let A be closed-interval Subset of REAL ; :: thesis: for Z being open Subset of REAL
for f being PartFunc of REAL ,REAL st A c= Z & Z c= ].(- 1),1.[ & ( for x being Real st x in Z holds
f . x = 1 / (1 + (x ^2 )) ) & Z = dom f & f | A is continuous holds
integral f,A = (arctan . (sup A)) - (arctan . (inf A))
let Z be open Subset of REAL ; :: thesis: for f being PartFunc of REAL ,REAL st A c= Z & Z c= ].(- 1),1.[ & ( for x being Real st x in Z holds
f . x = 1 / (1 + (x ^2 )) ) & Z = dom f & f | A is continuous holds
integral f,A = (arctan . (sup A)) - (arctan . (inf A))
let f be PartFunc of REAL ,REAL ; :: thesis: ( A c= Z & Z c= ].(- 1),1.[ & ( for x being Real st x in Z holds
f . x = 1 / (1 + (x ^2 )) ) & Z = dom f & f | A is continuous implies integral f,A = (arctan . (sup A)) - (arctan . (inf A)) )
assume that
A1: A c= Z and
A2: Z c= ].(- 1),1.[ and
A3: for x being Real st x in Z holds
f . x = 1 / (1 + (x ^2 )) and
A4: Z = dom f and
A5: f | A is continuous ; :: thesis: integral f,A = (arctan . (sup A)) - (arctan . (inf A))
A6: arctan is_differentiable_on Z by A2, SIN_COS9:81;
A7: for x being Real st x in dom (arctan `| Z) holds
(arctan `| Z) . x = f . x

proof

let x be Real; :: thesis: ( x in dom (arctan `| Z) implies (arctan `| Z) . x = f . x )
assume x in dom (arctan `| Z) ; :: thesis: (arctan `| Z) . x = f . x
then A8: x in Z by A6, FDIFF_1:def 8;
then (arctan `| Z) . x = 1 / (1 + (x ^2 )) by A2, SIN_COS9:81
.= f . x by A3, A8 ;
hence (arctan `| Z) . x = f . x ; :: thesis: verum

end;

dom (arctan `| Z) = dom f by A4, A6, FDIFF_1:def 8;
then A9: arctan `| Z = f by A7, PARTFUN1:34;
( f is_integrable_on A & f | A is bounded ) by A1, A4, A5, INTEGRA5:10, INTEGRA5:11;
hence integral f,A = (arctan . (sup A)) - (arctan . (inf A)) by A1, A2, A9, INTEGRA5:13, SIN_COS9:81; :: thesis: verum