The electric current through a wire varies with time as I = I _ 0 +...

Given that
Current $I = {I}_{0} + \betat$, ${I}_{0} = 20A$ and $\beta = 3A{s}^{- 1}$

The instantaneous current is given by

$I = 20 + 3t . . . \left(1\right)$

Integrating equation (1), we have

$\dfrac{dq}{dt} = 20 + 3t ⇒ {∫}_{0}^{q}dq = {∫}_{0}^{20} ( 20 + 3t ) dt ⇒ q = {∫}_{0}^{20}20dt + {∫}_{0}^{20}3tdt ⇒ q = {\left[20t + \dfrac{3{t}^{2}}{2}\right]}_{0}^{20} = 1000C$