Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 52

$$\frac{2}{3}$$

\begin{aligned} \int_{0}^{2}(t-[t])^{2} d t & \\ &=\int_{0}^{1} t^{2} d t+\int_{1}^{2}(t-1)^{2} d t \\ &=\left.\frac{1}{3} t^{3}\right|_{0} ^{1}+\left.\frac{1}{3}(t-1)^{3}\right|_{1} ^{2} \\ &=\frac{1}{3}+\frac{1}{3}=\frac{2}{3} \end{aligned}