 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=sin/exp_R
 & Z c= dom ((-1/2)(#)((sin+cos)/exp_R)) & Z = dom f & f|A is continuous
 implies integral(f,A)=((-1/2)(#)((sin+cos)/exp_R)).(upper_bound A)
                      -((-1/2)(#)((sin+cos)/exp_R)).(lower_bound A)
proof
  assume
A1:A c= Z & f=sin/exp_R
 & Z c= dom ((-1/2)(#)((sin+cos)/exp_R)) & Z = dom f & f|A is continuous;
  then
A2:f is_integrable_on A & f|A is bounded by INTEGRA5:10,11;
A3:(-1/2)(#)((sin+cos)/exp_R) is_differentiable_on Z by A1,Th7;
A4:for x st x in Z holds f.x=sin.x/exp_R.x by A1,RFUNCT_1:def 1;
A5:for x being Element of REAL
st x in dom (((-1/2)(#)((sin+cos)/exp_R))`|Z) holds
  (((-1/2)(#)((sin+cos)/exp_R))`|Z).x=f.x
  proof
  let x be Element of REAL;
  assume x in dom (((-1/2)(#)((sin+cos)/exp_R))`|Z);then
A6:x in Z by A3,FDIFF_1:def 7;then
   (((-1/2)(#)((sin+cos)/exp_R))`|Z).x=sin.x/exp_R.x by A1,Th7
   .=f.x by A6,A4;
   hence thesis;
   end;
  dom (((-1/2)(#)((sin+cos)/exp_R))`|Z)=dom f by A1,A3,FDIFF_1:def 7;
  then (((-1/2)(#)((sin+cos)/exp_R))`|Z)= f by A5,PARTFUN1:5;
  hence thesis by A1,A2,Th7,INTEGRA5:13;
end;
