dstefan
03-22-2008, 01:12 PM
Chapter 1, Problem 7:
Let f(x) be a continuous function. Show that
\lim_{h \to 0} \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ f(a), \quad \forall~~ a \in {\mathbb R}.
Solution:
Let F(x) = \int f(x) ~dx be the antiderivative of f(x). From the Fundamental Theorem of Calculus, it follows that \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ \frac{F(a+h) - F(a-h)}{2h}.
Using l'Hopital's rule and the fact that F'(x) = f(x), we find that
\lim_{h \to 0} \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ \lim_{h \to 0} \frac{F(a+h) - F(a-h)}{2h} ~=~ \frac{f(a+x) + f(a-h)}{2} ~=~ f(a)
since f(x) is a continuous function.
Let f(x) be a continuous function. Show that
\lim_{h \to 0} \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ f(a), \quad \forall~~ a \in {\mathbb R}.
Solution:
Let F(x) = \int f(x) ~dx be the antiderivative of f(x). From the Fundamental Theorem of Calculus, it follows that \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ \frac{F(a+h) - F(a-h)}{2h}.
Using l'Hopital's rule and the fact that F'(x) = f(x), we find that
\lim_{h \to 0} \frac{1}{2h} ~\int_{a-h}^{a+h} f(x) ~dx ~=~ \lim_{h \to 0} \frac{F(a+h) - F(a-h)}{2h} ~=~ \frac{f(a+x) + f(a-h)}{2} ~=~ f(a)
since f(x) is a continuous function.