Evaluate:

$\displaystyle \lim_{h \to 0} \frac{1}{h} \int_0^h {5e^{-x^2}dx} $

Let $\displaystyle f(x) = 5e^{-x^2} $, since f is continuous on [0, h] assuming h>0, I used the Fundamental Theorem of Calculus.

$\displaystyle F'(x) = \lim_{h\to0 } \frac{1}{h} \int_x^{x+h} f(t)dt = f(x) $

Let x = 0, we get

$\displaystyle \rightarrow \lim_{h\to0} \frac{1}{h} \int_0^h5e^{-x^2}dx = f(0) $

$\displaystyle \rightarrow f(0) = 5e^0 = 5$, correct?