Would you write a program for trapezoidal rule in c language?

Answer 1

yes. /* trapezoid.c */ #include #include float f(float); float a; float b; float x; float h; float sum; int n; int i; int main() { printf("Enter value for a: "); scanf("%f", &a); printf("Enter value for b: "); scanf("%f", &b); printf("Enter number of rectangles: "); scanf("%d", &n); h = (b - a) / n; sum = (0.5 * h) * (f(a) + f(b)); printf("%f\n", sum); for (i = 1; i < n; i++) { sum = sum + h * f(a + (i * h)); printf("%f\n", sum); } printf("The value of the integral is: %f\n", sum); } float f(float x) { float value; /* define function here */ value = x*x + 3; return value; }

Answer 2

// Source: Deepanshu Bhatti Solutions

#include<iostream>

#include<cmath>

using namespace std;

float f (float x)

{

return pow (x,3) + pow (x,2) - (4*x) - 5;

}

int main()

{

float a, b, d, i, n;

float I=0, J=0, A, K=0, E=0;

cout << " Given: f(x) = x^3 + 2x^2 - 4x - 5" << endl;

cout << "Enter lower limit: ";

cin >> a;

cout << "Enter upper Limit: ";

cin >> b;

cout << "Enter the number of intervals: ";

cin >> n;

d = (b - a) / n;

for (i=0; i<=n; ++i)

I = I + f (a + (i * d));

for (i=1; i<n; ++i)

J = J + f (a + (i * d));

A = (d / 2) * (I + J);

cout << "Definite integral: " << A << endl;

for (i=0; i<=n; ++i)

K = K + a + (i * d);

E = -((b - a) * d * d * ((6 * (K / n) + 4) / 12));

cout << "Total error: " << E << endl;

}

Answer 3

#include

#include

using namespace std;

float f (float x)

{

return pow (x,3) + pow (x,2) - (4*x) - 5;

}

int main()

{

float a, b, d, i, n;

float I=0, J=0, A, K=0, E=0;

cout << " Given: f(x) = x^3 + 2x^2 - 4x - 5" << endl;

cout << "Enter lower limit: ";

cin >> a;

cout << "Enter upper Limit: ";

cin >> b;

cout << "Enter the number of intervals: ";

cin >> n;

d = (b - a) / n;

for (i=0; i<=n; ++i)

I = I + f (a + (i * d));

for (i=1; i

J = J + f (a + (i * d));

A = (d / 2) * (I + J);

cout << "Definite integral: " << A << endl;

for (i=0; i<=n; ++i)

K = K + a + (i * d);

E = -((b - a) * d * d * ((6 * (K / n) + 4) / 12));

cout << "Total error: " << E << endl;

}

// Source: Deepanshu Bhatti Solutions