reserve f,f1,f2,g for PartFunc of REAL,REAL;
reserve A for non empty closed_interval Subset of REAL;
reserve p,r,x,x0 for Real;
reserve n for Element of NAT;
reserve Z for open Subset of REAL;

theorem
  A = [.0,PI/2.] implies integral(sin+cos,A) = 2
proof
  assume A = [.0,PI/2.];
  then upper_bound A=PI/2 & lower_bound A=0 by Th37;
  then integral(sin+cos,A) = (-cos).(PI/2) - (-cos).0 + sin.(PI/2) - sin.0 by
Th69
    .= -(cos.(PI/2)) - (-cos).0 + sin.(PI/2) - sin.0 by VALUED_1:8
    .= -0 - (-cos.0) + 1 - sin.0 by SIN_COS:76,VALUED_1:8
    .= - (-cos.0) + 1 - sin.(0+2 * PI) by SIN_COS:78
    .= - (-cos.(0+2*PI)) + 1 - 0 by SIN_COS:76,78
    .= 2 by SIN_COS:76;
  hence thesis;
end;
