Difficulty: Intermediate | Prerequisites: All differentiation rules (Sections 2.4–3.4)
Integration is the reverse of differentiation. Where a derivative answers "how fast is this changing?", an integral answers "how much has accumulated?" In business calculus, this means recovering a total cost from a marginal cost rate, or finding the total change in a quantity over a time interval. Definite integrals give a single number (the net accumulation between two bounds), while indefinite integrals give a family of functions (the general antiderivative).
An indefinite integral reverses differentiation: add one to the exponent, divide by the new exponent, and include + C. A definite integral evaluates the antiderivative at the upper bound minus the lower bound. The average value of a function on [a, b] is (1/(b–a)) times the definite integral from a to b. Applied problems often give you a rate (like C'(t)) and ask you to recover the original function using an initial condition.
Antiderivative
A function F(x) whose derivative is f(x). That is, F'(x) = f(x). Think of it as "undoing" differentiation.
Indefinite integral
∫ f(x) dx = F(x) + C, where F is any antiderivative of f and C is the constant of integration. The + C is required because many functions share the same derivative.
Definite integral
∫ from a to b of f(x) dx = F(b) – F(a). A single number representing the net accumulation of f(x) from x = a to x = b.
Fundamental Theorem of Calculus
Connects differentiation and integration: if F'(x) = f(x), then ∫ from a to b of f(x) dx = F(b) – F(a).
Average value of a function
(1/(b–a)) ∫ from a to b of f(x) dx. The single y-value that, if constant over [a, b], would give the same total area under the curve.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It accounts for the fact that infinitely many functions have the same derivative. Determined by an initial condition when one is given.
∫ x^n dx = x^(n+1)/(n+1) + C, provided n ≠ –1
∫ (1/x) dx = ln|x| + C
∫ e^x dx = e^x + C
∫ k dx = kx + C (constant)
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx
∫ k · f(x) dx = k ∫ f(x) dx
If ∫ from a to a of f(x) dx = 0 (same limits)
Rewrite everything in the form x^n before integrating. Add 1 to the exponent, divide by the new exponent, attach + C.
∫ (2x⁵ + 7x³ – 10e^x) dx = x⁶/3 + 7x⁴/4 – 10e^x + C.
∫ (3x⁴ – 5x⁷)/x⁵ dx. Simplify first: 3x⁻¹ – 5x². Integrate: 3ln|x| – 5x³/3 + C.
∫ (⁶√(x⁵) – 2/√x + 5/x³) dx. Rewrite: x^(5/6) – 2x^(–1/2) + 5x^(–3). Integrate: (6/11)x^(11/6) – 4x^(1/2) – (5/2)x^(–2) + C.
The key step is always the rewrite. Split fractions, convert radicals and reciprocals to power form, then apply the power rule term by term.
Find the antiderivative, then evaluate at the upper bound minus the lower bound. No + C needed.
∫ from –5 to 2 of 7 dx = 7x evaluated from –5 to 2 = 7(2) – 7(–5) = 14 + 35 = 49.
∫ from 2 to 4 of (x+1)(3x–4) dx. Expand: 3x² – x – 4. Antiderivative: x³ – x²/2 – 4x. Evaluate: [64 – 8 – 16] – [8 – 2 – 8] = 40 – (–2) = 42.
∫ from 55 to 55 of (4x³ – ³√(πx) – e^(πex)) dx = 0. Same upper and lower limits always give zero.
Average value = (1/(b–a)) ∫ from a to b of f(x) dx.
Example: f(x) = 2x⁴ – 3/x over [1, 5].
∫ from 1 to 5 of (2x⁴ – 3/x) dx = [2x⁵/5 – 3ln|x|] from 1 to 5.
At x = 5: 2(3125)/5 – 3ln(5) = 1250 – 3ln(5).
At x = 1: 2/5 – 0 = 0.4.
Integral = 1249.6 – 3ln(5) ≈ 1249.6 – 4.828 = 1244.77.
Average value = 1244.77 / 4 ≈ 311.19.
Given C'(t) = –2√t – 3 and C(0) = 850 (cost per unit at t = 0 months).
Integrate: C(t) = ∫ (–2t^(1/2) – 3) dt = –2 · (2/3)t^(3/2) – 3t + C = –(4/3)t^(3/2) – 3t + C.
Use the initial condition: C(0) = 0 – 0 + C = 850. So C = 850.
Cost function: C(t) = –(4/3)t^(3/2) – 3t + 850.
End of 4th year = 48 months: C(48) = –(4/3)(48)^(3/2) – 3(48) + 850 = –(4/3)(332.55) – 144 + 850 ≈ –443.4 – 144 + 850 = $262.60 per unit.
Students forget + C on indefinite integrals. Every indefinite integral must include it. Definite integrals do not.
Students try to integrate without simplifying first. Always split fractions, expand products, and rewrite roots as powers before applying the power rule.
Students apply the power rule to ∫ (1/x) dx incorrectly by writing x⁰/0. The power rule does not work when n = –1; the integral of x^(–1) is ln|x| + C.
Students confuse the initial condition with the constant of integration. Use the given condition (e.g. C(0) = 850) to solve for the constant after integrating.
⚠️ "Use proper integral notation" means write the integral sign, dx, and (for indefinite integrals) + C throughout your work.
⚠️ Expect at least one integral that requires simplification before you can integrate.
⚠️ The integral from a to a is always 0. This is a quick freebie if it appears.
⚠️ Applied problems give a rate function and an initial condition. Integrate, solve for C, then answer the specific question.
Fill in the blank: ∫ x⁴ dx = __________. (x⁵/5 + C)
True or False: ∫ from 3 to 3 of (x² + 7) dx = 0. (True. Same limits.)
Fill in the blank: ∫ (1/x) dx = __________. (ln|x| + C)
True or False: Definite integrals need + C. (False.)
Fill in the blank: The average value of f on [a, b] equals (1/(b–a)) times __________. (∫ from a to b of f(x) dx)
Q: Evaluate ∫ (2x⁵ + 7x³ – 10e^x) dx.
A: x⁶/3 + 7x⁴/4 – 10e^x + C.
Q: Evaluate ∫ from –5 to 2 of 7 dx.
A: 7x from –5 to 2 = 14 – (–35) = 49.
Q: C'(t) = –2√t – 3 and C(0) = 850. Find C(t) and C(48).
A: C(t) = –(4/3)t^(3/2) – 3t + 850. C(48) ≈ $262.60.
Q: Find the average value of f(x) = 2x⁴ – 3/x on [1, 5].
A: Average = (1/4)[∫ from 1 to 5 of (2x⁴ – 3/x) dx] = (1/4)(1250 – 3ln5 – 0.4) ≈ 311.19.
Integration reverses differentiation. Every derivative rule you learnt has a corresponding integration pattern. In more advanced courses, integration is used to find areas between curves, consumer and producer surplus, and present value of income streams. The connection between C'(t) and C(t) mirrors the relationship between marginal cost and total cost from the derivative chapters.
integration, integral, antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, power rule for integrals, constant of integration, +C, average value, net change, rate of change, initial condition, cost function recovery, MAC 2233, business calculus, FIU