$\therefore \quad I=\frac{d Q}{d t}=a-2 b t$

At $t=0, Q=0 \Rightarrow I=0$

Also, $I=0$ at $t=a / 2 b$

$\therefore \quad$ Total heat produced in resistance $R$

$H = \int\limits_0^{a/2b} {{I^2}Rdt = R\int\limits_0^{a/2b} {{{(a - 2bt)}^2}dt} } $

${ = R\int\limits_0^{a/2b} {\left( {{a^2} + 4{b^2}{t^2} - 4abt} \right)dt} }$

$=R\left[a^{2} t+4 b^{2} \frac{t^{3}}{3}-4 a b \frac{t^{2}}{2}\right]_{0}^{a / 2 b} $

$=R\left[a^{2} \times \frac{a}{2 b}+\frac{4 b^{2}}{3} \times \frac{a^{3}}{8 b^{3}}-\frac{4 a b}{2} \times \frac{a^{2}}{4 b^{2}}\right]$

$=\frac{a^{3} R}{b}\left[\frac{1}{2}+\frac{1}{6}-\frac{1}{2}\right]=\frac{a^{3} R}{6}$