Question

Q = at - bt²
$\therefore t \in \left[0, \frac{a}{b}\right]$
$i=\frac{dq}{dt}$= a - 2bt
Note: i is +ve t $\in \left(0,\frac{a}{2b}\right)$
And i is - ve t $\in \left(\frac{a}{2b},\frac{a}{b}\right)$
Positive current means current one direction and negative current means current in opposite direction.
$\therefore$ dH = i²Rdt
  = (a - 2bt)² R dt
H = $\underset{0}{\overset{a/b}{\int }}$(a - 2bt)² R dt
$={\overline{)\frac{(a-2bt{)}^{3}R}{3(-2b)}}}_0^{\frac{a}{b}}$
$=\frac{1}{-b}\left[{\left(a-2b\frac{a}{b}\right)}^3-(a{)}^3\right]R$
$=-\frac{1}{6b}$ [(- a)³ - a³]R
 H = $\frac{a^{3}R}{3b}$