Evaluating a definite integral of a piecewise function using FTC Part 2

Since $f$ is defined piecewise with a break at $x=2$, split the integral as $\int_0^3 f(x)\,dx = \int_0^2 x^2\,dx + \int_2^3 (6-x)\,dx$. For the first piece, an antiderivative of $x^2$ is $F(x) = \frac{x^3}{3}$, so $\int_0^2 x^2\,dx = F(2)-F(0) = \frac{8}{3} - 0 = \frac{8}{3}$.

For the second piece, an antiderivative of $6-x$ is $G(x) = 6x - \frac{x^2}{2}$, so $\int_2^3 (6-x)\,dx = G(3) - G(2) = \left(18 - \frac{9}{2}\right) - \left(12 - 2\right) = \frac{27}{2} - 10 = \frac{7}{2}$. Adding both pieces gives $\frac{8}{3} + \frac{7}{2} = \frac{16}{6} + \frac{21}{6} = \frac{37}{6}$.