The program evaluates the integral of three functions: y=x, y=x² and y=x³. After running the program you should see the following printed out in the Sim UART1 window:

Code Analysis

Lines 35, 41, and 47
These three lines make calls to the integral() function. Each one passes a different function’s address to the integral() function for evaluation. The address of a function is represented by the function’s name alone (no parentheses or parameters).
y1 = integral(0, 1, justx);
y2 = integral(0, 1, xsquared);
y3 = integral(0, 1, xcubed);

Lines 60-63
This function justx() simply returns the value of x (y = x)

float justx(float x)
{
return x;
}

Lines 72-75
This function xsquared() simply returns the value of x2 (y = x2)

float xsquared(float x)
{
return (x * x);
}

Lines 84-87
This function xcubed() simply returns the value of x3 (y = x3)

float xcubed(float x)
{
return (x * x * x);
}

Lines 99-112
This is the integral() function which will evaluate the integral of any mathematical function passed to it over the range specified by the lower bound a and the upper bound b. The third parameter of the function header is a function pointer. It expects to receive the address of a function, which may be passed simply as the name of a function.

float integral(float a, float b, float (*f)(float))
{
float sum = 0.0;
float x;
int n;

//Evaluate integral{a,b} f(x) dx
for (n = 0; n <= 100; n++)
{
x = ((n / 100.0) * (b-a)) + a;
sum += (f(x) * (b-a)) / 101.0;
}
return sum;
}

The algorithm used to evaluate the integral is beyond the scope of this course. But it should be noted that the function name that is passed to this function is accessed via its parameter name f. For example, the line that states:
sum += (f(x) * (b-a)) / 101.0;
invokes the function passed via f, and passes the parameter x to it. So, when the function xsquared() is passed, then f(x) evaluates xsquared(x), where x is a local variable defined within integral() and is defined in the line immediately above.