let A be non empty closed_interval Subset of REAL; :: thesis: ( A = [.0,(PI * 2).] implies integral ((- sin),A) = 0 )
assume A = [.0,(PI * 2).] ; :: thesis: integral ((- sin),A) = 0
then ( upper_bound A = PI * 2 & lower_bound A = 0 ) by Th37;
then integral ((- sin),A) = 1 - (cos . 0) by Th46, SIN_COS:76
.= 1 - (sin . ((PI / 2) - 0)) by SIN_COS:78
.= 1 - 1 by SIN_COS:76 ;
hence integral ((- sin),A) = 0 ; :: thesis: verum