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[SOLVED] Integrate a function between a and b

Hi!
I want to integrate a function between the lower boundary "a" and the upper boundary "b".
I have got:
header.h: inline double integrate(double(*f)(double), double a, double b) { double sum = 0; double n = 20; double result = (ba)/n; for (double k = 1.; k < n1; k+=1.) sum += f(a + k*result); return result * ((f(a) + f(b)) / 2 + sum); } /// and in mainwindow.cpp (in the void function): double cp = lineEdit>text().toDouble(&ok); double x; double a=.025; double b=.075; double function=1(cp*(1x)); double integrateresult = integrate(function, a, b);
But it does not integrate the function...
Does someone see the mistake?
Please share!
Thanks in advance!

integrate()
expects a pointer to function and you are giving it a double called "function". Naming a variable "function" does not make it a function :)In this code "function" variable is not a function. It's a variable of type double and undefined value (undefined because you never gave value to x so the result is garbage).

Hello @ChrisKawa !
Thanks for pointing out some problems.
I have done some improvements and the following code compiles without any problem:
in the headerfile:inline double f(double x) { return x*x; }
in mainwindow.cpp in the function which is emitted with a QPushButton:
double aB = .1; double bB = 2; int n = 10000; double h = (bBaB)/n; double area = (f(aB)+f(bB))/2; for (int i=1;i<n;i++) { area += f(aB+i*h); } QString resultString = ""; lineEdit>setText(resultString.setNum(area));
Furthermore, it even calculates the "area" and gives the result "14033.3". This is unrealistically high and incorrect!
Mathematica gives me for the integration of the same equation:
integral_0.1^2 x^2 dx = 2.66633
which is correct!If someone has a helpful idea, please share!

Solved!
area = area*h;
was missing after the forloop!
Now the result is: 2.66633
the same is with Mathematica: 2.66633
(For mathfriends: at least approximately, if not "too" many digits are compared; Nevertheless setting "n" from "n=10000" to a higher accuracy, the area will be approximated more correctly!)Thanks anyway! :)

@newe12 said:
double h = (bBaB)/n;
double area = (f(aB)+f(bB))/2;What is h for?
And why are you starting the area with the average value of the two boundary points?Shouldn't it be more like:
area += f(interval*i)
where i goes from boundary to boundary and interval is .1?