petyaGavrikov

Verified answer

#include <stdio.h>
#include <math.h>

int main(void) {
    const double pi=3.14159265359;
    double c, d, h, i1=0, i2=0;
    int n1=10, n2=100;
    printf("c = ");
    scanf("%lf",&c);
    printf("d = ");
    scanf("%lf",&d);
    h=(d-c)/n1;
    for (int i=0; i<n1; i++) i1 += pow(atan(c+i*h),2);
    i1 *= h;
    printf("I1 = %.4f\n", i1);
    
    h=(pi-0)/n2;
    for (int i=0; i<n2; i++) i2 += exp(10*(i*h));
    i2 *= h;
    printf("I2 = %.1f\n", i2);
    return 0;
}

Пример:
c = 2
d = 5
I1 = 4.8002
I2 = 3747660340620.0