Net displacement via the Net Change Theorem

By the Net Change Theorem, the net displacement equals $\int_0^5 v(t)\, dt = \int_0^5 (3t^2 - 12t + 9)\, dt$. First find an antiderivative: $\int (3t^2 - 12t + 9)\, dt = t^3 - 6t^2 + 9t + C$. This works because integrating each term with the power rule gives $t^3$, $-6t^2$, and $9t$ respectively.

Now apply the Fundamental Theorem of Calculus using $s(t) = t^3 - 6t^2 + 9t$: compute $s(5) - s(0)$. We get $s(5) = 125 - 150 + 45 = 20$ and $s(0) = 0$, so the net displacement is $20 - 0 = 20$ meters.