let A be closed-interval Subset of REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & Z = dom (cos + sin ) & (cos + sin ) | A is continuous holds
integral (cos + sin ),A = ((sin - cos ) . (upper_bound A)) - ((sin - cos ) . (lower_bound A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & Z = dom (cos + sin ) & (cos + sin ) | A is continuous implies integral (cos + sin ),A = ((sin - cos ) . (upper_bound A)) - ((sin - cos ) . (lower_bound A)) )
assume A1: ( A c= Z & Z = dom (cos + sin ) & (cos + sin ) | A is continuous ) ; :: thesis: integral (cos + sin ),A = ((sin - cos ) . (upper_bound A)) - ((sin - cos ) . (lower_bound A))
then A2: ( cos + sin is_integrable_on A & (cos + sin ) | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
Z = (dom cos ) /\ (dom sin ) by A1, VALUED_1:def 1;
then AB: Z c= dom (sin - cos ) by VALUED_1:12;
then A3: sin - cos is_differentiable_on Z by FDIFF_7:39;
A4: for x being Real st x in dom ((sin - cos ) `| Z) holds
((sin - cos ) `| Z) . x = (cos + sin ) . x

proof

let x be Real; :: thesis: ( x in dom ((sin - cos ) `| Z) implies ((sin - cos ) `| Z) . x = (cos + sin ) . x )
assume x in dom ((sin - cos ) `| Z) ; :: thesis: ((sin - cos ) `| Z) . x = (cos + sin ) . x
then A5: x in Z by A3, FDIFF_1:def 8;
then ((sin - cos ) `| Z) . x = (cos . x) + (sin . x) by AB, FDIFF_7:39
.= (cos + sin ) . x by A1, A5, VALUED_1:def 1 ;
hence ((sin - cos ) `| Z) . x = (cos + sin ) . x ; :: thesis: verum

end;

dom ((sin - cos ) `| Z) = dom (cos + sin ) by A1, A3, FDIFF_1:def 8;
then (sin - cos ) `| Z = cos + sin by A4, PARTFUN1:34;
hence integral (cos + sin ),A = ((sin - cos ) . (upper_bound A)) - ((sin - cos ) . (lower_bound A)) by A1, A2, AB, FDIFF_7:39, INTEGRA5:13; :: thesis: verum