let A be closed-interval Subset of REAL ; :: thesis: for f being PartFunc of REAL ,REAL
for Z being open Subset of REAL st A c= Z & f = sin / exp_R & Z c= dom ((- (1 / 2)) (#) ((sin + cos ) / exp_R )) & Z = dom f & f | A is continuous holds
integral f,A = (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (upper_bound A)) - (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (lower_bound A))
let f be PartFunc of REAL ,REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & f = sin / exp_R & Z c= dom ((- (1 / 2)) (#) ((sin + cos ) / exp_R )) & Z = dom f & f | A is continuous holds
integral f,A = (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (upper_bound A)) - (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( 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)) )
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 ) ; :: thesis: integral f,A = (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (upper_bound A)) - (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (lower_bound A))
then A2: ( f is_integrable_on A & f | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: (- (1 / 2)) (#) ((sin + cos ) / exp_R ) is_differentiable_on Z by A1, Th1;
B1: for x being Real st x in Z holds
f . x = (sin . x) / (exp_R . x) by RFUNCT_1:def 4, A1;
A4: for x being 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 Real; :: thesis: ( x in dom (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) implies (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) . x = f . x )
assume x in dom (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) ; :: thesis: (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) . x = f . x
then A5: x in Z by A3, FDIFF_1:def 8;
then (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) . x = (sin . x) / (exp_R . x) by A1, Th1
.= f . x by A5, B1 ;
hence (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) . x = f . x ; :: thesis: verum

end;

dom (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z) = dom f by A1, A3, FDIFF_1:def 8;
then ((- (1 / 2)) (#) ((sin + cos ) / exp_R )) `| Z = f by A4, PARTFUN1:34;
hence integral f,A = (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (upper_bound A)) - (((- (1 / 2)) (#) ((sin + cos ) / exp_R )) . (lower_bound A)) by A1, A2, Th1, INTEGRA5:13; :: thesis: verum