Difficulty: Intermediate to Advanced | Prerequisites: All derivative rules and applications (Ch. 2–3 notes).
This set covers Stewart sections 3.9 through 5.1: the reverse of differentiation (antiderivatives), how to compute area under a curve from first principles (Riemann sums), the precise definition of the definite integral, the Fundamental Theorem of Calculus connecting derivatives and integrals, the u-substitution technique for evaluating integrals, and areas between two curves. Integration is the second pillar of calculus. Where derivatives measure rates of change, integrals accumulate totals, and the Fundamental Theorem reveals that these two operations are inverses of each other.
An antiderivative reverses differentiation: if F'(x) = f(x), then F is an antiderivative of f. A definite integral is the limit of Riemann sums and computes the net area under a curve. The Fundamental Theorem of Calculus (FTC) ties these together: the definite integral from a to b of f(x) dx equals F(b) - F(a), where F is any antiderivative of f. U-substitution is the chain rule run backwards and is the primary technique for evaluating integrals.
Antiderivative
A function F(x) whose derivative is f(x): F'(x) = f(x). Also called a primitive. The general antiderivative is F(x) + C, where C is an arbitrary constant. Think of it as "undoing" differentiation.
Indefinite integral
The notation integral of f(x) dx = F(x) + C, representing the family of all antiderivatives of f. The + C is essential because many functions share the same derivative.
Riemann sum
An approximation of the area under y = f(x) from x = a to x = b, computed by slicing the region into n thin rectangles and summing their areas: sum of f(x_i) * Delta x. Think of it as stacking narrow bars under the curve.
Definite integral
The limit of Riemann sums as the number of rectangles goes to infinity: integral from a to b of f(x) dx = lim(n->infinity) sum f(x_i) Delta x. This gives the exact net signed area between the curve and the x-axis.
Sigma notation
The symbol Sigma (capital Greek S) used to write sums compactly: sum from i = m to n of q_i = q_m + q_{m+1} + ... + q_n.
Fundamental Theorem of Calculus, Part 1 (FTC1)
If f is continuous on [a, b], then I(x) = integral from a to x of f(t) dt is an antiderivative of f: I'(x) = f(x). In simple terms, the derivative of a running integral of f gives you f back.
Fundamental Theorem of Calculus, Part 2 (FTC2)
If f is continuous on [a, b] and F is any antiderivative of f, then integral from a to b of f(x) dx = F(b) - F(a). In simple terms, to evaluate a definite integral, find an antiderivative and subtract its values at the endpoints.
U-substitution
The integration technique that reverses the chain rule. You set u = g(x), compute du = g'(x) dx, rewrite the integral in terms of u, integrate, and substitute back. Think of it as undoing the chain rule's "inside function."
Area between curves
The area between y = f(x) (top) and y = g(x) (bottom) from x = a to x = b is integral from a to b of [f(x) - g(x)] dx, provided f(x) >= g(x) on the interval.
