Toady Posted April 2, 2007 Share Posted April 2, 2007 (edited) expandcollapse popup;======================================================= ; Function: _AreaUnderCurve ; Author: Toady ; ; Uses Trapezoidal rule to give move accurate results ; Action: Performs an evaluated calculus integral over ; a given range on a single variable expression ; Read NOTES below before using ; ; $expr: String - Correctly formatted math expression ; $var: String - Independent variable ; $lower_limit: Float - starting location ; $upper_limit: Float - ending location ; $numIntvs: Integer - Number of trapezoids under curve ; - Higher number = more accurate ; $Places: Integer - Number of places to round result ; Returns: Area in float ; ; NOTE: upper_limit must be greator than lower_limit ; NOTE: Does not take into account integration x-intercepts ; Therefore must integrate on those intercepts 1 by 1 ; NOTE: A higher $numIntvls is more accurate but too high will ; take longer to calculate ; ; Ex 1) Calculate Pi ; 4*_AreaUnderCurve("Sqrt(1-((x)^2))","x",0,1,10000) ; Ex 2) Quadatic polynomial 2(x)^2 + 7x + 3 ; _AreaUnderCurve("2*(x)^2 + 7*x + 3","x",0,1,10000) ; Ex 3) Logorithm ; _AreaUnderCurve("Log(x)","x",0,1,10000) ;======================================================= Func _AreaUnderCurve($expr,$var,$lower_limit,$upper_limit,$numIntvs,$Places = 4) If Not StringInStr($expr,$var) Then Return "Variable not found in expression" ElseIf Not IsInt($numIntvs) Or $numIntvs < 0 Then return "Interval is < 0 or not an integer" EndIf Local $dx = ($upper_limit-$lower_limit)/$numIntvs Local $area, $x, $y = 0 If $upper_limit < $lower_limit Then return "upper limit must be greater than lower limit" ElseIf $lower_limit >= 0 Or $Upper_Limit <= 0 Then $y = Execute(StringReplace($expr,$lower_limit,"$x")) For $i = 1 To $numIntvs - 1 $x = ($i*$dx) $y += 2*Execute(StringReplace($expr,$var,"$x")) Next $y += Execute(StringReplace($expr,$upper_limit,"$x")) return Round(Abs($dx*$y)/2,$Places) ElseIf $lower_limit < 0 And $upper_limit >= 0 Then Local $y_Neg, $y_Pos = 0 $y_Neg = Execute(StringReplace($expr,0,"$x")) For $i = 1 To $numIntvs - 1 $x = ($i*((Abs($lower_limit))/$numIntvs)) $y_Neg += 2*Execute(StringReplace($expr,$var,"$x")) Next $y_Neg += Execute(StringReplace($expr,$lower_limit,"$x")) $y_Pos = Execute(StringReplace($expr,0,"$x")) For $i = 1 To $numIntvs - 1 $x = ($i*(($upper_limit)/$numIntvs)) $y_Pos += 2*Execute(StringReplace($expr,$var,"$x")) Next $y_Pos = Execute(StringReplace($expr,$upper_limit,"$x")) return Round(Abs(($y_Pos+$y_Neg)*$dx)/2,$Places) EndIf EndFunc ;_AreaUnderCurve()==>Had to write this script for one of my courses in college. It performs an integration of an expression over a specified interval and returnsthe area under the curve. Calculate Pi: _AreaUnderCurve("4*Sqrt(1-((x)^2))","x",0,1,10000)Polynomial: _AreaUnderCurve("2*(x)^2 + 7*x + 3","x",0,1,10000)Logorithm: _AreaUnderCurve("Log(x)","x",0,1,10000) Edited April 4, 2007 by Toady www.itoady.com A* (A-star) Searching Algorithm - A.I. Artificial Intelligence bot path finding Link to comment Share on other sites More sharing options...

jvanegmond Posted April 2, 2007 Share Posted April 2, 2007 Good job. I really had no idea how to do this programming-wise.. github.com/jvanegmond Link to comment Share on other sites More sharing options...

tmo Posted April 2, 2007 Share Posted April 2, 2007 Good job. I really had no idea how to do this programming-wise.. just use the definition of an integral. like he did Link to comment Share on other sites More sharing options...

jvanegmond Posted April 2, 2007 Share Posted April 2, 2007 (edited) just use the definition of an integral. like he did I would never add values when you take tiny steps in a formula, lol. I'd just calculate it the way everyone does it. Edited April 2, 2007 by Manadar github.com/jvanegmond Link to comment Share on other sites More sharing options...

Toady Posted April 2, 2007 Author Share Posted April 2, 2007 I would never add values when you take tiny steps in a formula, lol. I'd just calculate it the way everyone does it.Why not? rofl www.itoady.com A* (A-star) Searching Algorithm - A.I. Artificial Intelligence bot path finding Link to comment Share on other sites More sharing options...

jvanegmond Posted April 3, 2007 Share Posted April 3, 2007 Why not? rofl:|Ok... Integral for f(x) = x between 0 and 1 .. We all know this is 1. Now, what you do is this:Result = 0Take a tiny step.. Get the value of x there (0.00001)Result = 0.00001Take a tiny step.. Get the value of x there (0.00001)Result = 0.00002Take a tiny step.. Get the value of x there (0.00001)Result = 0.00003Take a tiny step.. Get the value of x there (0.00001)Result = 0.00004Take a tiny step.. Get the value of x there (0.00001)Result = 0.00005Take a tiny step.. Get the value of x there (0.00001)Result = 0.00006Take a tiny step.. Get the value of x there (0.00001)Result = 0.00007Take a tiny step.. Get the value of x there (0.00001)Result = 0.00008Take a tiny step.. Get the value of x there (0.00001)Do I have to continue? github.com/jvanegmond Link to comment Share on other sites More sharing options...

Toady Posted April 3, 2007 Author Share Posted April 3, 2007 Yes I know I was joking. This function takes advantage of CPU power not integration rules as iin calculus. I just found programming this function fun.. Not the most efficient way to calculate PI though, to get PI accuratly to 5 decimal places it takes my PC 38 seconds lol. But still, neat concept.. www.itoady.com A* (A-star) Searching Algorithm - A.I. Artificial Intelligence bot path finding Link to comment Share on other sites More sharing options...

tmo Posted April 3, 2007 Share Posted April 3, 2007 actually the integral for f(x) = x between 0 and 1 is 0.5 Link to comment Share on other sites More sharing options...

Toady Posted April 3, 2007 Author Share Posted April 3, 2007 actually the integral for f(x) = x between 0 and 1 is 0.5 I see you been using the function. lol jk www.itoady.com A* (A-star) Searching Algorithm - A.I. Artificial Intelligence bot path finding Link to comment Share on other sites More sharing options...

jvanegmond Posted April 3, 2007 Share Posted April 3, 2007 actually the integral for f(x) = x between 0 and 1 is 0.5 Lol! I can't believe I had that wrong... :"> I know it is 0.5 ... github.com/jvanegmond Link to comment Share on other sites More sharing options...

Toady Posted April 17, 2007 Author Share Posted April 17, 2007 Is this useful? or do people not do this type of math? www.itoady.com A* (A-star) Searching Algorithm - A.I. Artificial Intelligence bot path finding Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now