Casio fx-991EX vs Hp 50g speed difference

[Albert Chan]

(07-25-2018 06:32 PM)Albert Chan Wrote:  Just integrate above transformed f, from -1 to 1:

\(\int_0^{200}sin(x) dx \) = \(\int_{-1}^{1}150(1-u^2) sin[50u (3 - u^2) + 100] du \)

(07-25-2018 02:44 AM)Carsen Wrote:  When I enter the integral in approx. mode, I get the answer of 0.512812324889 in 111.36 seconds with an error of 1.3E-9.

(07-25-2018 08:07 PM)Carsen Wrote:  If I evaluate the transformed integral in approx. mode, I get the answer of .512812325001 in 61.45 seconds.

For transformed integral, it only take half the time ?
And, compare against 1 - cos(200), more accurate ?

Is it tested on the same calculator ?

What if it were applied not just to periodic function, but others ?
Say, this thread original post ?

\(\int_0^{500}e^{-x}dx \) = \(\int_{-1}^{1}375(1-u^2) e^{-(125u (3 - u^2) + 250)} du \)