Вычисление определенного интеграла по функции методом прямоугольников MPI

#include "stdafx.h"
#include "math.h"
#include "mpi.h"
int _tmain(int argc, char* argv[]) {
int numbIntervals = (int)((b-a)/IntervalWidth);
double IntervalHeight = 0.0;
double x = 0.0;
int Interval = 0;
double summaLocal = 0.0;
double summaGlobal;
int ProcNumb;
int ProcRank;
int namelen;
double t1, t2;
double IntervalWidth = 0.00003;
double a = Pi;
printf("a = %.16f\n", a);
double b = Pi*3/2;
printf ("b = %.16f\n", b);
MPI_Init (&argc &argv);
MPI_Comm_size(MPI_COMM_WORLD,&ProcNumb);
MPI_Comm_rank(MPI_COMM_WORLD &ProcRank);
printf ("Process %d of %d running on %s\n", ProcRank, ProcNumb );
if ( ProcRank == 0 ) {
printf ("numbIntervals = %i\n", numbIntervals);
t1 = MPI_Wtime();
}
for ( Interval=1+ProcRank; Interval <= numbIntervals; Interval +=ProcNumb ) {
x = a + (IntervalWidth * ((double)Interval-0.5));
IntervalHeight += (1.5*x*x+x)/(x*x*x+2);
}
summaLocal = IntervalWidth * IntervalHeight;
printf ("summaLocal is approximately %.16f\n", summaLocal);
MPI_Reduce (&summaLocal, &summaGlobal, 1, MPI_DOUBLE, MPI_SUM, 0,MPI_COMM_WORLD );
if ( ProcRank == 0 ) {
t2 = MPI_Wtime();
printf ("Integral is approximately %.16f\n", summaGlobal);
printf ("Time is %f seconds \n", t2-t1);
}
MPI_Finalize();
return 0;
}