let A be non empty closed_interval Subset of REAL; for f being PartFunc of REAL,REAL
for Z being open Subset of REAL st A c= Z & Z = dom f & f = (exp_R * sin) (#) cos holds
integral (f,A) = ((exp_R * sin) . (upper_bound A)) - ((exp_R * sin) . (lower_bound A))
let f be PartFunc of REAL,REAL; for Z being open Subset of REAL st A c= Z & Z = dom f & f = (exp_R * sin) (#) cos holds
integral (f,A) = ((exp_R * sin) . (upper_bound A)) - ((exp_R * sin) . (lower_bound A))
let Z be open Subset of REAL; ( A c= Z & Z = dom f & f = (exp_R * sin) (#) cos implies integral (f,A) = ((exp_R * sin) . (upper_bound A)) - ((exp_R * sin) . (lower_bound A)) )
assume A1:
( A c= Z & Z = dom f & f = (exp_R * sin) (#) cos )
; integral (f,A) = ((exp_R * sin) . (upper_bound A)) - ((exp_R * sin) . (lower_bound A))
then
Z = (dom (exp_R * sin)) /\ (dom cos)
by VALUED_1:def 4;
then A2:
Z c= dom (exp_R * sin)
by XBOOLE_1:18;
then A3:
exp_R * sin is_differentiable_on Z
by FDIFF_7:37;
cos is_differentiable_on Z
by FDIFF_1:26, SIN_COS:67;
then
f | Z is continuous
by A1, A3, FDIFF_1:21, FDIFF_1:25;
then
f | A is continuous
by A1, FCONT_1:16;
then A4:
( f is_integrable_on A & f | A is bounded )
by A1, INTEGRA5:10, INTEGRA5:11;
A5:
for x being Real st x in Z holds
f . x = (exp_R . (sin . x)) * (cos . x)
A7:
for x being Element of REAL st x in dom ((exp_R * sin) `| Z) holds
((exp_R * sin) `| Z) . x = f . x
dom ((exp_R * sin) `| Z) = dom f
by A1, A3, FDIFF_1:def 7;
then
(exp_R * sin) `| Z = f
by A7, PARTFUN1:5;
hence
integral (f,A) = ((exp_R * sin) . (upper_bound A)) - ((exp_R * sin) . (lower_bound A))
by A1, A2, A4, FDIFF_7:37, INTEGRA5:13; verum