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Common Integration Techniques

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Why This Matters

Integration is the backbone of applied calculus in business and management—it's how you move from rates of change back to total quantities. When you see a marginal cost function, integration gives you total cost. When you're given a rate of revenue growth, integration tells you accumulated revenue. Every technique you learn here is a tool for answering the question: "What's the total?"

You're being tested on your ability to recognize which technique fits which integral and execute it correctly under time pressure. The exam won't just ask you to integrate—it'll give you integrals that look tricky until you spot the right approach. Don't just memorize formulas; know when each method applies and why it works. Master the pattern recognition, and you've got this.


Foundational Rules: Your Starting Point

Every integration problem begins here. These rules handle the simplest cases directly and form the building blocks for every other technique. Think of these as your default settings—try them first before reaching for fancier methods.

Power Rule

  • xndx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C for n1n \neq -1—this handles any polynomial term and most root expressions when rewritten as fractional exponents
  • Works for negative and fractional exponents too; rewrite 1x2\frac{1}{x^2} as x2x^{-2} and x\sqrt{x} as x1/2x^{1/2} before applying
  • The "+C" is non-negotiable—indefinite integrals always include the constant of integration; forgetting it costs points

Constant Multiple Rule

  • kf(x)dx=kf(x)dx\int k \cdot f(x) \, dx = k \int f(x) \, dx—pull constants outside the integral to simplify before integrating
  • Combine with the sum/difference rule to break apart complex expressions term by term
  • Essential for management applications where coefficients represent prices, rates, or quantities (e.g., integrating 5x35x^3 for a cost function)

Compare: Power Rule vs. Constant Multiple Rule—both simplify integrals, but the power rule handles the variable structure while the constant multiple rule handles scalar coefficients. On exams, apply constant multiple first, then power rule to each term.


Substitution Methods: Undoing the Chain Rule

When you see a composite function—a function inside another function—substitution is your go-to technique. The core idea: replace a complicated inner expression with a single variable uu, integrate the simpler form, then substitute back.

U-Substitution

  • Let u=g(x)u = g(x) where g(x)g(x) is the "inner function"—then du=g(x)dxdu = g'(x) \, dx, and you rewrite the entire integral in terms of uu
  • Look for a function and its derivative appearing together; the derivative (or a constant multiple of it) must be present for substitution to work cleanly
  • Common in business applications like integrating 3x2ex3dx\int 3x^2 e^{x^3} dx where u=x3u = x^3 transforms this into eudu\int e^u \, du

Trigonometric Substitution

  • Used for integrals with a2x2\sqrt{a^2 - x^2}, a2+x2\sqrt{a^2 + x^2}, or x2a2\sqrt{x^2 - a^2}—substitute x=asinθx = a\sin\theta, x=atanθx = a\tan\theta, or x=asecθx = a\sec\theta respectively
  • Leverages Pythagorean identities to eliminate square roots; less common in management calculus but appears in optimization problems
  • Always draw a reference triangle to convert back to xx after integrating in terms of θ\theta

Compare: U-Substitution vs. Trigonometric Substitution—both involve changing variables, but u-sub handles composite functions while trig sub specifically targets radical expressions with quadratics. For MAC2233, u-substitution appears far more frequently.


Integration by Parts: Products of Different Function Types

When you're integrating a product of two different types of functions—like xexx \cdot e^x or xln(x)x \cdot \ln(x)—substitution won't help. Integration by parts reverses the product rule for derivatives.

Integration by Parts Formula

  • udv=uvvdu\int u \, dv = uv - \int v \, du—choose uu as the function that simplifies when differentiated, and dvdv as the function that's easy to integrate
  • Use LIATE to choose uu: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—pick uu from earlier in this list
  • May require multiple applications—integrals like x2exdx\int x^2 e^x \, dx need parts applied twice; watch for patterns that cycle back

Compare: U-Substitution vs. Integration by Parts—substitution works when you see a function composed with another (chain rule reverse), while parts works when you see functions multiplied together (product rule reverse). If substitution fails on a product, try parts.


Special Function Integrals: Exponentials and Logarithms

These functions appear constantly in management contexts—growth models, depreciation, compound interest, elasticity. Memorize these forms; they're tested directly.

Exponential Functions

  • exdx=ex+C\int e^x \, dx = e^x + C—the only function that is its own antiderivative; for ekxdx\int e^{kx} \, dx, the answer is 1kekx+C\frac{1}{k}e^{kx} + C
  • axdx=axln(a)+C\int a^x \, dx = \frac{a^x}{\ln(a)} + C for any base a>0a > 0, a1a \neq 1—this handles growth/decay at non-natural rates
  • Management applications: continuous compounding (erte^{rt}), population growth, radioactive decay, and continuous income streams

Logarithmic Functions

  • ln(x)dx=xln(x)x+C\int \ln(x) \, dx = x\ln(x) - x + C—derived using integration by parts with u=ln(x)u = \ln(x) and dv=dxdv = dx
  • 1xdx=lnx+C\int \frac{1}{x} \, dx = \ln|x| + C—this is the "missing case" from the power rule where n=1n = -1
  • Appears in economic models involving logarithmic utility functions and percentage growth calculations

Compare: exdx\int e^x \, dx vs. ln(x)dx\int \ln(x) \, dx—exponentials integrate directly with no change in form, while logarithms require integration by parts. Know both cold; they're exam staples.


