Difficulty: Intermediate | Prerequisites: Limits and Continuity, Derivatives and Differentiation Rules study notes.
Integration is the reverse of differentiation. Where the derivative asks "what is the rate of change?", the integral asks "given the rate of change, what is the total accumulation?" This topic closes the Calculus I loop: limits give you derivatives, derivatives give you tools for analysis, and integration brings you back to the original function (plus applications like area under a curve). The Fundamental Theorem of Calculus is the bridge connecting derivatives and integrals, and it is one of the most important results in all of mathematics.
An antiderivative reverses differentiation. The indefinite integral gives a family of functions (with + C). The definite integral gives a number (area or net accumulation). The Fundamental Theorem of Calculus links the two. U-substitution is the chain rule in reverse.
Antiderivative
A function F(x) whose derivative is f(x), i.e. F'(x) = f(x). Also called a primitive. In simple terms, it is the function you get by "undoing" differentiation.
Indefinite integral
The general antiderivative of f(x), written as the integral of f(x) dx = F(x) + C, where C is the constant of integration. Think of it as: the integral sign asks "what function, when differentiated, gives me f(x)?" The + C accounts for the fact that many functions share the same derivative.
Definite integral
The integral of f(x) dx from a to b, which gives a number: the net signed area between the graph of f and the x-axis from x = a to x = b. In simple terms, it adds up infinitely many infinitely thin slices to find total accumulation.
Fundamental Theorem of Calculus (FTC), Part 1
If F(x) = integral from a to x of f(t) dt, then F'(x) = f(x). Differentiation undoes integration. In simple terms, the derivative of an accumulation function gives you back the original function.
Fundamental Theorem of Calculus, Part 2
If f is continuous on [a, b] and F is any antiderivative of f, then the integral from a to b of f(x) dx = F(b) - F(a). This is the evaluation formula: find an antiderivative, plug in the endpoints, subtract.
U-substitution
A technique for reversing the chain rule when integrating. Let u = g(x), then du = g'(x) dx, and the integral transforms into a simpler form in terms of u. Think of it as: spot a function and its derivative sitting together in the integrand, and relabel.
Constant of integration (C)
The arbitrary constant added to every indefinite integral, reflecting the fact that two antiderivatives of the same function can differ by a constant.
Displacement
The net change in position over a time interval, found by integrating velocity: displacement = integral from a to b of v(t) dt.
Integral of x^n dx = x^(n+1)/(n+1) + C, for n != -1.
Integral of e^x dx = e^x + C.
Integral of cos(x) dx = sin(x) + C.
Integral of sin(x) dx = -cos(x) + C.
Integral of 1/x dx = ln(x) + C for x > 0.
Constants factor out: integral of k * f(x) dx = k * integral of f(x) dx.
Integral of (3x^2 + 2x) dx. Apply the power rule term by term: x^3 + x^2 + C.
Integral of (e^x + cos(x)) dx = e^x + sin(x) + C.
Use FTC Part 2: integral from a to b of f(x) dx = F(b) - F(a).
Example: integral from 1 to 2 of 2x dx. Antiderivative: x^2. Evaluate: 2^2 - 1^2 = 4 - 1 = 3.
Example: integral from 0 to 1 of (4x^3 - 6x^2) dx. Antiderivative: x^4 - 2x^3. Evaluate: (1 - 2) - (0 - 0) = -1.
If F(x) = integral from a to x of f(t) dt, then F'(x) = f(x).
When the upper limit is a function of x (like x^2), use the chain rule.
Example: g(x) = integral from 0 to x^2 of sin(t) dt. By FTC Part 1 + chain rule: g'(x) = sin(x^2) * 2x = 2x sin(x^2).
Spot a composite function in the integrand. Let u = inner function, compute du.
Example: integral of 2x * sqrt(x^2 + 1) dx. Let u = x^2 + 1, then du = 2x dx. The integral becomes integral of sqrt(u) du = (2/3)u^(3/2) + C = (2/3)(x^2 + 1)^(3/2) + C.
For definite integrals, either convert the limits to u-values or substitute back to x before evaluating.
Displacement over [a, b] = integral from a to b of v(t) dt.
Example: v(t) = 3t^2 - 4t on [0, 2]. Antiderivative: t^3 - 2t^2. Evaluate: (8 - 8) - (0 - 0) = 0. The particle returns to its starting position.
Integral of 1/x dx = ln|x| + C (use absolute value for the general case).
For x > 0, this simplifies to ln(x) + C.
