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

[Albert Chan]

Valentin just post the solution, which did include the area of the tiny speck.
Although I knew of the speck, I thought it were not OP were asking ... I guessed wrong.

Anyway, I tried this with HP Prime emulator.
integrate() use Adaptive Gaussian Quadrature, switching to Romberg if that failed.

For the main area, it work wonderfully, even with singularity inside interval.
Adpative part just zeroed in to it automatically, as needed.

CAS> s := sqrt(-ln(y*y/30.07+exp(-sin(y)))) :;
CAS> f := surd(y+s,3) - surd(y-s,3) :;
CAS> int(f, y, 0, 2.82740261413)

2.07662636775

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Assuming speck is an ellipse, we already have 5+ digits accuracy.

CAS> c := -4.06717950603 // center of speck interval
CAS> a, b := 0.017967241585, f(y=c)/2 // major, minor axis
CAS> pi*a*b // ellipse area

7.19760850546e−5

The shape is simple, but Adaptive Gaussian Quadrature failed, switched to Romberg.
Romberg result (no u-substituion) is worse than above ellipse estimate.

CAS> int(f, y, c-a, c+a)

[7.19759235247e−5, 7.19761930451e−5]

The reason it failed is because the gap (2a), relative to center (c), is just too small.
Internally, integrate limit is from -1 to 1, with gaussian quadrature weights and abscissa.
When it get mapped to this tight domain, there is just not enough room. (think pigeonhole principle)

The inaccurate mapped abscissa generated fuzziness to f value, which mess up the area.
All we need is to "push out" f fuzziness. (1e-4 > area of speck)

CAS> int(f + 1e-4/(2a), y, c-a, c+a) - 1e-4

7.19761930451e−5