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Tony's Intertwingled Memex • Reference Manual

Substitution Rule

Formulas & Definitions
Core Definition

A technique for integration that reverses the Chain Rule. If u=g(x)u = g(x) is a differentiable function whose range is an interval, and ff is continuous on that interval:

∫f(g(x))g′(x) dx=∫f(u) du\int f(g(x)) g'(x) \, dx = \int f(u) \, du

Concept Explanation

Substitution simplifies integration by introducing a new variable uu that maps a complex composite function into a basic integration form. For definite integrals, you must transform the integration boundaries (aa and bb) to match the variable uu using the relation u=g(x)u = g(x).

Visual / Geometric Intuition

Interactive Desmos Graph

Engineering Applications

Used to analyze frequency responses in signal processing, calculate electrical charging curves in RC circuits, and scale kinetic properties in thermodynamic simulations.

Example Problem

Evaluate the definite integral:

∫0ln⁡2xex2 dx\int_{0}^{\sqrt{\ln 2}} x e^{x^2} \, dx

Solution

Let u=x2u = x^2. Then du=2x dx  ⟹  x dx=12dudu = 2x \, dx \implies x \, dx = \frac{1}{2} du.

Step 1: Translate the integration boundaries

  • When x=0  ⟹  u=02=0x = 0 \implies u = 0^2 = 0
  • When x=ln⁡2  ⟹  u=(ln⁡2)2=ln⁡2x = \sqrt{\ln 2} \implies u = (\sqrt{\ln 2})^2 = \ln 2

Step 2: Substitute and integrate

∫0ln⁡2xex2 dx=∫0ln⁡2eu(12 du)\int_{0}^{\sqrt{\ln 2}} x e^{x^2} \, dx = \int_{0}^{\ln 2} e^u \left(\frac{1}{2} \, du\right)
=12[eu]0ln⁡2= \frac{1}{2} \Big[ e^u \Big]_0^{\ln 2}
=12(eln⁡2−e0)= \frac{1}{2} (e^{\ln 2} - e^0)
=12(2−1)=12= \frac{1}{2} (2 - 1) = \frac{1}{2}

Connections & References