# Thread: [SOLVED] Limit of an Integral

1. ## [SOLVED] Limit of an Integral

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?

2. Originally Posted by Paperwings
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?

It looks dandy to me.

Tonio

3. Ah, ok, thank you. I wasn't sure if I was applying the FTC properly.