let A be non empty 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 . (upper_bound A)) - (arctan . (lower_bound 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 . (upper_bound A)) - (arctan . (lower_bound 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 . (upper_bound A)) - (arctan . (lower_bound 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 . (upper_bound A)) - (arctan . (lower_bound A))
A6: arctan is_differentiable_on Z by A2, SIN_COS9:81;
A7: for x being Element of REAL st x in dom (arctan `| Z) holds
(arctan `| Z) . x = f . x

proof

let x be Element of 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 7;
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 7;
then A9: arctan `| Z = f by A7, PARTFUN1:5;
( f is_integrable_on A & f | A is bounded ) by A1, A4, A5, INTEGRA5:10, INTEGRA5:11;
hence integral (f,A) = (arctan . (upper_bound A)) - (arctan . (lower_bound A)) by A1, A2, A9, INTEGRA5:13, SIN_COS9:81; :: thesis: verum