HP71B Integral Questions

[Albert Chan]

(02-02-2020 03:31 PM)Albert Chan Wrote:  From the user's manual, about the algoritm, it uses Romberg's method, after a non-linear transform.

There was an error in the manual. x = ½(3u-u³) substitution *required* integrand limit -1 to 1, not a to b

\(\large \int _{-1} ^1 f(x) dx = {3 \over 2}\int _{-1} ^ 1 (1-u^2) f \left(
{ u (3-u^2) \over 2} \right) du \)

For a general case, with integrand limit a to b, we have:

\(\large \int _a ^b f(x) dx = {3(b-a) \over 4}\int _{-1} ^ 1 (1-u^2) f \left(
{ 2(b+a) + (b-a) u (3-u^2) \over 4} \right) du \)

From the manual:

Quote:INTEGRAL uses extended precision. Internally, sums are accumulated in 15-digit numbers ...
up to 65,535 points can be sampled on each subinterval, thus computing the integral to greater precision.

Trivia:
For 2 intervals, h = 1 → 2^16 intervals, h = 1/2^15 = 0.00003 05175 78125 (exact)
With 15 digits precision, and |u| ≤ 1, this implied INTEGRAL internally calculated u's are all exact.