[tex]\displaystyle\int_0^5R(t)\,\mathrm dt\approx\sum_{i=1}^4\frac{R(t_i)+R(t_{i-1})}2\Delta t_i[/tex]
where [tex]t_0=0,t_1=1,t_2=2,t_3=4,t_4=5[/tex], and we define [tex]\Delta t_i=t_i-t_{i-1}[/tex] for [tex]i\ge1[/tex]. The sum is adding up the areas of several trapezoids, each with dimensions determined by the values of [tex]t_i[/tex] and [tex]R(t_i)[/tex]. (See the picture below)
So the number of deliveries made is approximately
[tex]\displaystyle\frac{(250+100)(1-0)+(200+250)(2-1)+(350+200)(4-2)+(100+350)(5-4)}2=1175[/tex]
