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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.}]

?

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3
  • $\begingroup$ I am getting very large values for functions with reasonable plot $\endgroup$
    – user54562
    Commented Jan 8, 2018 at 7:10
  • $\begingroup$ 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] $\endgroup$
    – Mariusz Iwaniuk
    Commented Jan 8, 2018 at 7:22
  • 1
    $\begingroup$ See:mathematica.stackexchange.com/questions/32782/… $\endgroup$
    – Mariusz Iwaniuk
    Commented Jan 8, 2018 at 7:28

1 Answer 1

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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    *)
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