let A be non empty closed_interval Subset of REAL; :: thesis: for f, f1 being PartFunc of REAL,REAL
for Z being open Subset of REAL st A c= Z & Z c= ].(- 1),1.[ & f = () / (f1 + (#Z 2)) & ( for x being Real st x in Z holds
f1 . x = 1 ) & Z = dom f & f | A is continuous holds
integral (f,A) = ((- ()) . ()) - ((- ()) . ())

let f, f1 be PartFunc of REAL,REAL; :: thesis: for Z being open Subset of REAL st A c= Z & Z c= ].(- 1),1.[ & f = () / (f1 + (#Z 2)) & ( for x being Real st x in Z holds
f1 . x = 1 ) & Z = dom f & f | A is continuous holds
integral (f,A) = ((- ()) . ()) - ((- ()) . ())

let Z be open Subset of REAL; :: thesis: ( A c= Z & Z c= ].(- 1),1.[ & f = () / (f1 + (#Z 2)) & ( for x being Real st x in Z holds
f1 . x = 1 ) & Z = dom f & f | A is continuous implies integral (f,A) = ((- ()) . ()) - ((- ()) . ()) )

assume A1: ( A c= Z & Z c= ].(- 1),1.[ & f = () / (f1 + (#Z 2)) & ( for x being Real st x in Z holds
f1 . x = 1 ) & Z = dom f & f | A is continuous ) ; :: thesis: integral (f,A) = ((- ()) . ()) - ((- ()) . ())
then A2: ( f is_integrable_on A & f | A is bounded ) by ;
Z = (dom ()) /\ ((dom (f1 + (#Z 2))) \ ((f1 + (#Z 2)) " )) by ;
then A3: ( Z c= dom () & Z c= (dom (f1 + (#Z 2))) \ ((f1 + (#Z 2)) " ) ) by XBOOLE_1:18;
then A4: Z c= dom ((f1 + (#Z 2)) ^) by RFUNCT_1:def 2;
dom ((f1 + (#Z 2)) ^) c= dom (f1 + (#Z 2)) by RFUNCT_1:1;
then A5: Z c= dom (f1 + (#Z 2)) by A4;
A6: - () is_differentiable_on Z by A1, A3, Th50;
A7: for x being Real st x in Z holds
f . x = (exp_R . ()) / (1 + (x ^2))
proof
let x be Real; :: thesis: ( x in Z implies f . x = (exp_R . ()) / (1 + (x ^2)) )
assume A8: x in Z ; :: thesis: f . x = (exp_R . ()) / (1 + (x ^2))
then (() / (f1 + (#Z 2))) . x = (() . x) / ((f1 + (#Z 2)) . x) by
.= (exp_R . ()) / ((f1 + (#Z 2)) . x) by
.= (exp_R . ()) / ((f1 . x) + ((#Z 2) . x)) by
.= (exp_R . ()) / (1 + ((#Z 2) . x)) by A1, A8
.= (exp_R . ()) / (1 + (x #Z 2)) by TAYLOR_1:def 1
.= (exp_R . ()) / (1 + (x ^2)) by FDIFF_7:1 ;
hence f . x = (exp_R . ()) / (1 + (x ^2)) by A1; :: thesis: verum
end;
A9: for x being Element of REAL st x in dom ((- ()) `| Z) holds
((- ()) `| Z) . x = f . x
proof
let x be Element of REAL ; :: thesis: ( x in dom ((- ()) `| Z) implies ((- ()) `| Z) . x = f . x )
assume x in dom ((- ()) `| Z) ; :: thesis: ((- ()) `| Z) . x = f . x
then A10: x in Z by ;
then ((- ()) `| Z) . x = (exp_R . ()) / (1 + (x ^2)) by A1, A3, Th50
.= f . x by ;
hence ((- ()) `| Z) . x = f . x ; :: thesis: verum
end;
dom ((- ()) `| Z) = dom f by ;
then (- ()) `| Z = f by ;
hence integral (f,A) = ((- ()) . ()) - ((- ()) . ()) by ; :: thesis: verum