An antiderivative of f(x) is any function F(x) with F'(x) = f(x)
The general antiderivative is F(x) + C, because adding a constant does not change the derivative
Basic antiderivatives (reverse the power rule and trig derivatives):
integral of x^p dx = x^(p+1) / (p+1) + C, provided p is not -1
integral of cos(x) dx = sin(x) + C
integral of sin(x) dx = -cos(x) + C
integral of sec^2(x) dx = tan(x) + C
integral of sec(x)tan(x) dx = sec(x) + C
Linearity: integral of [af(x) + bg(x)] dx = a * integral of f(x) dx + b * integral of g(x) dx
Initial value problems: given f'(x) and a starting condition like f(0) = 5, find f(x) by integrating and solving for C
To compute the area under y = f(x) >= 0 from x = a to x = b, divide [a, b] into n subintervals of width Delta x = (b - a)/n
Choose a sample point x_i in each subinterval. The Riemann sum is: sum from i = 1 to n of f(x_i) * Delta x
Common choices: left endpoints, right endpoints, or midpoints. All give the same limit as n -> infinity for integrable functions
The distance problem motivates this: given velocity v(t), the distance from t = a to t = b is approximated by summing v(t_i) * Delta t, and the exact distance is the limit
Sigma notation shortcuts for computing sums: sum of i from 1 to n = n(n+1)/2, sum of i^2 = n(n+1)(2n+1)/6, sum of i^3 = [n(n+1)/2]^2
Formal definition: integral from a to b of f(x) dx = lim(n->infinity) sum from i = 1 to n of f(x_i) Delta x
The function is integrable if this limit exists for every choice of sample points. Continuous functions are integrable, as are functions with finitely many jump or removable discontinuities
Key properties:
integral from a to a of f(x) dx = 0
integral from a to b of f(x) dx = -integral from b to a of f(x) dx
integral from a to b of c * f(x) dx = c * integral from a to b of f(x) dx
integral from a to b of [f(x) + g(x)] dx = integral from a to b of f(x) dx + integral from a to b of g(x) dx
integral from a to b = integral from a to c + integral from c to b (splitting the interval)
Comparison: if f(x) >= g(x) on [a, b], then integral of f >= integral of g
FTC Part 1: if f is continuous on [a, b], define I(x) = integral from a to x of f(t) dt. Then I'(x) = f(x). The running integral is an antiderivative
With the chain rule: d/dx [integral from a to g(x) of f(t) dt] = f(g(x)) * g'(x)
FTC Part 2: if F'(x) = f(x) and f is continuous on [a, b], then integral from a to b of f(x) dx = F(b) - F(a)
This is the evaluation theorem: find any antiderivative, plug in the limits, subtract
Example: integral from 0 to 2 of x^2 dx = [x^3/3] from 0 to 2 = 8/3 - 0 = 8/3
The two parts together say: differentiation and integration are inverse processes
The technique reverses the chain rule: if the integrand looks like f'(g(x)) * g'(x), then substituting u = g(x) simplifies to integral of f'(u) du
Procedure:
Choose u = (inside function), compute du = g'(x) dx
Rewrite everything in the integral in terms of u and du
Integrate with respect to u
Substitute back to express the answer in terms of x
Example: integral of x * cos(x^2) dx. Let u = x^2, du = 2x dx, so x dx = du/2. The integral becomes (1/2) * integral of cos(u) du = (1/2) sin(u) + C = (1/2) sin(x^2) + C
For definite integrals, you can either substitute back to x and use the original limits, or change the limits to u-values: when x = a, u = g(a); when x = b, u = g(b)
Area between y = f(x) (top) and y = g(x) (bottom) from x = a to x = b: A = integral from a to b of [f(x) - g(x)] dx
If the curves cross within [a, b], split the integral at the crossing points, because which function is on top changes
To find crossing points, solve f(x) = g(x)
Example: area between y = x^2 + x + 1 and y = 2x^2 - 1 from x = 0 to x = 1. Top minus bottom = (x^2 + x + 1) - (2x^2 - 1) = -x^2 + x + 2. Integrate: [-x^3/3 + x^2/2 + 2x] from 0 to 1 = -1/3 + 1/2 + 2 = 13/6
Basic antiderivatives:
integral of x^p dx = x^(p+1)/(p+1) + C (p not equal to -1)
integral of cos x dx = sin x + C
integral of sin x dx = -cos x + C
integral of sec^2 x dx = tan x + C
Sigma sum formulas:
sum i = n(n+1)/2
sum i^2 = n(n+1)(2n+1)/6
sum i^3 = [n(n+1)/2]^2
\int_a^b f(x)\,dx = F(b) - F(a) \quad \text{where } F'(x) = f(x)\frac{d}{dx}\int_a^x f(t)\,dt = f(x)U-substitution: set u = g(x), du = g'(x) dx, transform and integrate.
Area between curves: A = integral from a to b of [f(x) - g(x)] dx, with f on top.
Integration computes total quantities from rates: total distance from velocity, total revenue from marginal revenue, total energy from power. Riemann sums are the basis for numerical integration methods used in engineering when exact antiderivatives do not exist. Areas between curves appear in economics (consumer and producer surplus), physics (work done by a variable force), and probability (area under density curves).
