## Calculus (3rd Edition)

Let I= $\int_{0}^{4} (6t-3) dt$. Since $\int (6t-3) dt= \frac{6t^{2}}{2}-3t= 3t^{2}-3t=F(x)$, Therefore, by the second fundamental theorem, we get I= F(4)-F(0)= $(3\times4^{2}-3\times4)-0= 36$.