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Write a program to find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.
#include float start_point, /* GLOBAL VARIABLES */ end_point, total_area; int numtraps; main( ) { void input(void ); float find_area(float a,float b,int n); /* prototype */ print(“AREA UNDER A CURVE”); input( ); total_area = find_area(start_point, end_point, numtraps); printf(“TOTAL AREA = %f”, total_area); } void input(void ) { printf(“\n Enter lower limit:”); scanf(“%f”, &start_point); printf(“Enter upper limit:”); scanf(“%f”, &end_point); printf(“Enter number of trapezoids:”); scanf(“%d”, &numtraps); } float find_area(float a, float b, int n) { float base , lower, h1, h2; /* LOCAL VARIABLES */ float function_x(float x); /* prototype */ float trap_area(float h1,float h2,float base );/*prototype*/ base = (b-1)/n; lower = a; for (lower =a; lower <= b-base ; lower = lower + base ) { h1 = function_x(lower); h1 = function_x(lower + base ); total_area += trap_area(h1, h2, base ); } return (total_area); float trap_area(float height_1,float height_2,float base ) { float area; /* LOCAL VARIABLE */ area = 0.5 * (height_1 + height_2) * base ; return (area); } float function_x(float x) { /* F(X) = X * X + 1 */ return (x*x + 1); } Output AREA UNDER A CURVE Enter lower limit: 0 Enter upper limit: 3 Enter number of trapezoids: 30 TOTAL AREA = 12.005000 AREA UNDER A CURVE Enter lower limit: 0 Enter upper limit: 3 Enter number of trapezoids: 100 TOTAL AREA = 12.000438
how to implement a state transition diagram?
#include using namespace std; void print(int marks_arr[],int cnt) { int ind[cnt]; int i=0; int j=0; int k=0; int s=0; for(k=0;k { ind[k]=0; } int cnt1=0; for(i=0;i
i can work
Recently social media has been flooded by fb posts, tweets, news articles about only thing demonetization.A great debate is ongoing over it. Most of the people discussing in social
Dynamic Initialization of objects: It is initializing the objects by passing the valued to the constructor from the user input or other means. Through cin operator a value
C Program for DIVISER void main() { int result,number,min; clrscr(); printf("ENTER THE NUMBER="); flushall();
to design a car that travels along the room and gives the length of the room
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