reserve i,n,m for Nat,
        r,s for Real,
        A for non empty closed_interval Subset of REAL;

theorem Th6:
  integral( ((-1)|^i) (#) ( #Z (2*n) / ( #Z 0 + #Z 2)),A)
   = ((-1)|^i) * (((1/(2*n+1))*((upper_bound A) |^ (2*n+1)))
   - ((1/(2*n+1))*((lower_bound A) |^ (2*n+1))))
   + integral((-1)|^(i+1) (#) ( #Z (2*(n+1)) / ( #Z 0 + #Z 2)),A)
proof
  set II =(-1)|^i,i1=i+1,n1=n+1,II1=(-1)|^i1;
  set Z0= #Z 0,Z2=#Z 2,Z2n=#Z (2*n);
  set f=II (#) Z2n;
  set g = II1 (#) ( #Z (2*n1) / ( Z0 + Z2));
A1:dom (Z2n / ( Z0 + Z2)) = dom (II (#) ( Z2n / ( Z0 + Z2)))
    by VALUED_1:def 5;
  dom ( #Z (2*n1) / ( Z0 + Z2)) = dom (II1 (#) ( #Z (2*n1) / ( Z0 + Z2)))
    by VALUED_1:def 5;
  then
A2:dom g=REAL by Th4;
A3:dom f=REAL by FUNCT_2:def 1;
  then
A4:dom ( f+g )=REAL/\REAL by A2, VALUED_1:def 1;
  for x be Element of REAL holds
    (II (#) ( Z2n / ( Z0 + Z2))).x = (f+g).x
  proof
    let x be Element of REAL;
    x|^(2*n+1+1) = x* x|^(2*n+1) by NEWTON:6
                .=x* (x*(x|^(2*n))) by NEWTON:6;
    then x|^(2*n) + x|^(2*n1) =
    x|^(2*n) *(1 + x^2);
    then
A5:  ( x|^(2*n) + x|^(2*n1))/(1+x^2) = x|^(2*n) by XCMPLX_1:89;
A6: - II*((x|^(2*n1))/(1+x^2)) = ((-1) * II)*((x|^(2*n1))/(1+x^2))
                              .= II1 * ((x|^(2*n1))/(1+x^2)) by NEWTON:6;
    x|^(2*n) = x #Z (2*n) by PREPOWER:36
            .= Z2n.x by TAYLOR_1:def 1;
    then
A7: II*(x|^(2*n)) = f.x by VALUED_1:6;
A8: II1*((x|^(2*n1))/(1+x^2)) = II1 * ( #Z (2*n1) / ( Z0 + Z2)).x by Th4
    .= g.x by VALUED_1:6;
    thus (II (#) ( Z2n / ( Z0 + Z2))).x
       = II * ( Z2n / ( Z0 + Z2)).x by VALUED_1:6
      .= II * ((x|^(2*n))/(1+x^2)) by Th4
      .= II * (x|^(2*n) - (x|^(2*n1))/(1+x^2)) by A5
      .= II * (x|^(2*n)) + II1*((x|^(2*n1))/(1+x^2)) by A6
      .= (f+g).x by A7,A8,A4,VALUED_1:def 1;
  end;
  then
A9:f+g = II (#) ( Z2n / ( Z0 + Z2)) by A1,Th4,A4;
A10:dom Z2n=REAL by FUNCT_2:def 1;
  f | A is continuous & Z2n|A is continuous;
  then
A11: f is_integrable_on A & f|A is bounded &
  Z2n is_integrable_on A & Z2n|A is bounded by A10,INTEGRA5:10,11,A3;
  #Z (2*n1) / ( Z0 + Z2) is continuous by Th4;
  then g | A is continuous;
  then
A12: g is_integrable_on A & g|A is bounded by INTEGRA5:10,11,A2;
A13:2*n in NAT by ORDINAL1:def 12;
  thus integral(II (#) ( Z2n / ( Z0 + Z2)),A) = integral(f,A) + integral(g,A)
       by A3,A2,A11,A12,A9,INTEGRA6:11
     .= II*integral(Z2n,A) + integral(g,A) by A10,A11,INTEGRA6:9
     .= II*(((1/(2*n+1))*((upper_bound A) |^ (2*n+1))) - ((1/(2*n+1)  )*
        ((lower_bound A) |^ (2*n+1)))) + integral(g,A)
         by INTEGRA9:19,A13;
end;
