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.] implies integral(sin-cos,A) = 2
proof
  assume A = [.0,PI.];
  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
Th78
    .= -(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)-0) by SIN_COS:76,78
    .= - (-cos.(0+2*PI)) + 1 - 0 by SIN_COS:76,78
    .= 2 by SIN_COS:76;
  hence thesis;
end;
