Difficulty: Intermediate to Advanced. Prerequisites: All differentiation material (Sections 2.1–4.6).
Integration is the reverse of differentiation. The indefinite integral finds a family of antiderivatives (plus a constant C), and the definite integral computes a net quantity between two bounds. The Fundamental Theorem of Calculus links the two: evaluate the antiderivative at the upper and lower limits and subtract. This section also covers simple differential equations and the average value of a function over an interval.
Antiderivative
A function F(x) whose derivative is f(x). In simple terms, the antiderivative answers: "What function, when differentiated, gives me f(x)?"
Indefinite integral
∫ f(x) dx = F(x) + C. The family of all antiderivatives of f(x), plus an arbitrary constant C.
Definite integral
∫ from a to b of f(x) dx = F(b) − F(a). A specific number, not a function. Geometrically, it represents the net signed area under the curve from x = a to x = 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). Think of it as: integration and differentiation undo each other.
Differential equation
An equation involving an unknown function and its derivatives. A first-order differential equation involves only the first derivative, e.g. dy/dx = f(x).
Average value of a function
(1/(b − a)) ∫ from a to b of f(x) dx. The single height that, spread across the interval, gives the same total area as the function.
Power rule for integrals: ∫ 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.
Constant multiple: ∫ k·f(x) dx = k ∫ f(x) dx.
Sum/difference: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
Rewrite roots as fractional powers: ∫ ³√(x²) dx = ∫ x^(2/3) dx.
Rewrite fractions: ∫ (x⁴ − 3x⁷)/x² dx = ∫ (x² − 3x⁵) dx.
Rewrite reciprocals: ∫ 1/⁴√(x³) dx = ∫ x^(−3/4) dx.
Split fractions with multiple terms in the numerator before integrating.
Integrate to get the general solution (with + C), then use the initial condition to solve for C.
Example: df/dx = 3/x² + 9/x − 1, f(1) = 1. Integrate: f(x) = −3/x + 9 ln|x| − x + C. Plug in x = 1: 1 = −3 + 0 − 1 + C, so C = 5.
Given C'(x) (the marginal cost), integrate to find C(x). Use C(0) = fixed cost to find C.
Example: C'(x) = 4x² − 5x, fixed cost = 3000. C(x) = (4/3)x³ − (5/2)x² + 3000.
A first-order differential equation dy/dx = g(x) is solved by integrating both sides.
General solution: y = ∫ g(x) dx (includes + C).
Particular solution: use an initial condition y(a) = b to find C.
∫ from a to b of f(x) dx = F(b) − F(a), where F is any antiderivative of f.
Use the notation F(x)|from a to b = F(b) − F(a).
A definite integral with equal limits is always zero: ∫ from a to a = 0.
Example: ∫ from 0 to 2 of 3e^x dx = 3e^x | from 0 to 2 = 3e² − 3.
Example: ∫ from e to e² of (1/x) dx = ln|x| from e to e² = ln(e²) − ln(e) = 2 − 1 = 1.
The increase in cost from producing a to b units is ∫ from a to b of C'(x) dx.
Example: C'(x) = 500 − x/3. Increase from 300 to 900: ∫ from 300 to 900 of (500 − x/3) dx.
Average value of f on [a, b] = (1/(b − a)) ∫ from a to b of f(x) dx.
Example: g(t) = 8t² − 3t on [−2, 3]. Average = (1/5) ∫ from −2 to 3 of (8t² − 3t) dt.
\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{1}{x}\,dx = \ln|x| + C\int e^x\,dx = e^x + C\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{(FTC)}\text{Average value} = \frac{1}{b-a}\int_a^b f(x)\,dxStudents forget the + C on indefinite integrals. Every indefinite integral must include it.
Students try to integrate a product by integrating each factor separately. That does not work; you must expand or use substitution.
Students confuse the power rule for integration with the power rule for differentiation. Integration adds 1 to the exponent and divides; differentiation subtracts 1 and multiplies.
Students sometimes apply the integration power rule when n = −1. That case gives ln|x|, not x⁰/0.
⚠️ Basic indefinite integrals (rewrite and apply the power rule) are tested heavily. Practise rewriting roots, reciprocals and split fractions.
⚠️ The FTC (evaluate a definite integral) is a guaranteed exam topic. Know the procedure cold.
⚠️ Finding a particular antiderivative from an initial condition (e.g. C(0) = fixed cost) is a common exam problem.
⚠️ Average value of a function over an interval is a short, self-contained problem that appears often.
Fill in the blank: ∫ x⁴ dx = ________. x⁵/5 + C.
True or False: ∫ from 4 to 4 of (5x⁴ − 12)⁹ dx = 0. True. Same upper and lower limit.
Fill in the blank: ∫ (1/x) dx = ________. ln|x| + C.
True or False: The particular antiderivative has no + C. True. The initial condition determines C, so it becomes a specific number.
Fill in the blank: The average value formula is ________. (1/(b − a)) ∫ from a to b of f(x) dx.
Q: Evaluate ∫ (3x² − 5x + 1) dx.
A: x³ − (5/2)x² + x + C.
Q: Evaluate ∫ from 0 to 2 of 3e^x dx.
A: 3e^x | from 0 to 2 = 3e² − 3 ≈ 19.17.
Q: A company has marginal cost C'(x) = 4x² − 5x and fixed cost $3000. Find C(x).
A: Integrate: C(x) = (4/3)x³ − (5/2)x² + C. C(0) = 3000, so C = 3000. C(x) = (4/3)x³ − (5/2)x² + 3000.
Q: Find the average value of g(t) = 8t² − 3t on [−2, 3].
A: (1/5) ∫ from −2 to 3 of (8t² − 3t) dt = (1/5)[(8/3)t³ − (3/2)t²] from −2 to 3 = (1/5)[(72 − 13.5) − (−64/3 − 6)] = (1/5)[58.5 + 27.33] = (1/5)(85.83) ≈ 17.17.
Q: Solve the differential equation dy/dx = √x with y(1) = 2.
A: y = ∫ x^(1/2) dx = (2/3)x^(3/2) + C. y(1) = 2: 2 = 2/3 + C, so C = 4/3. y = (2/3)x^(3/2) + 4/3.
Integration reverses differentiation (Sections 2.4–3.4). Definite integrals compute net change, which connects to the marginal analysis idea (Section 2.7): integrating the marginal cost from a to b gives the total cost increase over that production range. Average value problems are a stepping stone toward probability and statistics applications in later courses.
Integration, antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, FTC, constant of integration, power rule for integrals, ln|x|, e^x, differential equation, first-order ODE, particular solution, general solution, initial condition, average value, net change, MAC 2233, business calculus, FIU