# Geometric area integration of 1/x

1. May 31, 2009

### morrobay

given:
ln (x) = integral 1/t dt
from 1 to x
and x=30

Without a calculator and only a graph of y=1/x
How could you show that this geometric area under this curve (with any type of unit)
is equal to 3.4 area units, the ln (30)

not homework, I am looking for a tangible/physical example of above definition

2. May 31, 2009

### HallsofIvy

I'm not sure what you question is. If there were a purely geometric way to show that the area under y= 1/x, from x= 1 to x= 30, we wouldn't need integral calculus!

You could, of course, approximate it, using "Rieman sums". If we divide the x-axis, from x= 1 to x= 30, into N equal intervals, the right end of each interval is x= 1+ 29i/N with i from 1 to 30 and each interval has length 29/N. If we set up a rectangle with f(x)= f(1+ i/30)= 1/(1+ i/30)= 30/(i+30), then the area of each rectangle is (30/(i+ 30))(29/N). The area under the curve, then, is approximated by $\sum 30(29)/(N(i+ 30)$. I suppose you could do that without using a calculator but I wouldn't want to! I will stick with calculus.