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

theorem
 A c= Z & f=2(#)(exp_R(#)sin)
 & Z c= dom (exp_R(#)(sin-cos)) & Z = dom f & f|A is continuous implies
 integral(f,A)=(exp_R(#)(sin-cos)).(upper_bound A)-
 (exp_R(#)(sin-cos)).(lower_bound A)
proof
  assume
A1:A c= Z & f=2(#)(exp_R(#)sin)
   & Z c= dom (exp_R(#)(sin-cos)) & Z = dom f & f|A is continuous;then
A2:f is_integrable_on A & f|A is bounded by INTEGRA5:10,11;
A3:exp_R(#)(sin-cos) is_differentiable_on Z by A1,FDIFF_7:40;
A4:for x st x in Z holds f.x=2 * exp_R.x * sin.x
   proof
   let x;
   assume x in Z;
   (2(#)(exp_R(#)sin)).x=2*(exp_R(#)sin).x by VALUED_1:6
                       .=2*(exp_R.x * sin.x) by VALUED_1:5;
   hence thesis by A1;
   end;
A5:for x being Element of REAL
st x in dom ((exp_R(#)(sin-cos))`|Z) holds
   ((exp_R(#)(sin-cos))`|Z).x=f.x
   proof
   let x be Element of REAL;
   assume x in dom ((exp_R(#)(sin-cos))`|Z);then
A6:x in Z by A3,FDIFF_1:def 7;then
   ((exp_R(#)(sin-cos))`|Z).x=2 * exp_R.x * sin.x by A1,FDIFF_7:40
   .=f.x by A4,A6;
   hence thesis;
   end;
   dom ((exp_R(#)(sin-cos))`|Z)=dom f by A1,A3,FDIFF_1:def 7;
   then ((exp_R(#)(sin-cos))`|Z)= f by A5,PARTFUN1:5;
   hence thesis by A1,A2,FDIFF_7:40,INTEGRA5:13;
end;
