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HomeCalculus II: Integral Calculus - Integration, Series, and Parametric EquationsIntegration by Substitution
Study Guide680 words

Integration by Substitution

Substitution

Integration by Substitution: Study Guide

Learning Objectives

  • Identify the appropriate "inner function" uuu to simplify complex integrands.
  • Apply algebraic manipulation (constant alteration) to match the differential dududu.
  • Solve integrals requiring substitution by eliminating original variables (solving for xxx).
  • Evaluate definite integrals by accurately mapping the limits of integration from xxx-space to uuu-space.

Key Terms & Glossary

  • uuu-Substitution: A mathematical technique used to evaluate integrals by reversing the chain rule.
  • Integrand: The mathematical expression or function that is being integrated.
  • Definite Integral: An integral with a defined start and end point (limits of integration), computing net area.
  • **Differential (du)∗∗:Aninfinitesimallysmallchangeinthevariableudu)**: An infinitesimally small change in the variable udu)∗∗:Aninfinitesimallysmallchangeinthevariableu, crucial for variable transformation.

The "Big Idea"

[!IMPORTANT] uuu-Substitution is the Chain Rule in reverse. When you take the derivative of a composite function, you multiply by the derivative of the inside function. uuu-substitution seeks out that "inside function" and its derivative to un-do the process, transforming an impossible integral into a fundamental one.

Formula / Concept Box

ConceptFormula / EquationPurpose
Indefinite Integral Substitution∫f(g(x))g′(x) dx=∫f(u) du\int f(g(x))g'(x) \, dx = \int f(u) \, du∫f(g(x))g′(x)dx=∫f(u)duReplaces composite functions to find general antiderivatives.
Definite Integral Substitution∫abf(g(x))g′(x) dx=∫g(a)g(b)f(u) du\int_{a}^{b} f(g(x))g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du∫ab​f(g(x))g′(x)dx=∫g(a)g(b)​f(u)duEvaluates exact area without needing to convert back to xxx.
Change of Variablesu=g(x)  ⟹  x=g−1(u)u = g(x) \implies x = g^{-1}(u)u=g(x)⟹x=g−1(u)Used when xxx terms remain in the integrand after standard substitution.

Hierarchical Outline

  • 1. Basics of Substitution
    • Identify uuu: Usually the quantity under a root, raised to a power, or in a denominator.
    • Find dududu: Differentiate uuu with respect to xxx.
  • 2. Substitution with Alteration
    • Scaling constants: Multiply or divide dududu to match the xxx terms exactly.
    • Pulling constants out: Move constant multipliers outside the ∫\int∫ sign.
  • 3. Eliminating the Original Variable
    • **Leftover x′s∗∗:Sometimessubstitutionleavesextraxx's**: Sometimes substitution leaves extra xx′s∗∗:Sometimessubstitutionleavesextrax terms.
    • Solve for xxx: Rearrange the u=...u = ...u=... equation to find xxx in terms of uuu.
  • 4. Definite Integrals
    • Mapping limits: Evaluate u(a)u(a)u(a) and u(b)u(b)u(b) to find new integration bounds.
    • Efficiency: With new limits, you never have to substitute back to xxx.

Visual Anchors

1. The uuu-Substitution Decision Process

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2. Transforming Limits of Integration

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Definition-Example Pairs

  • uuu-Substitution
    • Definition: An integration technique replacing a function g(x)anditsderivativewithasinglevariableug(x) and its derivative with a single variable ug(x)anditsderivativewithasinglevariableu and dududu.
    • Real-World Example: Like translating a document from English to Spanish to make it easier for a Spanish speaker to summarize, then translating the summary back to English.
  • Definite Integral Bound Change
    • Definition: Recalculating the start and end points of an area calculation to match a new variable framework.
    • Real-World Example: If you switch your speedometer from miles to kilometers, you must also change your trip's start and end markers from miles to kilometers.
  • Constant Alteration
    • Definition: Multiplying or dividing the dududu equation by a constant to precisely match the remaining integrand.
    • Real-World Example: Adjusting a recipe. If a recipe calls for 2 eggs but you only have 1, you scale the entire recipe by a factor of $1/2$.

