08-24-2019, 03:34 PM

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:

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

(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