Area under the curve

Firstly, we find where the curve cuts the $x$ -axis.

$4x-x^{2} = x(4-x) = 0 \implies x = 0 \text{or} 4$

Hence the area under the curve

$= \int_0^4 (4x-x^{2}) dx$

$= \Bigg [2x^{2} - \frac{x^{3}}{3}\Bigg ]_0^4$

$= \Bigg (32 - \frac{64}{3}\Bigg) - 0$

$= \frac{32}{3}$

Hence $a = 32, b = 3$ , and $a + b + ab = 32 + 3 + 32 \cdot 3 = \boxed{131}$ .