Rational Functions: Breaking Down Fractions

When the integrand is a ratio of polynomials, you need algebraic manipulation before integrating. The strategy depends on comparing the degrees of numerator and denominator.

Long Division for Rational Functions

  • Apply when degree of numerator ≥ degree of denominator—divide first to get a polynomial plus a proper fraction
  • Simplifies the integral into manageable pieces: integrate the polynomial directly, then handle the remaining fraction
  • Example: x3+2x+1dx\int \frac{x^3 + 2}{x + 1} \, dx requires division before any other technique applies

Partial Fraction Decomposition

  • Breaks complex fractions into simpler terms—works when the denominator factors and the numerator's degree is smaller
  • Express as sum of fractions with unknown coefficients: 1(x1)(x+2)=Ax1+Bx+2\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}, then solve for AA and BB
  • Each resulting fraction integrates to a logarithm or simple power; essential for demand/supply equilibrium problems

Compare: Long Division vs. Partial Fractions—use long division first if the numerator's degree is too high, then apply partial fractions to the remainder. They're sequential tools, not alternatives.


The Fundamental Theorem: Connecting Everything

This theorem is the conceptual heart of calculus—it links differentiation and integration as inverse operations and makes definite integrals computable.

Fundamental Theorem of Calculus

  • If F(x)=f(x)F'(x) = f(x), then abf(x)dx=F(b)F(a)\int_a^b f(x) \, dx = F(b) - F(a)—find any antiderivative, evaluate at bounds, subtract
  • Gives net accumulated quantity between two points; in management, this means total cost, total revenue, or total profit over an interval
  • No "+C" needed for definite integrals—the constant cancels when you subtract F(a)F(a) from F(b)F(b)

Trigonometric Functions

  • sin(x)dx=cos(x)+C\int \sin(x) \, dx = -\cos(x) + C and cos(x)dx=sin(x)+C\int \cos(x) \, dx = \sin(x) + C—note the sign change for sine
  • Less common in management calculus but may appear in seasonal business models or cyclical demand patterns
  • For complex trig integrals, use identities like sin2(x)=1cos(2x)2\sin^2(x) = \frac{1 - \cos(2x)}{2} to simplify before integrating

Compare: Indefinite vs. Definite Integrals—indefinite integrals give a family of functions (include +C), while definite integrals give a number (no +C). FRQs often ask you to set up the indefinite form, then evaluate as definite.


Quick Reference Table

TechniqueWhen to UseKey Formula/Pattern
Power RulePolynomial terms, roots as fractional exponentsxn+1n+1+C\frac{x^{n+1}}{n+1} + C
U-SubstitutionComposite functions; function + its derivative presentLet u=g(x)u = g(x), replace dxdx with dudu
Integration by PartsProducts of different function typesudv=uvvdu\int u \, dv = uv - \int v \, du
Exponential Integrationexe^x or axa^x termsex+Ce^x + C or axlna+C\frac{a^x}{\ln a} + C
Logarithm Integrationln(x)\ln(x) or 1x\frac{1}{x}xln(x)x+Cx\ln(x) - x + C or $$\ln
Long DivisionRational function, numerator degree ≥ denominatorDivide first, then integrate pieces
Partial FractionsFactorable denominator, proper fractionDecompose, then integrate each term
Fundamental TheoremDefinite integrals, total quantitiesF(b)F(a)F(b) - F(a)

Self-Check Questions

  1. You see 4x(x2+5)3dx\int 4x(x^2 + 5)^3 \, dx. Which technique applies, and what would you choose for uu?

  2. Compare the integrals xexdx\int x e^x \, dx and xex2dx\int x e^{x^2} \, dx. Why does one require integration by parts while the other uses u-substitution?

  3. What's the first step when integrating x2+3x+7x+2dx\int \frac{x^2 + 3x + 7}{x + 2} \, dx, and why?

  4. If a marginal cost function is MC(x)=6x24x+10MC(x) = 6x^2 - 4x + 10, write the integral that gives total cost from producing 0 to 100 units. What theorem justifies evaluating this?

  5. You need to integrate 3x(x1)dx\int \frac{3}{x(x-1)} \, dx. Explain how partial fraction decomposition transforms this into two simpler integrals.