let A be closed-interval Subset of REAL ; :: thesis: ( A = [.0 ,PI .] implies integral (sin + cos ),A = 2 )
assume A = [.0 ,PI .] ; :: thesis: integral (sin + cos ),A = 2
then ( upper_bound A = PI & lower_bound A = 0 ) by Th37;
then integral (sin + cos ),A = ((((- cos ) . PI ) - ((- cos ) . 0 )) + (sin . PI )) - (sin . 0 ) by Th69
.= (((- (cos . PI )) - ((- cos ) . 0 )) + (sin . PI )) - (sin . 0 ) by VALUED_1:8
.= (((- (cos . PI )) - (- (cos . 0 ))) + (sin . PI )) - (sin . 0 ) by VALUED_1:8
.= ((- (- (cos . 0 ))) + 1) - (sin . (0 + (2 * PI ))) by SIN_COS:81, SIN_COS:83
.= ((- (- (cos . (0 + (2 * PI ))))) + 1) - 0 by SIN_COS:81, SIN_COS:83
.= 2 by SIN_COS:81 ;
hence integral (sin + cos ),A = 2 ; :: thesis: verum