This is mostly straight-forward, a matter of allocating the slack of
e-2.7 = 0.01828... or in relative terms 0.0067255...
For example, we can allocate roughly a third into discarding the inifinite tail. Conveniently, the integral from -inf to the cutoff equals the function value at the cutoff and we find that
c = -5 ; e^c = 0.0067378...
is in the right ballpark.
We will chop the rest into equal intervals and and approximate the integral by tangents to the midpoint (these are relatively easy to cut out of rectangular blocks). if s is half the interval length, then the relative error is
which, conveniently, doesn't depend on location, only on interval width. Further, there is a closed form expression for the integral of the approximation function
because the areas under the tangents form a geometric series. Using this formula and limiting ourselves to s of the form 1/(2n) (for easy cutting) we find
s = 1/8
is just small enough.
we cut the interval [-5:1] into bins of width 1/4 and in each bin we approximate the integral by the area under the tangent to its midpoint.
Using formula (*) we find that this cutting friendly lower approximation has integral
exact function in red, piecewise linear lower approximation in black
and will therefore accomodate our two squares and one rectangle with some area to spare.