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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