Total charge from a current function via the Net Change Theorem

By the Net Change Theorem, the total charge is $Q = \int_0^4 I(t)\,dt = \int_0^4 (3t^2 + 2t)\, dt.$ This directly applies $\int_a^b F'(t)\,dt = F(b)-F(a)$, treating $I(t)$ as the rate of change of charge.

An antiderivative of $3t^2 + 2t$ is $F(t) = t^3 + t^2$. Evaluating gives $Q = F(4) - F(0) = (4^3 + 4^2) - (0^3+0^2) = 64 + 16 - 0 = 80.$ So the total charge that flows through the wire over $[0,4]$ seconds is $80$ coulombs.