let A be non empty 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 A3: Z c= dom (sin - cos) by VALUED_1:12;
then A4: sin - cos is_differentiable_on Z by FDIFF_7:39;
A5: for x being Element of REAL st x in dom ((sin - cos) `| Z) holds
((sin - cos) `| Z) . x = (cos + sin) . x

proof

let x be Element of 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 A6: x in Z by A4, FDIFF_1:def 7;
then ((sin - cos) `| Z) . x = (cos . x) + (sin . x) by A3, FDIFF_7:39
.= (cos + sin) . x by A1, A6, VALUED_1:def 1 ;
hence ((sin - cos) `| Z) . x = (cos + sin) . x ; :: thesis: verum

end;

dom ((sin - cos) `| Z) = dom (cos + sin) by A1, A4, FDIFF_1:def 7;
then (sin - cos) `| Z = cos + sin by A5, PARTFUN1:5;
hence integral ((cos + sin),A) = ((sin - cos) . (upper_bound A)) - ((sin - cos) . (lower_bound A)) by A1, A2, A3, FDIFF_7:39, INTEGRA5:13; :: thesis: verum