Sunday, July 10, 2022

Calculus: Scaled Integration

Calculus:  Scaled Integration


Introduction


The take the integral:


∫( f(x) dx, A, B) 


The integral can be transformed to new limits C and D via linear transformation:



∫( f(x) dx, A, B) → ∫( g(y) dy, C, D)


Where the interval [ C, D ] has the smaller range than [ A, B ].


Set the transformation to:


y = m*x + β


C = m*A + β

D = m*B + β


Solving for m, β:


(C - D) = (A - B) * m

m = (C - D) / (A - B)


and


β = C - A * m = D - B * m


Solving y = m*x + β for x:


x = 1/m * (y - β)


Taking the derivative of both sides:


dx = 1/m  dy


The transformed integral:


∫( f(x) dx, A, B) → 1/m * ∫( f(1/m * (y - β)) dy, C, D)


Examples


Example 1:  


∫(x^2 - 5 dx, 10 ,16) but scale the integration interval to [1, 2].


A = 10, B = 16, C = 1, D = 2

m = (1 - 2)/(10 - 16) = 1/6,  1/m = 6

β = 1 - 10 * 1/6 = -2/3

x = 6 * (y + 2/3) = 6 * y + 4


Transformed Integral:

6 * ∫( (6*y + 4)^2 - 5 dy, 1, 2) = 1008


Example 2:


∫( e^x * ln(x + 2) dx, 0, 5) but scale the integration to [0, 1].


A = 0, B = 5, C = 0, D = 1

m = -1/-5 = 1/5,  1/m = 5

β = 0 - 0 * 1/5 = 0

x = 5 * y


Transformed Integral:

5 * ∫( e^(5 * y) * ln(5 * y + 2) dy, 0, 1) ≈ 262.8586594


Numerical integrals were calculated with the TI-36X Pro.  


When trying the Simpson's rule or Trapezoid rule, I find the smaller range does not really give better estimates.  But I am presenting the technique and if this helps in the future, great.  


Coming up:


July 11 - July 15, 2022:  TI-58 and TI-59 Week

Next Regular Blog:  July 23, 2022


Eddie  


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