That is largely straight-forward, a matter of allocating the slack of
e-2.7 = 0.01828… or in relative phrases 0.0067255…
For instance, we will allocate roughly a 3rd into discarding the inifinite tail. Conveniently, the integral from -inf to the cutoff equals the perform worth on the cutoff and we discover that
c = -5 ; e^c = 0.0067378…
is in the appropriate ballpark.
We’ll chop the remaining into equal intervals and and approximate the integral by tangents to the midpoint (these are comparatively straightforward to chop out of rectangular blocks). if s is half the interval size, then the relative error is
1-2s/(e^s-e^-s)
which, conveniently, does not depend upon location, solely on interval width. Additional, there’s a closed type expression for the integral of the approximation perform
(*) 2s(e^1-e^c)/(e^s-e^-s)
as a result of the areas underneath the tangents type a geometrical sequence. Utilizing this formulation and limiting ourselves to s of the shape 1/(2n) (for simple slicing) we discover
s = 1/8
is simply sufficiently small.
In abstract:
we reduce the interval [-5:1] into bins of width 1/4 and in every bin we approximate the integral by the world underneath the tangent to its midpoint.
Utilizing formulation (*) we discover that this slicing pleasant decrease approximation has integral
2.704495…
actual perform in pink, piecewise linear decrease approximation in black
and can subsequently accomodate our two squares and one rectangle with some space to spare.
