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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. The area under a curve betw
#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
Arithmetic Pipeline The complex arithmetic operations like floating point and multiplication operations use a lot of the time of the ALU. These operations can also be pipelined
Add a byte number from one memory location to a byte from subsequent memory location and put sum in the third memory location. Also save carry flag in the least significant bit of
What is guard bits? Guard bits are extra bits which are produced while the intermediate steps to yield maximum accuracy in the final results.
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