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write a program to find the area under the curve y=f(x) between x=a and y=b integrate y=f(x) between the limits of a and b. the area under a curve between two points can be found by doing a definite integral between a to points.
#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
Multi-Threaded Processors In unit 2, we have gone through the use of distributed shared memory in parallel computer architecture. Although the use of distributed shared memory
The 2's complement of the number 1101110 is ? Ans. 1's complement of 1101110 is = 0010001 ans hence 2's complement of 1101110 is = 0010001 + 1 = 0010010.
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