let A be non empty closed_interval Subset of REAL; :: thesis: ( A = [.0,((PI * 3) / 2).] implies integral ((- sin),A) = - 1 )
assume A = [.0,((PI * 3) / 2).] ; :: thesis: integral ((- sin),A) = - 1
then ( upper_bound A = (PI * 3) / 2 & lower_bound A = 0 ) by Th37;
then integral ((- sin),A) = 0 - (cos . 0) by Th46, SIN_COS:76
.= 0 - (sin . ((PI / 2) - 0)) by SIN_COS:78
.= 0 - 1 by SIN_COS:76 ;
hence integral ((- sin),A) = - 1 ; :: thesis: verum