theorem
  A = [.0,PI*2.] implies integral(-sin,A) = 0
proof
  assume A=[.0,PI*2.];
  then upper_bound A=PI*2 & lower_bound A=0 by Th37;
  then integral(-sin,A) = 1 - cos.0 by Th46,SIN_COS:76
    .= 1 - sin.(PI/2 - 0) by SIN_COS:78
    .= 1 - 1 by SIN_COS:76;
  hence thesis;
end;
