Eep. Let's see, so the function equals zero whenever $\displaystyle x^{3}$ is an integer. $\displaystyle 1^{3}=1, 2^{3}=8$, so I can split this up into

$\displaystyle -\int_{1}^{\sqrt[3]{2}}f(x)+\int_{\sqrt[3]{2}}^{\sqrt[3]{3}}f(x) - \int_{\sqrt[3]{3}}^{\sqrt[3]{4}}f(x) + \int_{\sqrt[3]{4}}^{\sqrt[3]{5}}f(x) - \int_{\sqrt[3]{5}}^{\sqrt[3]{6}}f(x) + \int_{\sqrt[3]{6}}^{\sqrt[3]{7}}f(x) - \int_{\sqrt[3]{7}}^{2}f(x)$

and evaluate that. The antiderivative looks like it should be $\displaystyle -\frac{1}{3\pi}\cos(\pi x^{3})$, so this won't be super difficult, just tedious. Thanks!