with $2018\times\frac{1}{4} = \boxed{504.5}$ .