It was necessary to develop a program for finding the value of a specific integral using the Gauss method. Function for integration and integration interval are shown below

Information on the Gauss method was taken from here:

http://ums.physics.usu.ru/st/num_03.pdf

http://aco.ifmo.ru/el_books/numerical_methods/lectures /glava2_3.html

y = cos (x) · (x +1) ^ - 1 + 2x [1; 25]

# include & lt; iostream & gt;
#Include & lt; conio.h & gt;
#Include & lt; math.h & gt;
#Define N 3.
Using Namespace STD;
Double F (Double X)
{
  RETURN COS (X) * (1./(x+1)) +2**;
}
Double Gauss (Double A, Double B)
{
  Double xi [n] = { - 0.7745967,0,0,7745967};
  Double Ci [n] = {0.5555556,0.8888889,0.5555556};
  Double Ra = (B-A) / 2;
  Double su = (a + b) / 2;
  Double Q, S, I;
  for (int i = 0; i & lt; n; i ++)
  {
    Q = SU + RA * XI [I - 1];
    S = Ci [I - 1] * F (Q);
  }
  I = ra * s;
  Return I;
}
INT MAIN ()
{
  Double A;
  Double B;
  COUT & LT; & LT; "VVedite A" & lt; & lt; endl;
  CIN & GT; & GT; A;
  COUT & LT; & LT; "VVedite B" & lt; & lt; endl;
  CIN & GT; & GT; B;
  cout & lt; & lt; "i =" & lt; & lt; gauss (a, b);
  Return 0;
}

An invalid response is displayed on 278.025, although Wolfram and other online calculators are calculating 623.7358

Answer 1, Authority 100%

Let’s start with the fact that in the cycle

for (int i = 0; i & lt; n; i ++)
{
  Q = SU + RA * XI [I - 1];
  S = Ci [I - 1] * F (Q);
}

What do you like xi [-1] and ci [-1] – well, when i = 0 ?

Next – calculated for i = 0 and i = 1 values ​​you just throw away. Generally. Why then do they even count them? 🙂

Next to ask or enough? 🙂

p.s. Just count on this range – [1; 25] – on all three points – quiet nonsense. Get something similar on the answer, but no more. Of course, not three times the difference, but … you need to break the entire interval for smaller segments, and only then on each smaller segment to use Gauss formula.

p.p.s. Okay, the pre-Christmas sale 🙂 Keep …

# include & lt; iostream & gt;
#Include & lt; conio.h & gt;
#Include & lt; math.h & gt;
Const int n = 3;
const int n = 100;
Using Namespace STD;
Double F (Double X)
{
  RETURN COS (X) * (1./(x+1)) +2**;
}
Double Gauss (Double A, Double B)
{
  const double xi [n] = { - 0.7745967,0,0,7745967};
  const double ci [n] = {0.5555556,0.8888889,0.5555556};
  Double Ra = (B-A) / 2;
  Double su = (a + b) / 2;
  Double Q, S = 0.0;
  for (int i = 0; i & lt; n; i ++)
  {
    Q = SU + RA * XI [i];
    S + = ci [i] * f (q);
  }
  RETURN RA * S;
}
INT MAIN ()
{
  double a = 1.0;
  Double B = 25.0;
  Double S = 0.0;
  for (int i = 0; i & lt; n; ++ i)
  {
    S + = Gauss (A + I * (B - A) / N, A + (I + 1) * (B - A) / N);
  }
  COUT & LT; & LT; "I =" & lt; & lt; S & LT; & LT; Endl;
  Return 0;
}