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Homework task
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#include <stdio.h> | |
#include <stdlib.h> | |
#include <math.h> | |
#define F(x) (f(x)-g(x)) | |
#define F1(x) (f1(x)-g1(x)) | |
typedef double (*func)(double); | |
double root(func f, func g, func f1, func g1, double a, double b, double eps1) | |
{ | |
int right = F1(a) * (F1(b)-F1(a)) > 0; | |
if(right) | |
eps1 = -eps1; | |
double c = right ? a : b; | |
while(F(c)*F(c+eps1) > 0) | |
c = c - F(c)/F1(c); | |
return c; | |
} | |
double integral(func f, func g, double a, double b, double eps2) | |
{ | |
int N = 1; | |
double oldI, I = -1.0; | |
do | |
{ | |
oldI = I; | |
I = F(a) + F(b); | |
double h = (b-a) / N; | |
int i; | |
for(i = 1; i < N; i++) | |
if(i % 2) | |
I += 2 * F(a+i*h); | |
else | |
I += 4 * F(a+i*h); | |
I *= h/3; | |
N *= 2; | |
} | |
while((N == 1) || (fabs(oldI - I) >= eps2 * 15)); | |
return I; | |
} | |
double ff(double x) | |
{ | |
return 0.35*x*x - 0.95*x + 2.7; | |
} | |
double ff1(double x) | |
{ | |
return 0.35*2*x - 0.95; | |
} | |
double gg(double x) | |
{ | |
return pow(3, x) + 1; | |
} | |
double gg1(double x) | |
{ | |
return log(3) * pow(3, x); | |
} | |
double hh(double x) | |
{ | |
return 1/(x+2); | |
} | |
double hh1(double x) | |
{ | |
return -1/(x+2)/(x+2); | |
} | |
int main() | |
{ | |
double | |
x1 = root(hh, ff, hh1, ff1, -1.95, -1.5, 0.000001), // about -1.8 | |
x2 = root(gg, hh, gg1, hh1, -1.75, -1.0, 0.000001), // about -1.2 | |
x3 = root(ff, gg, ff1, gg1, -2.00, 1.00, 0.000001); // about 0.32 | |
printf("%f %f %f\n", x1, x2, x3); | |
double res = | |
integral(ff, hh, x1, x2, 0.000001) + | |
integral(ff, gg, x2, x3, 0.000001); // about 3.94 | |
printf("%f\n", res); | |
return 0; | |
} |
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Results:
-1.821137 -1.209384 0.324613
3.948069