let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,(PI / 2).] implies integral (sin - cos ),A = 0 )
assume A = [.0 ,(PI / 2).] ; :: thesis: integral (sin - cos ),A = 0
then ( upper_bound A = PI / 2 & lower_bound A = 0 ) by Th37;
then integral (sin - cos ),A = (((- cos ) . (PI / 2)) - ((- cos ) . 0 )) - ((sin . (PI / 2)) - (sin . 0 )) by Th78
.= ((- (cos . (PI / 2))) - ((- cos ) . 0 )) - ((sin . (PI / 2)) - (sin . 0 )) by VALUED_1:8
.= ((- 0 ) - (- (cos . 0 ))) - (1 - (sin . 0 )) by SIN_COS:81, VALUED_1:8
.= (- (- (cos . 0 ))) - (1 - (sin . (0 + (2 * PI )))) by SIN_COS:83
.= (- (- (cos . (0 + (2 * PI ))))) - (1 - 0 ) by SIN_COS:81, SIN_COS:83
.= 0 by SIN_COS:81 ;
hence integral (sin - cos ),A = 0 ; :: thesis: verum