theorem
  A = [.0,PI*3/2.] implies integral(sin-cos,A) = 2
proof
  assume A = [.0,PI*3/2.];
  then upper_bound A=PI*3/2 & lower_bound A=0 by Th37;
  then
  integral(sin-cos,A) = (-cos).(PI*3/2) - (-cos).0 - (sin.(PI*3/2) - sin.0
  ) by Th78
    .= -(cos.(PI*3/2)) - (-cos).0 - (sin.(PI*3/2) - sin.0) by VALUED_1:8
    .= -(cos.(PI*3/2)) - (-cos.0) - (sin.(PI*3/2) - sin.0) by VALUED_1:8
    .= -(cos.(PI*3/2)) - (-cos.(0+2*PI)) - (sin.(PI*3/2) - sin.0) by SIN_COS:78
    .= -0 +1 - (-1 - 0) by SIN_COS:76,78;
  hence thesis;
end;
