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[Albert Chan]

(08-16-2023 11:55 AM)Albert Chan Wrote:  We could estimate ∫(e^x^3, x = 0 .. 6) without calculator.

Let y = e^x^3 , dy = e^x^3 * (3*x^2) dx

∫(e^x^3, x = 0 .. 6) = 1/3 * ∫(ln(y)^(-2/3), y = 1 .. e^6^3)

Integrand f(y) = ln(y)^(-2/3) is decreasing, *very flat* on the right, shaped like └------

∫(y dx) = ∫(1/(3x^2) dy)

A high growth function, going "vertical", viewed from y-axis is very flat!

We flatten it a bit more, before integrate, with y=z^3, dy = 3*z^2 dz
HP71B INTEGRAL internal u-transform should turn this into nice bell shape.

∫(e^x^3, x = a .. b) = ∫((3*ln(z))^(-2/3)*z^2, z = e^(a^3/3) .. e^(b^3/3))

10 DEF FNZ(Z) @ N=N+1 @ FNZ=(3*LN(Z))^(-2/3)*Z*Z @ END DEF
20 INPUT "P,A,B = ";P,A,B
30 N=0 @ DISP INTEGRAL(EXP(A^3/3),EXP(B^3/3),P,FNZ(IVAR)), N
40 GOTO 20

>DESTROY ALL @ RUN

P,A,B = 1e-3, 0,6
 5.96393763345E91      15
P,A,B = 1e-6, 0,6
 5.96393808826E91      31
P,A,B = 1e-9, 0,6
 5.96393809181E91      63

(10-28-2023 01:54 AM)Gerson W. Barbosa Wrote:  Try integrating from x = 5 to x = 6 and compare the results and times.

Interestingly, integrate from 5 .. 6 produce curve of same shape!
In other words, with 5 .. 6, same sample points, same integral result.