Why is the discrete formulation of the fundamental theorem of integral calculus correct?

Question

Define Diff$_hf = \frac{f(x+h)-f(x)}{h}$.

Define Av$_hf(a)=\frac{1}{h}\int_a^{a+h} f$

Why is the following correct?

$\int_a^b$Diff$_hf = Avf(b) -Avf(a) $.

Answer 1

Consider the function $$g(x):={\rm Av}_hf(x)={1\over h}\int_x^{x+h} f(t)\>dt\ .$$ Then $$g'(x)={1\over h}\bigl(f(x+h)-f(x)\bigr)={\rm Diff}_hf(x)\ .$$ It follows that $${\rm Av}_hf(b)-{\rm Av}_hf(a)=g(b)-g(a)=\int_a^b g'(x)\>dx=\int_a^b{\rm Diff}_hf(x)\>dx\ .$$