Integral of x^n dx = x^(n+1)/(n+1) + C (n != -1)
Integral of 1/x dx = ln|x| + C
Integral of e^x dx = e^x + C
Integral of cos(x) dx = sin(x) + C
Integral of sin(x) dx = -cos(x) + C
FTC Part 1: d/dx [integral from a to x of f(t) dt] = f(x)
FTC Part 1 with chain rule: d/dx [integral from a to g(x) of f(t) dt] = f(g(x)) * g'(x)
FTC Part 2: integral from a to b of f(x) dx = F(b) - F(a)
Displacement = integral from a to b of v(t) dt
Integration computes total quantities from rates: total distance from speed, total revenue from marginal revenue, total energy from power. In physics, the work done by a variable force is the integral of force over distance. In probability, the area under a probability density curve gives the likelihood of an event. Engineers integrate stress distributions to find the total load on a structure.
Students forget the + C on indefinite integrals. Every indefinite integral has a constant of integration. Definite integrals do not (the constants cancel in F(b) - F(a)).
Students confuse the integral of 1/x with the power rule for n = -1. The power rule gives x^0/0, which is undefined. The correct antiderivative is ln|x| + C.
Students apply FTC Part 1 without the chain rule when the upper limit is a function of x. If the upper limit is x^2, you must multiply by the derivative of x^2 (which is 2x).
Students mix up displacement and total distance. Displacement is the integral of velocity (which can be negative). Total distance is the integral of |v(t)|.
FTC Part 1 is tested directly: "If F(x) = integral from a to x of f(t) dt, then F'(x) = ?" The answer is f(x). Know this cold.
The long-answer section includes a u-substitution problem and a FTC + chain rule problem (g(x) = integral from 0 to x^2 of sin(t) dt). Both are worth 6 marks.
Basic antiderivatives (power rule, e^x, trig, 1/x) appear in multiple-choice and short-answer. Speed matters here.
Displacement from velocity is a short-answer staple. Set up the definite integral and evaluate.
Forgetting + C on an indefinite integral costs a mark in short-answer sections.
Fill in the blank: integral of (3x^2 + 2x) dx = ___. (x^3 + x^2 + C.)
True or False: integral from 1 to 2 of 2x dx = 4. (False. It equals 3.)
Fill in the blank: FTC Part 1 says d/dx [integral from a to x of f(t) dt] = ___. (f(x).)
True or False: integral of 1/x dx = x^0 + C. (False. It equals ln|x| + C.)
Fill in the blank: If g(x) = integral from 0 to x^2 of sin(t) dt, then g'(x) = ___. (2x sin(x^2).)
Q: Determine the integral of (3x^2 + 2x) dx.
A: x^3 + x^2 + C.
Q: Evaluate integral from 1 to 2 of 2x dx.
A: Antiderivative: x^2. Evaluate: 4 - 1 = 3.
Q: Find the general antiderivative of f(x) = e^x + cos(x).
A: e^x + sin(x) + C.
Q: The Fundamental Theorem of Calculus Part 1 states that if F(x) = integral from a to x of f(t) dt, then F'(x) = ?
A: f(x).
Q: Evaluate integral of (1/x) dx for x > 0.
A: ln(x) + C.
Q: A particle moves with velocity v(t) = 3t^2 - 4t m/s. Find its displacement over [0, 2].
A: Integral from 0 to 2 of (3t^2 - 4t) dt. Antiderivative: t^3 - 2t^2. Evaluate: (8 - 8) - (0) = 0 m.
Q: Evaluate integral from 0 to 1 of (4x^3 - 6x^2) dx.
A: Antiderivative: x^4 - 2x^3. Evaluate: (1 - 2) - (0) = -1.
Q: Use u-substitution to evaluate integral of 2x * sqrt(x^2 + 1) dx.
A: Let u = x^2 + 1, du = 2x dx. Integral becomes integral of sqrt(u) du = (2/3)u^(3/2) + C = (2/3)(x^2 + 1)^(3/2) + C.
Q: Evaluate g'(x) where g(x) = integral from 0 to x^2 of sin(t) dt.
A: By FTC Part 1 + chain rule: g'(x) = sin(x^2) * d/dx[x^2] = 2x sin(x^2).
Integration is the reverse of differentiation, so the Derivatives notes are the direct prerequisite. U-substitution is the chain rule running backwards. The FTC connects the integral (an accumulation) to the derivative (a rate), unifying the two halves of calculus. Displacement from velocity ties integration to the position-velocity-acceleration framework in the Applications of Derivatives notes. In Calculus II, these ideas extend to integration by parts, partial fractions, and improper integrals.
integration, antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, FTC, u-substitution, substitution rule, constant of integration, displacement, area under a curve, power rule for integrals, e^x integral, ln(x), 1/x, accumulation function, net area, Calculus I, MSU, Michigan State University, Calculus 1