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\int_{0}^{1}(3 x-1)^{4} \mathrm{dx}=\left.\frac{1}{15}(3 x-1)^{5}\right|_{0} ^{1}=\frac{1}{15}(3 \cdot 1-1)^{5}-\frac{1}{15}(3 \cdot 0-1)^{5}=\frac{1}{15}(2)^{5}-\frac{1}{15}(-1)^{5}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}