Given, $f(x)=\left\{\begin{array}{ll}2 x^{2}+1, & x \leq 1 \\ 4 x^{3}-1, & x>1\end{array}\right.$
$\begin{aligned} \therefore \int_{0}^{2} f(x) d x &=\int_{0}^{1} f(x) d x+\int_{1}^{2} f(x) d x \\ &=\int_{0}^{1}\left(2 x^{2}+1\right) d x+\int_{1}^{2}\left(4 x^{3}-1\right) d x \\ &=\left[\frac{2 x^{3}}{3}+x\right]_{0}^{1}+\left[\frac{4 x^{4}}{4}-x\right]_{1}^{2} \\ &=\frac{2}{3}(1)^{3}+1-(0+0) \\ &+\left[(2)^{4}-2-\left\{(1)^{4}-1\right)\right.\end{aligned}$
$$
\begin{array}{l}
=\frac{2}{3}+1+[16-2-0] \\
=\frac{2}{3}+15=\frac{2+45}{3} \\
=\frac{47}{3} \mathrm{sq} \text { units }
\end{array}
$$