The
method of integration comes from calculus.
In general, if you have two well-behaved functions, and you'd like to know an area between their graphs, integration will do it.
I had already determined the functions needed, and I used a CAS (computer algebra system) to complete each integral's "number-crunching" for me. (I'm lazy.)
Pieces of functions A(x), B(x), C(x), D(x), and F(x)=2.25 can be spliced together, to form the "top" boundary of the shape.
Pieces of functions E(x) and G(x)=0 can be spliced together, to form the "bottom" boundary of the shape.
Basically, we're filling in the shape with an infinite number of vertical "slices", and integration provides the means to sum up these many slices' areas, to get a total area. (You may think of each slice as simply a vertical line segment, with infinitely-thin width.)
We get the height of each slice by taking the difference between the top function value and the bottom function value, and integration does this at each x (from left to right) over the interval from x=0 to x=1.938.
In the image below, there are points along both the top (green) and bottom (blue) boundaries where the respective piece-wise functions change (eg: from A(x) to B(x) on top, from E(x) to G(x) on bottom) . This necessitates having to integrate over sub-intervals (red). We add up the results.
\(\displaystyle \displaystyle \int_{0}^{0.063} \! 2.25\!-\!E(x) \; \mathrm{d}x = 0.07954\)
\(\displaystyle \displaystyle \int_{0.063}^{0.438} \! A(x)\!-\!E(x) \; \mathrm{d}x = 0.59346\)
\(\displaystyle \displaystyle \int_{0.438}^{1.179} \! B(x)\!-\!E(x) \; \mathrm{d}x = 0.564391\)
\(\displaystyle \displaystyle \int_{1.179}^{1.250} \! C(x)\!-\!E(x) \; \mathrm{d}x = 0.04273\)
\(\displaystyle \displaystyle \int_{1.250}^{1.603} \! C(x) \; \mathrm{d}x = 0.19300\)
\(\displaystyle \displaystyle \int_{1.603}^{1.938} \! D(x) \; \mathrm{d}x = 0.09439\)
I hope that my hastily-written summary is helpful. :cool: