Difficulty: Intermediate | Prerequisites: Sections 2.1–3.4 (all derivative rules, algebra with exponents).
Integration reverses differentiation. If the derivative tells you how fast something is changing, the integral recovers the total quantity from that rate of change. This is the payoff for all the derivative work: you can go from a marginal cost function back to the total cost, compute the total increase in cost over a production range, and find the average value of a function. The Fundamental Theorem of Calculus connects the two halves of calculus into a single computation.
An antiderivative F(x) is a function whose derivative is f(x). Indefinite integrals give the general antiderivative plus a constant C. Definite integrals compute F(b) − F(a) and give a number representing accumulated change. The average value of f on [a, b] is (1/(b − a)) times the definite integral.
Antiderivative
A function F(x) such that F′(x) = f(x). The reverse of differentiation.
Indefinite integral
∫ f(x) dx = F(x) + C. The family of all antiderivatives of f, differing by a constant.
Constant of integration (C)
Because the derivative of any constant is zero, an antiderivative always includes an arbitrary constant C. A particular value of C is found when an initial condition is given.
Definite integral
∫ from a to b of f(x) dx = F(b) − F(a). A number (not a function) representing the net accumulated value of f from a to b.
Fundamental Theorem of Calculus (FTC)
If f is continuous on [a, b] and F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) − F(a).
In simple terms: to evaluate a definite integral, find the antiderivative, plug in the top bound, plug in the bottom bound, and subtract.
Differential equation
An equation involving a derivative, such as dy/dx = f(x). Solving it means finding y as a function of x. A general solution has +C; a particular solution uses an initial condition to determine C.
Average value of a function
(1/(b − a)) ∫ from a to b of f(x) dx. The height of a rectangle that has the same area as the region under f on [a, b].
Power Rule: ∫ x^n dx = x^(n+1)/(n+1) + C, where n ≠ −1.
Reciprocal: ∫ (1/x) dx = ln|x| + C, where x ≠ 0.
Exponential: ∫ e^x dx = e^x + C.
Properties: For constant k:
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
∫ k f(x) dx = k ∫ f(x) dx.
Algebra first, then integrate. Rewrite roots, fractions and products into power-rule form before integrating.
Worked examples:
∫ (3x² − 5x + 2) dx = x³ − (5/2)x² + 2x + C.
∫ 4/x⁷ dx = ∫ 4x^(−7) dx = 4 · x^(−6)/(−6) + C = −2/(3x⁶) + C.
∫ (x⁴ − 3x⁷)/x² dx. Split: ∫ (x² − 3x⁵) dx = x³/3 − x⁶/2 + C.
∫ [³√(x²) + 1/⁴√(x³)] dx = ∫ [x^(2/3) + x^(−3/4)] dx = (3/5)x^(5/3) + 4x^(1/4) + C.
∫ (x⁸e^x − x⁷)/x⁸ dx = ∫ (e^x − 1/x) dx = e^x − ln|x| + C. Note: x⁸e^x is one term, so x⁸ cancels leaving e^x.
Finding a particular antiderivative: Integrate to get the general form F(x) + C, then use the given condition to solve for C.
Example: f′(x) = 3/x² + 9/x − 1, f(1) = 1. Rewrite: f′(x) = 3x^(−2) + 9/x − 1. Integrate: f(x) = −3/x + 9 ln|x| − x + C. Use f(1) = 1: −3 + 0 − 1 + C = 1, so C = 5. Answer: f(x) = −3/x + 9 ln|x| − x + 5.
From marginal cost to cost function: If C′(x) = 4x² − 5x and fixed cost is $3000, integrate C′(x) to get C(x) = (4/3)x³ − (5/2)x² + K. Use C(0) = 3000: K = 3000.
Differential equations: An equation like dy/dx = f(x) is solved by bringing dx to the right side and integrating both sides.
General solution: the answer includes +C.
Particular solution: an initial condition y(a) = b is used to find C.
Example: dy/dx = √x, y(1) = 2. Integrate: y = (2/3)x^(3/2) + C. Use y(1) = 2: 2/3 + C = 2, so C = 4/3. Answer: y = (2/3)x^(3/2) + 4/3.
Your instructor awards full marks only when you show the proper technique: separate, integrate both sides, then solve for C.
If f is continuous on [a, b] and F is any antiderivative of f, then:
∫ from a to b of f(x) dx = F(x) | from a to b = F(b) − F(a).
The vertical bar notation F(x)|_a^b means "antiderivative evaluated at b minus antiderivative evaluated at a." It is a straight line, not a limit symbol. No +C is needed for definite integrals.
Worked examples:
∫ from 0 to 3 of 4e^x dx = 4e^x |_0^3 = 4e³ − 4e⁰ = 4e³ − 4.
∫ from 1 to 3 of 5/√x dx = ∫ 5x^(−1/2) dx = 10x^(1/2) |_1^3 = 10√3 − 10.
∫ from −1 to 4 of (4x³ + 9x² − 6x) dx = [x⁴ + 3x³ − 3x²] |_{−1}^4 = (256 + 192 − 48) − (1 − 3 − 3) = 400 − (−5) = 405.
∫ from e to e² of 1/x dx = ln|x| |_e^{e²} = ln(e²) − ln(e) = 2 − 1 = 1.
∫ from 4 to 4 of (5x⁴ − 12)⁹ dx = 0 (same limits).
Application: cost increase. C′(x) = 500 − x/3 for 0 ≤ x ≤ 900. The increase in cost from producing 300 to 900 bikes per month is:
∫ from 300 to 900 of (500 − x/3) dx = [500x − x²/6] |_{300}^{900}.
Average value of a function:
(1/(b − a)) ∫ from a to b of f(x) dx.
Example: f(x) = 8x² − 3x on [−2, 3]. Average = (1/5) ∫ from −2 to 3 of (8x² − 3x) dx. Integrate: (1/5)[8x³/3 − 3x²/2] |_{−2}^3. Evaluate at each bound and subtract.
Integration rules:
∫ x^n dx = x^(n+1)/(n+1) + C (n ≠ −1)
∫ 1/x dx = ln|x| + C
∫ e^x dx = e^x + C
∫ k f(x) dx = k ∫ f(x) dx
∫ [f ± g] dx = ∫ f dx ± ∫ g dx
FTC: ∫_a^b f(x) dx = F(b) − F(a)
Average value: (1/(b − a)) ∫_a^b f(x) dx
Students forget the +C on indefinite integrals. Every indefinite integral must include it.
Students try to integrate rational functions or products directly without rewriting. Always split fractions and expand products into power-rule form first.
Students confuse the definite integral bar notation with the limit symbol. The bar is a straight vertical line, not a limit.
Students add +C to definite integrals. Definite integrals evaluate to a number; C cancels in the subtraction.
⚠️ Solve differential equations using proper technique: bring dx over, integrate both sides. Show this process for full credit.
⚠️ Average value of a function problems are common exam questions.
⚠️ Applications like "compute the increase in cost from production level a to b" require setting up and evaluating a definite integral.
⚠️ Always rewrite the integrand using algebra before integrating.
Fill in the blank: ∫ x⁴ dx = ________.
True or false: ∫ 1/x dx = ln(x) + C.
Fill in the blank: ∫ from a to a of f(x) dx = ________.
True or false: A definite integral needs +C.
Fill in the blank: The average value formula is ________.
Answers: 1. x⁵/5 + C. 2. False (it is ln|x| + C, with absolute value). 3. Zero. 4. False. 5. (1/(b − a)) ∫_a^b f(x) dx.
Q: Evaluate ∫ (3x² − 5x + 2) dx.
A: x³ − (5/2)x² + 2x + C.
Q: A company’s marginal cost is C′(x) = 4x² − 5x with fixed cost $3000. Find C(x).
A: C(x) = (4/3)x³ − (5/2)x² + 3000.
Q: Evaluate ∫ from 0 to 3 of 4e^x dx.
A: 4e^x |_0^3 = 4e³ − 4.
Q: Evaluate ∫ from e to e² of (1/x) dx.
A: ln|x| |_e^{e²} = ln(e²) − ln(e) = 2 − 1 = 1.
Q: Find the average value of f(x) = 8x² − 3x on [−2, 3].
A: (1/5) ∫ from −2 to 3 of (8x² − 3x) dx = (1/5)[(8x³/3 − 3x²/2)]_{−2}^3 = (1/5)[(72 − 27/2) − (−64/3 − 6)] = compute to get 97/6 ≈ 16.17.
Integration is the inverse of differentiation (Sections 2.4–3.4). The definite integral computes area, total accumulated change, and connects directly to the marginal analysis from Section 2.7. U-substitution (if covered later) extends these rules to handle composite functions.
antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, FTC, integration, power rule for integrals, constant of integration, differential equation, general solution, particular solution, initial condition, average value of a function, marginal cost, area under curve, MAC 2233, business calculus, FIU, Sections 5.1, 5.5