The charge flowing through a resistance R varies with time t as Q = at – bt², where a and b are positive constants. The total heat produced in R is 

$\text{ Given, }\mathrm{Q}=\mathrm{at}-{\mathrm{bt}}^2\therefore \mathrm{i}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{a}-2\mathrm{bt}$

To find total heat produced, charge flowing is to be maximum.

$\begin{array}{r}\text{ i.e, }\frac{\mathrm{di}}{\mathrm{dt}}=0\Rightarrow \mathrm{a}-2\mathrm{bt}=0\\ \mathrm{a}=2\mathrm{bt}\Rightarrow \mathrm{t}=\frac{\mathrm{a}}{2\mathrm{b}} \end{array}$

∴ Heat produced in this time 

$\begin{array}{r}\mathrm{H}={\int }_{{\mathrm{i}}^2\mathrm{Rdt}} ={\int }_0^{\mathrm{a}/2\mathrm{b}} (\mathrm{a}-2\mathrm{bt}{)}^2\mathrm{Rdt} \\ ={\int }_0^{\mathrm{a}/2\mathrm{b}} \left({\mathrm{a}}^2+4{\mathrm{b}}^2{\mathrm{t}}^2-4\mathrm{abt}\right)\mathrm{Rdt} \\ =\mathrm{R}{\left({\mathrm{a}}^2\mathrm{t}+4{\mathrm{b}}^2\frac{{\mathrm{t}}^3}{3}-4\mathrm{ab}\cdot \frac{{\mathrm{t}}^2}{2}\right)}_0^{\mathrm{a}/2\mathrm{b}} \\ =\mathrm{R}\left({\mathrm{a}}^2\cdot \frac{\mathrm{a}}{2\mathrm{b}}+\frac{4{\mathrm{b}}^2}{3}\cdot \frac{{\mathrm{a}}^3}{8{\mathrm{b}}^3}-2\mathrm{ab}\cdot \frac{{\mathrm{a}}^2}{4{\mathrm{b}}^2}\right)\\ =\mathrm{R}\left(\frac{{\mathrm{a}}^3}{2\mathrm{b}}+\frac{{\mathrm{a}}^3}{6\mathrm{b}}-\frac{{\mathrm{a}}^3}{2\mathrm{b}}\right)=\frac{{\mathrm{a}}^3\mathrm{R}}{6\mathrm{b}} \end{array}$