[VA] SRC #012d - Then and Now: Area

[Albert Chan]

exp(-((x-d)^3-y)^2) > (R = y^2/M + exp(-sin(y)))      // Given: M=30.07, d=1.596

Let s = sqrt(-log(R)) ≥ 0

((x-d)^3-y)^2 < s^2
y-s < (x-d)^3 < y+s

x is real if s is real --> R ≤ 1 --> 0 ≤ y ≤ 2.82740261413

Height, \(\displaystyle f(y) = x_2 - x_1 = \sqrt[3]{y+s} - \sqrt[3]{y-s}\)

f slope is infinite when y = s --> R = exp(-y^2) --> y = 0 or 0.831971149978

Let a = 0.831971149978, a+b = a+B/2 = 2.82740261413
Infinite slopes at y = 0 and a, moved to z = 0

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

Let g(z) = RHS integrand, and substitute z = x³/2, to make z=0 infinite slope, down to 0.
INTEGRAL built-in u-transform should turn curve to bell-shaped, easy to integrate.      (*)

Area = \(\displaystyle \int_0^{1/2} g(z)\;dz
= \int_0^1 g\!\left(\frac{x^3}{2}\right)·\left(\frac{3}{2}x^2\;dx \right)
\)

10 DESTROY ALL @ M=30.07 @ A=.831971149978 @ B=2.82740261413-A @ P=1E-6
20 T=1/3 @ DEF FND(Y,S)=(Y+S)^T-SGN(Y-S)*ABS(Y-S)^T
30 DEF FNF(Y)=FND(Y,SQR(-LN(Y*Y/M+EXP(-SIN(Y)))))
40 B=B*2 @ DEF FNG(Z)=(FNF(A*Z)+FNF(A-A*Z))*A+FNF(A+B*Z)*B
50 SETTIME 0 @ DISP INTEGRAL(0,1,P,FNG(.5*IVAR^3)*IVAR^2)*1.5,TIME

>run
 2.07662636748      .49      ! @200× --> HP71B = 98 sec
>p=1e-9 @ run 50
 2.07662636775      .92      ! @200× --> HP71B = 184 sec

---

(*) at y = a+b, f slope is infinite too.
Above code assumed u-transform able to fix this edge.
If not, we move all edges to z=0, with this revised g(z).

>40 DEF FNG(Z)=(FNF(A*Z)+FNF(A-A*Z))*A+(FNF(A+B*Z)+FNF(A+B-B*Z))*B
>run
 2.07662636769      .6        ! @200× --> HP71B = 120 sec
>p=1e-9 @ run 50
 2.07662636775      1.2      ! @200× --> HP71B = 240 sec