One can easily find the integral $\int_{0}^{\infty}\exp(-x)dx$. It is equal to 1. But is there a way to understand this geometrically without integration?
If i rotate the picture i see that $\int_{0}^{\infty}\exp(-x)dx=-\int_{0}^{1}\ln(t)dt$. Maybe there is some property of exp or log which allows to avoid integration?
PS:
I would like to accept the Mamikon's method pointed out by Jim Belk. But it is impossible to accept comments... So I accept the second best.
The following argument uses only basic properties of the exponential function and the integral, but not the fundamental theorem of calculus:
Let $\int_0^\infty e^{-x}dx=:I$. By looking at a figure we see that for any $c>0$ we have $$I=\int_0^c e^{-x}dx +\int_c^\infty e^{-x}dx= \int_0^c e^{-x}dx + e^{-c} I$$ or $$(1-e^{-c}) I =\int_0^c e^{-x}dx\ .$$ Using $e^{-c}\leq e^{-x}\leq1 \ \ (0\leq x\leq c)$ we conclude that $$c \ e^{-c} \leq (1-e^{-c}) I \leq c\ .$$ Now divide by $1-e^{-c}$ and let $c\to 0+$ to get the desired result.