reserve r,p,x for Real;
reserve n for Element of NAT;
reserve A for non empty closed_interval Subset of REAL;
reserve Z for open Subset of REAL;
reserve a,b,x for Real;
reserve n for Element of NAT;
reserve A for non empty closed_interval Subset of REAL;
reserve f,f1,f2 for PartFunc of REAL,REAL;
reserve Z for open Subset of REAL;

theorem
  A c= Z & (for x st x in Z holds f.x = a-x & f.x > 0) & dom (-(ln*f)) =
Z & dom (-(ln*f)) = dom f2 & (for x st x in Z holds f2.x = 1/(a-x) ) & f2|A is
  continuous implies integral(f2,A) = (-(ln*f)).(upper_bound A)-(-(ln*f)).(
  lower_bound A)
proof
  assume that
A1: A c= Z and
A2: for x st x in Z holds f.x = a-x & f.x > 0 and
A3: dom (-(ln*f)) = Z and
A4: dom (-(ln*f)) = dom f2 and
A5: for x st x in Z holds f2.x = 1/(a-x) and
A6: f2|A is continuous;
A7: f2 is_integrable_on A by A1,A3,A4,A6,INTEGRA5:11;
A8: (-(ln*f)) is_differentiable_on Z by A2,A3,FDIFF_4:3;
A9: for x being Element of REAL st x in dom ((-(ln*f))`|Z)
holds ((-(ln*f))`|Z).x = f2.x
  proof
    let x be Element of REAL;
    assume x in dom ((-(ln*f))`|Z);
    then
A10: x in Z by A8,FDIFF_1:def 7;
    then ((-(ln*f))`|Z).x = 1/(a-x) by A2,A3,FDIFF_4:3
      .= f2.x by A5,A10;
    hence thesis;
  end;
  dom ((-(ln*f))`|Z) = dom f2 by A3,A4,A8,FDIFF_1:def 7;
  then ((-(ln*f))`|Z) = f2 by A9,PARTFUN1:5;
  hence thesis by A1,A3,A4,A6,A7,A8,INTEGRA5:10,13;
end;
