## Calculus (3rd Edition)

$2.7552 \ kg$
The total mass $(M)$ of the rod can be calculated as: $Mass, M= \int_{0}^{1.2} (1+2x+\dfrac{2x^3}{9}) \ dx \\=[x+x^2+\dfrac{x^4}{18}]_0^{1.2} \\=1.2+(1.2)^2+\dfrac{(1.2)^4}{18} \approx 2.7552 \ kg$