How do I find a natural log without a calculator?

For example, to approximate #ln(7)#, split the interval #[1, 7]# into a number of strips of equal width, and sum the areas of the trapezoids with vertices:

#(x_n, 0)#, #(x_n, 1/x_n)#, #(x_(n+1), 0)#, #(x_(n+1), 1/(x_(n+1)))#

If we use strips of width #1#, then we get six trapezoids with average heights: #3/4#, #5/12#, #7/24#, #9/40#, #11/60#, #13/84#.

If you add these, you find:

#3/4+5/12+7/24+9/40+11/60+13/84 ~~ 2.02#

If we use strips of width #1/2#, then we get twelve trapezoids with average heights: #5/6#, #7/12#, #9/20#,...,#25/156#,#27/182#

Then the total area is:

#1/2 xx (5/6+7/12+9/20+...+25/156+27/182) ~~ 1.97#

Actually #ln(7) ~~ 1.94591#, so these are not particularly accurate approximations.

Simpson's rule approximates the area under a curve using a quadratic approximation. For a given #h > 0#, the area under the curve of #f(x)# between #x_0# and #x_0 + 2h# is given by:

#h/3(f(x_0) + 4f(x_0+h) + f(x_0+2h))#

Try improving the approximation using Simpson's rule, using #h = 1/2# then we get six approximate areas to sum:

#h/3(25/6+73/30+145/84+241/180+361/330+505/546)~~1.947#

That's somewhat better.