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=n+1xn+1+C for n=−1—this handles any polynomial term and most root expressions when rewritten as fractional exponents
Works for negative and fractional exponents too; rewrite x21 as x−2 and x as x1/2 before applying
The "+C" is non-negotiable—indefinite integrals always include the constant of integration; forgetting it costs points
Constant Multiple Rule
∫k⋅f(x)dx=k∫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 5x3 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 u, integrate the simpler form, then substitute back.
U-Substitution
Let u=g(x) where g(x) is the "inner function"—then du=g′(x)dx, and you rewrite the entire integral in terms of u
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 where u=x3 transforms this into ∫eudu
Trigonometric Substitution
Used for integrals with a2−x2, a2+x2, or x2−a2—substitute x=asinθ, x=atanθ, or x=asecθ 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 x after integrating in terms of θ
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 x⋅ex or x⋅ln(x)—substitution won't help. Integration by parts reverses the product rule for derivatives.
Integration by Parts Formula
∫udv=uv−∫vdu—choose u as the function that simplifies when differentiated, and dv as the function that's easy to integrate
Use LIATE to choose u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential—pick u from earlier in this list
May require multiple applications—integrals like ∫x2exdx 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—the only function that is its own antiderivative; for ∫ekxdx, the answer is k1ekx+C
∫axdx=ln(a)ax+C for any base a>0, a=1—this handles growth/decay at non-natural rates
Management applications: continuous compounding (ert), population growth, radioactive decay, and continuous income streams
Logarithmic Functions
∫ln(x)dx=xln(x)−x+C—derived using integration by parts with u=ln(x) and dv=dx
∫x1dx=ln∣x∣+C—this is the "missing case" from the power rule where n=−1
Appears in economic models involving logarithmic utility functions and percentage growth calculations
Compare:∫exdx vs. ∫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: ∫x+1x3+2dx 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: (x−1)(x+2)1=x−1A+x+2B, then solve for A and B
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), then ∫abf(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) from F(b)
Trigonometric Functions
∫sin(x)dx=−cos(x)+C and ∫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)=21−cos(2x) 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
Technique
When to Use
Key Formula/Pattern
Power Rule
Polynomial terms, roots as fractional exponents
n+1xn+1+C
U-Substitution
Composite functions; function + its derivative present
Let u=g(x), replace dx with du
Integration by Parts
Products of different function types
∫udv=uv−∫vdu
Exponential Integration
ex or ax terms
ex+C or lnaax+C
Logarithm Integration
ln(x) or x1
xln(x)−x+C or $$\ln
Long Division
Rational function, numerator degree ≥ denominator
Divide first, then integrate pieces
Partial Fractions
Factorable denominator, proper fraction
Decompose, then integrate each term
Fundamental Theorem
Definite integrals, total quantities
F(b)−F(a)
Self-Check Questions
You see ∫4x(x2+5)3dx. Which technique applies, and what would you choose for u?
Compare the integrals ∫xexdx and ∫xex2dx. Why does one require integration by parts while the other uses u-substitution?
What's the first step when integrating ∫x+2x2+3x+7dx, and why?
If a marginal cost function is MC(x)=6x2−4x+10, write the integral that gives total cost from producing 0 to 100 units. What theorem justifies evaluating this?
You need to integrate ∫x(x−1)3dx. Explain how partial fraction decomposition transforms this into two simpler integrals.