The difference between the greatest and least values of the function F(x)=∫0x (t+1)dt on [1,3] is

F′(x)=x+1>0 for x∈[1,3] Therefore, the greatest value is F(3) and the least value is F(1). The required difference is ∫03 (t+1)dt−∫01 (t+1)dt=∫13 (t+1)dt=6.