## 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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