Solving Integral Equations

[Eddie W. Shore]

The program INTEGRALSOLVE solve the following equation:
(Format: ∫( integrand dvar, lower, upper)

∫( f(t) dt, 0, x) = a

∫( f(t) dt, 0, x) - a = 0

It is assumed that x>0.

We can use the Second Theorem of Calculus which takes the derivative of the integral:

d/dx ∫( f(t) dt, a, x) = f(x)

We don't have to worry about lower limit a at all for the theorem to work.

∫( f(t) dt, 0, x) - a

Take the derivative with respect to x on both sides (d/dx):

= d/dx ∫( f(t) dt, 0, x) - a

= d/dx ∫( f(t) dt, 0, x) - d/dx a

Let F(t) be the anti-derivative of f(t):

= d/dx (F(x) - F(0)) - 0

= d/dx F(x) - d/dx F(0)

F(0) is a constant.

= f(x)


Newton's Method to find the roots of f(x) can be found by the iteration:

x_(n+1) = x_n - f(x_n) / f'(x_n)

Applying that to find the roots of ∫( f(t) dt, 0, x) - a:

x_(n+1) = x_n - (∫( f(t) dt, 0, x_n) - a) / f(x_n)

Program:

Code:

EXPORT INTEGRALSOLVE(f,a,x)
BEGIN
// f(X) as a string, area, guess
// ∫(f(X) dX,0,x) = a
// EWS 2019-07-26
// uses Function app
LOCAL x1,x2,s,i,w;
F0:=f;
s:=0;
x1:=x;
WHILE s==0 DO
i:=AREA(F0,0,x1)-a;
w:=F0(x1);
x2:=x1-i/w;
IF ABS(x1-x2)<1ᴇ−12 THEN
s:=1;
ELSE
x1:=x2;
END;
END;

RETURN approx(x2);
END;

Example 1:

∫( 2*t^3 dt, 0, x) = 16
Guess = 2
Root ≈ 2.3784

Example 2:

∫( sin^2 t dt, 0, x) = 1.4897
Guess = 1
(Radians Mode)
Root ≈ 2.4999

Source:

Green, Larry. "The Second Fundamental Theorem of Calculus" Differential Calculus for Engineering and other Hard Sciences. Lake Tahoe Community College.
http://www.ltcconline.net/greenl/courses...ECFUND.HTM
Retrieved July 25, 2019

Blog entry: http://edspi31415.blogspot.com/2019/08/h...lving.html