Comments and discussion on Valentin's 4th "Then and Now" - Area

[Albert Chan]

First we write as 3 mini-integrals, with bad region move to the left (lower limit)

\(\displaystyle\quad\int_0^{a+b} = \int_0^{a/2} - \int_{a}^{a/2} + \int_a^{a+b}\)

Second, let y = c1+c2*z, dy = c2 dz, so that integrand limits matched.
We could use any limits. I picked 0 .. 0.5, thus needed B = 2b.

\(\quad\,\int_0^{a/2} f(y)\;dy = \int_0^{1/2} f(a\,z)\, (a\,dz)\)

\(-\int_{a}^{a/2} f(y)\;dy = \int_0^{1/2} -f(a-a\,z)\, (-a\,dz)\)

\(\int_{a}^{a+\frac{B}{2}} f(y)\;dy = \int_0^{1/2} f(a+B\,z)\, (B\,dz)\)

Add it all up, we have:

\(\displaystyle \int_0^{a+b} f(y)\;dy
= \int_0^{1/2} \Big[\big(f(a\,z) + f(a- a\,z)\big)·a \;+\; f(a+B\,z)·B\Big] \,dz
\)