The integration problem is often impossible to be handled by analytical methods (studied in mathematical analysis or calculus). In such cases we have to rely on approximating methods. Monte Carlo methods are some of the several approaches which may be used to approximate the values of the definite integrals. Moreover there are several ways that the Monte Carlo methods may be applied to solve the definite integration problem. In this problem we will show one of the basic applications of Monte Carlo methods and we will apply it in the approximation of various integrals. The key idea in our approach is to construct an enclosing rectangle for the required area and to randomly generate points inside that rectangle. Finally we approximate the required area by checking the ratio of the generated points which are inside it.

It is known that the area under the curve of some continuous function (as shown in the following figure) is expressed by the definite integral:

The key idea in our approach is to construct an enclosing rectangle for the required area and to randomly generate points inside that rectangle. Finally we approximate the required area by checking the percentage of the generated points which are inside it, so we can express it by the formula:

Here S is denoting the required area (under the curve) depicted in grey in the above figures, 0 is the area of the enclosing rectangle and p is a value between 0 and 1 representing the ratio of the generated points which fall inside the S area (grey zone in the figure) divided by the total number of points. For example in the figure above we would have = 19/27 0.7037

The function whose definite integral will be calculated, will be given in hardcoded way in the form of a function (but notice that our basic method works only for the cases when the values of the function in the integration interval are always non-negative):

In this problem you are required to write a parallel program which will estimate the value of a definite integral. The program will create several parallel units and each of them will generate random points inside the enclosing rectangle. Finally the definite integral will be estimated as:

0

Here, N is the total number of generated random points by all threads and N0 is the number of

random points which are located below the curve.

Write a program using POSIX threads which receives as command line parameters the total number of random points (N) that will be generated, the number of threads that will be used, the bounds of the definite integral, an upper bound on the values of the function in this interval, and will estimate the value of the definite integral using the Monte Carlo method described above. A sample execution may be:

 ./MonteCarloIntegralEstimator 2000000 10 0 1.5 2.5 

Write a program using OpenMP which receives as command line parameters the total number of random points (N) that will be generated, the number of threads that will be used, the bounds of the definite integral, an upper bound on the values of the function in this interval, and will estimate the value of the definite integral using the Monte Carlo method described above. A sample execution may be:

 ./MonteCarloIntegralEstimator 2000000 10 0 1.5 2.5 

Write a program using MPI which reads as input from the user (only once) the total number of random points (N) that will be generated, the bounds of the definite integral, an upper bound on the values of the function in this interval, and will estimate the value of the definite integral using the Monte Carlo method described above. The number of processes will be provided to the mpiexec command. A sample execution may be:

 mpiexec -n 10 ./MonteCarloIntegralEstimator 

An example that may be used for testing purposes:

() = 2, from 1 to 3, as an upper bound may use 10 (any value greater than 9 is fine), and the expected result is 8.66666 . (In this case the enclosing rectangle is determined by the lines x = 1, x = 3, y = 0, y = 10, so its area is 20. If we generate 100000 points and 43345 points are below the curve, then the approximation of the integral for this case is:

43345 20 = 8.669 100000

double f(double x){ return ...; //may fill it according to your preferences

}