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 b³R/6a. True or False?

Given: Q = at − bt² ⇒ i = dQ/dt = a − 2bt
Therefore, current will be zero at t = a/2b
We know that: dH = i²Rdt ⇒ H = ∫₀^(a/2b) (a − 2bt)²Rdt ⇒ H = [(a − 2b(a/2b))³/R ÷ 2b] − [(a − 2b(0))³/R ÷ 2b] ⇒ H = a³R/6b