let A be non empty 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:76, SIN_COS:78
.= ((- (- (cos . (0 + (2 * PI))))) + 1) - 0 by SIN_COS:76, SIN_COS:78
.= 2 by SIN_COS:76 ;
hence integral ((sin + cos),A) = 2 ; :: thesis: verum