The difference between the greatest and he least value of the functionF(x)=∫0x (t+1)dt on [2, 3] is

Differentiating the given function, we getF′(x)=t+1t=xdxdx−[t+1]t=0d0dx=x+1This is positive for all x ∈ [2, 3], so F is an increasing function in this interval. Therefore its greatest value isF(3)=∫03 (t+1)dt and its least value is F(2)=∫02 (t+1)dt,so that the required difference between these values is∫03 (t+1)dt−∫02 (t+1)dt=∫23 (t+1)dt=72.