The change following through a resistance R varies with time ‘t’ as ${\text{Q\hspace{0.17em}=\hspace{0.17em}at\hspace{0.17em}-\hspace{0.17em}bt}}^{\text{2}}$ . The total heat produced in R is

$$\begin{gathered} {\text{Q\hspace{0.17em}=\hspace{0.17em}at\hspace{0.17em}-\hspace{0.17em}bt}}^{\text{2}} \\ \text{⇒i}=\frac{\text{d}Q}{\text{d}t}=\text{a\hspace{0.17em}-\hspace{0.17em}2bt} \end{gathered}$$

$now, \text{i\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em}\hspace{0.17em}at\hspace{0.17em}\hspace{0.17em}t\hspace{0.17em}=\hspace{0.17em}}\frac{\text{a}}{\text{2b}}$

$Total heat produced is given by$

$$\begin{gathered} \text{H}=\underset{0}{\overset{a/2b}{\int }}{i}^{2}Rdt=R\underset{0}{\overset{a/2b}{\int }}{\left(a-2bt\right)}^{2}dt \\ =\text{R}\underset{0}{\overset{a/2b}{\int }}\left(a^2-4abt+4b^2{t}^2\right)dt \\ =\text{R}{\left[a^{2}t-4ab\frac{t^2}{2}+\frac{4b^2{t}^3}{3}\right]}_0^{a/2b} \\ =\text{R}\left[a^2.\frac{a}{2b}-\frac{4ab}{2}.\frac{a^2}{4b^2}+\frac{4b^2}{3}.\frac{a^3}{8b^3}\right] \\ =\frac{a^{3}R}{b}\left[\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right]=\frac{a^{3}R}{6b} \end{gathered}$$