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.
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)
HP Prime Program INTEGRALSOLVE
Enter f(X) as a string as it will be stored in Function App variable F0. Use X as the independent variable.
// f(X) as a string, area, guess
// ∫(f(X) dX,0,x) = a
// EWS 2019-07-26
// uses Function app
WHILE s==0 DO
TI NSpire CX CAS Program INTEGRALSOLVE
(Caution: This program needs to be typed in)
Use t as the independent variable.
Define LibPub integralsolve(f,a,x)=
:© f(x), area, guess: ∫(f(t) dt,0,x = a)
: If abs(x2-x1)≤1E−12 Then
∫( 2*t^3 dt, 0, x) = 16
Guess = 2
Root ≈ 2.3784
∫( sin^2 t dt, 0, x) = 1.4897
Guess = 1
Root ≈ 2.4999
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/105/Antiderivatives/SECFUND.HTM Retrieved July 25, 2019
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