Worked Examples

▶Example 1: Substitution with Constant Alteration

Problem: Evaluate ∫x(x2+1)4 dx\int x(x^2 + 1)^4 \, dx∫x(x2+1)4dx

Step-by-Step Breakdown:

  1. Choose uuu: Let u=x2+1u = x^2 + 1u=x2+1 (the "inner" function).
  2. Find dududu: du=2x dxdu = 2x \, dxdu=2xdx.
  3. Alteration: The integrand has x dxx \, dxxdx, not 2x dx2x \, dx2xdx. So, divide by 2: 12du=x dx\frac{1}{2}du = x \, dx21​du=xdx.
  4. Substitute: ∫u4(12du)=12∫u4 du\int u^4 \left(\frac{1}{2}du\right) = \frac{1}{2}\int u^4 \, du∫u4(21​du)=21​∫u4du.
  5. Integrate: 12⋅u55+C=u510+C\frac{1}{2} \cdot \frac{u^5}{5} + C = \frac{u^5}{10} + C21​⋅5u5​+C=10u5​+C.
  6. Back-Substitute: (x2+1)510+C\frac{(x^2 + 1)^5}{10} + C10(x2+1)5​+C.
▶Example 2: Eliminating the Original Variable

Problem: Evaluate ∫xx−1 dx\int x \sqrt{x - 1} \, dx∫xx−1​dx

Step-by-Step Breakdown:

  1. Choose uuu: Let u=x−1u = x - 1u=x−1.
  2. Find dududu: du=dxdu = dxdu=dx.
  3. Handle Leftovers: We still have an xoutsidetheroot.Solveourux outside the root. Solve our uxoutsidetheroot.Solveouru equation for xxx: x=u+1x = u + 1x=u+1.
  4. Substitute Everything: ∫(u+1)u du=∫(u+1)u1/2 du\int (u + 1)\sqrt{u} \, du = \int (u + 1)u^{1/2} \, du∫(u+1)u​du=∫(u+1)u1/2du.
  5. Distribute: ∫(u3/2+u1/2) du\int (u^{3/2} + u^{1/2}) \, du∫(u3/2+u1/2)du.
  6. Integrate: 25u5/2+23u3/2+C\frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C52​u5/2+32​u3/2+C.
  7. Back-Substitute: 25(x−1)5/2+23(x−1)3/2+C\frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C52​(x−1)5/2+32​(x−1)3/2+C.
▶Example 3: Changing Definite Integral Bounds

Problem: Evaluate ∫012xex2 dx\int_{0}^{1} 2x e^{x^2} \, dx∫01​2xex2dx

Step-by-Step Breakdown:

  1. Choose uuu: Let u=x2u = x^2u=x2.
  2. Find dududu: du=2x dxdu = 2x \, dxdu=2xdx. (Perfect match!)
  3. Change Bounds:
    • Lower bound: x=0  ⟹  u=02=0x = 0 \implies u = 0^2 = 0x=0⟹u=02=0.
    • Upper bound: x=1  ⟹  u=12=1x = 1 \implies u = 1^2 = 1x=1⟹u=12=1.
  4. Substitute: ∫01eu du\int_{0}^{1} e^u \, du∫01​eudu. (Note: these are uuu-bounds now!)
  5. Integrate & Evaluate: [eu]01=e1−e0=e−1[e^u]_0^1 = e^1 - e^0 = e - 1[eu]01​=e1−e0=e−1.

Checkpoint Questions

  1. Why must we sometimes solve for xxx in terms of uuu after making our initial substitution?
  2. When altering the dududu equation, why are we only allowed to multiply or divide by constants (and not variables)?
  3. If you perform a u−substitutiononadefiniteintegralandupdateyourlimitsofintegration,doyouneedtoback−substitutexu-substitution on a definite integral and update your limits of integration, do you need to back-substitute xu−substitutiononadefiniteintegralandupdateyourlimitsofintegration,doyouneedtoback−substitutex at the very end? Why or why not?
  4. What is the fundamental difference between evaluating an indefinite integral using substitution versus a definite integral?

[!TIP] Check your answers: Try solving the Checkpoint Questions without looking back at the text. If you're stuck on question 3, review the "Definite Integrals" section in the Hierarchical Outline!

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