Question

If \({\rm{f}}\) is a continuous function, what is the limit as \({\rm{h}} \to {\rm{0}}\)of the average value of \({\rm{f}}\)on the interval \({\rm{(x,x + h)}}\) ?

Short answer

Average value of a function \(f\)on the interval \((x,x + h)\)is \(f(x)\).

Step-by-step solution

Definition of the average value function

Average value of\(f\)an interval \((a,b)\)is

\({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\)

Applying average value formula to get required result

Average value of \(f\) on interval \((x,x + h)\)is

\(\begin{aligned}{c}{f_{ave}} &= \frac{1}{{(x + h) - x}}\int_x^{x + h} f (x)dx\\ &= \frac{1}{h}\int_x^{x + h} f (x)dx\end{aligned}\)

Taking limit as \({\rm{h}} \to {\rm{0}}\)

Let us simplify,

\(\begin{aligned}{c}\mathop {\lim }\limits_{h \to 0} {f_{ave}} &= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\int_x^{x + h} f (x)dx\\ &= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}(F(x))_x^{x + h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{1}{h}(F(x + h) - F(x))\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{F(x + h) - F(x)}}{h}\end{aligned}\)

This is the limit definition of the derivative.

\(\begin{array}{c}\mathop {\lim }\limits_{h \to 0} {f_{ave}} = {F^'}(x)\\ = f(x)\end{array}\(

Therefore, limit as \(h \to 0\(of the average value of \(f\(on the interval \((x,x + h(\(is \(f(x)\(.