let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,(PI / 2).] implies integral (- sin ),A = - 1 )
assume A = [.0 ,(PI / 2).] ; :: thesis: integral (- sin ),A = - 1
then ( upper_bound A = PI / 2 & lower_bound A = 0 ) by Th37;
then integral (- sin ),A = 0 - (cos . 0 ) by Th46, SIN_COS:81
.= 0 - (sin . ((PI / 2) - 0 )) by SIN_COS:83
.= 0 - 1 by SIN_COS:81 ;
hence integral (- sin ),A = - 1 ; :: thesis: verum