Integration of nonnegative function

Question

Can someone explain the results of

Plot[
  ((E^(-.6 *x))^(-1 + 38) (1 - E^(-.6*x))^(-1 + 160) *.6)/Beta[-1 + 38., 160.], 
  {x, 1/.6, 4.5}]

and

1/Beta[-1 + 38., 160.]
  Integrate[0.6*(E^(-0.6*x))^(-1 + 38)*(1 - E^(-0.6*x))^(-1 + 160), {x, 2, 4.}]

?

I am getting very large values for functions with reasonable plot — user54562, Jan 08 '18 at 07:10
Try N[1/Beta[-1 + 38, 160]* Integrate[(E^(-6/10 *x))^(-1 + 38) (1 - E^(-6/10*x))^(-1 + 160) *6/ 10, {x, 2, 4}], 30] — Mariusz Iwaniuk, Jan 08 '18 at 07:22
See:https://mathematica.stackexchange.com/questions/32782/very-different-results-from-evaluating-same-expression-with-different-precisions/32787#32787 — Mariusz Iwaniuk, Jan 08 '18 at 07:28

score 1 · Answer 1 · answered Jan 08 '18 at 07:50

This is a problem of precision with large numbers.

nint = 1/Beta[-1 + 38., 160.]*
          NIntegrate[(E^(-.6 *x))^(-1 + 38) (1 - E^(-.6*x))^(-1 + 
                      160) *.6, {x, 2, 4}]

(*    0.999858    *)

int = 1/Beta[-1 + 38, 160]* Integrate[
            Rationalize[(E^(-.6 *x))^(-1 + 38) (1 - E^(-.6*x))^(-1 + 160) *.6, 
            0], {x, 2, 4}]

(*  A very large output ...   *)

N[int]

(*    -1.35051*10^66    *)


Block[{$MaxExtraPrecision = 500}, N[int, 10]]

(*    0.9998583622    *)

Integration of nonnegative function

1 Answers1