Students often forget the + C in indefinite integrals. Every antiderivative has a family of solutions differing by a constant.
Students confuse the integral of x^(-1) with the power rule. The power rule gives x^0/0, which is undefined. The antiderivative of 1/x is ln|x| + C (covered in Calculus II).
Students apply u-substitution but forget to change the dx to du, or forget to change the limits when computing a definite integral with the substitution.
Students assume the integral from a to b of f(x) dx is always positive. It is not. When f(x) < 0, the integral counts area as negative. Net signed area can be zero or negative.
Students forget to check which curve is on top when computing area between curves. If you subtract the wrong way, you get a negative area.
⚠️ Setting up and evaluating Riemann sums (with sigma notation) for specific functions is a standard exam problem, especially for f(x) = x^2 or x^3 on simple intervals.
⚠️ FTC Part 2 (the evaluation theorem) is the workhorse of integration: find an antiderivative and plug in the limits. Practise until it is automatic.
⚠️ FTC Part 1 problems ask you to differentiate an integral: d/dx integral from a to x of f(t) dt = f(x). Variations include d/dx integral from a to g(x) of f(t) dt = f(g(x)) * g'(x).
⚠️ U-substitution is heavily tested. Common patterns: x*cos(x^2), sin(x)*cos(x), (2x+1)*sqrt(x^2+x). Recognise the inner function and its derivative.
⚠️ Areas between curves: know how to find intersection points and set up the correct integral with top minus bottom.
What is the general antiderivative of x^4? (x^5/5 + C.)
True or false: integral from a to b of f(x) dx = -integral from b to a of f(x) dx. (True.)
By FTC Part 1, d/dx integral from 0 to x of cos(t) dt = ? (cos(x).)
In u-substitution, if u = x^2, then du = ? (2x dx.)
True or false: the area between two curves can be computed as a single integral even if the curves cross. (False. You must split at the crossing points.)
Q: Evaluate integral from 1 to 3 of (2x + 1) dx.
A: Antiderivative: x^2 + x. Evaluate: (9 + 3) - (1 + 1) = 12 - 2 = 10.
Q: Find d/dx integral from 0 to x^3 of sin(t) dt.
A: By FTC1 with the chain rule: sin(x^3) * 3x^2.
Q: Evaluate integral of (2x + 1)^5 dx using u-substitution.
A: Let u = 2x + 1, du = 2 dx. Integral = (1/2) integral of u^5 du = (1/2)(u^6/6) + C = (2x+1)^6 / 12 + C.
Q: Evaluate integral from 0 to 1 of x * e^(x^2) dx. (Treat e^u as its own antiderivative.)
A: Let u = x^2, du = 2x dx. When x = 0, u = 0; when x = 1, u = 1. Integral = (1/2) integral from 0 to 1 of e^u du = (1/2)(e^1 - e^0) = (e - 1)/2.
Q: Find the area between y = x and y = x^2 from x = 0 to x = 1.
A: On [0, 1], x >= x^2 (since 0 <= x <= 1). Area = integral from 0 to 1 of (x - x^2) dx = [x^2/2 - x^3/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
The FTC connects the two halves of Calculus I: everything about derivatives (Ch. 2 and 3) feeds into finding antiderivatives, and the Riemann-sum machinery gives integrals their independent meaning as accumulated totals. In Calculus II, you will learn more integration techniques (integration by parts, partial fractions, trig substitution), compute volumes of revolution, arc lengths, and surface areas using integrals, and extend integrals to improper integrals (infinite limits or asymptotes). The sigma-notation Riemann sums also lead into series and sequences.
antiderivative, indefinite integral, definite integral, Riemann sum, sigma notation, summation, area under a curve, Fundamental Theorem of Calculus, FTC, FTC1, FTC2, evaluation theorem, u-substitution, change of variables, area between curves, net signed area, initial value problem, + C, constant of integration, integrand, limits of integration, integrable, MATH 132, Stewart Chapters 3.9 through 5.1, Calculus I