Difficulty: Intermediate to Advanced | Prerequisites: All derivative rules from Sections 10.1 through 12.6
Integration is the reverse of differentiation. Where derivatives answer "how fast is it changing?," integrals answer "how much has accumulated?" The indefinite integral gives you a family of antiderivatives. The definite integral gives you a number: the net area under a curve, the total cost, the total distance. The Fundamental Theorem of Calculus is the bridge between the two. Once the basics are in place, the course extends integration to business applications (consumer surplus, producer surplus, income streams) and to improper integrals, where the interval or the function is unbounded.
An antiderivative undoes a derivative. The definite integral computes accumulated change (net area). Substitution and integration by parts are the two main techniques for finding antiderivatives. Applications include area between curves, average value, consumer and producer surplus, and improper integrals.
Antiderivative
A function F(x) whose derivative is f(x): F'(x) = f(x). In simple terms, it is a function that "undoes" the derivative.
Indefinite integral
∫ f(x) dx = F(x) + C, where F is any antiderivative and C is the constant of integration. The result is a family of functions differing by a constant.
Constant of integration (C)
Because the derivative of a constant is zero, every antiderivative has an arbitrary constant added. Forgetting +C on an indefinite integral is a common mark lost on exams.
Substitution (u-substitution)
The integration counterpart of the chain rule. Replace a piece of the integrand with u, express dx in terms of du, integrate, then substitute back.
Definite integral
∫ from a to b of f(x) dx. It gives the net signed area between f(x) and the x-axis over the interval [a, b]. Positive where f is above the axis, negative where f is below.
Riemann sum
An approximation of the definite integral by summing areas of rectangles. As the number of rectangles increases, the Riemann sum approaches the definite integral.
Fundamental Theorem of Calculus (FTC)
∫ from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f. This connects the area problem to antidifferentiation.
Integration by parts
∫ u dv = uv - ∫ v du. Used when the integrand is a product and substitution does not work. Think of it as the product rule in reverse.
Area between two curves
∫ from a to b of [f(x) - g(x)] dx, where f(x) ≥ g(x) on [a, b]. It gives the area of the region enclosed between the two functions.
Average value of a function
(1/(b - a)) · ∫ from a to b of f(x) dx. The "average height" of the function over an interval.
Consumer surplus
The total amount consumers save by paying the market price rather than the maximum they would have paid. Computed as the area between the demand curve and the market price line.
Producer surplus
The total benefit producers receive by selling at the market price rather than the minimum they would have accepted. Area between the market price line and the supply curve.
Improper integral
An integral where the interval is infinite (∫ from a to ∞) or the integrand has a vertical asymptote in the interval. Evaluated using limits. If the limit exists and is finite, the integral converges; otherwise it diverges.
∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, provided n ≠ -1. (Power rule for integrals.)
∫ x⁻¹ dx = ∫ (1/x) dx = ln |x| + C.
∫ eˣ dx = eˣ + C.
∫ aˣ dx = aˣ / ln a + C.
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx. (Sum/difference rule.)
∫ k · f(x) dx = k · ∫ f(x) dx. (Constant multiple rule.)
Always include +C for indefinite integrals.
Check your work: differentiate your answer. You should get back the original integrand.
Substitution reverses the chain rule. Look for a composite structure: an "inner function" whose derivative (or a constant multiple of it) also appears in the integrand.
Steps:
Choose u = (inner function).
Compute du = u'(x) dx.
Rewrite the integral entirely in terms of u and du.
Integrate with respect to u.
Substitute back to express the answer in terms of x.
For definite integrals with substitution, you can either change the limits of integration to u-values or substitute back and use the original x-limits.
Common substitutions: u = exponent in e^(something), u = argument of ln, u = expression inside a power.
The definite integral ∫ from a to b of f(x) dx represents the net signed area between the curve and the x-axis.
Riemann sums approximate this area by dividing [a, b] into n subintervals, each of width Δx = (b - a)/n, and summing f(xₖ) · Δx.
Left, right, and midpoint sums use different sample points within each subinterval. As n → ∞, all converge to the definite integral (for continuous functions).
Area above the x-axis counts as positive; area below counts as negative. Total (unsigned) area requires splitting the integral where f crosses zero.
Fundamental Theorem of Calculus: If F is any antiderivative of f, then ∫ from a to b of f(x) dx = F(b) - F(a).
This means you compute a definite integral by finding an antiderivative, then evaluating it at the upper limit minus the lower limit.
No +C is needed for definite integrals (the constants cancel).
Properties:
∫ from a to a of f(x) dx = 0.
∫ from a to b of f(x) dx = -∫ from b to a of f(x) dx.
∫ from a to b of [f(x) + g(x)] dx = ∫ from a to b of f(x) dx + ∫ from a to b of g(x) dx.
∫ from a to c of f(x) dx = ∫ from a to b of f(x) dx + ∫ from b to c of f(x) dx.
Formula: ∫ u dv = uv - ∫ v du.
Choose u as the part that becomes simpler when differentiated, and dv as the part you can integrate. A common mnemonic for choosing u: LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).
Sometimes you need to apply integration by parts more than once.
Example: ∫ x · eˣ dx. Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. Result: xeˣ - ∫ eˣ dx = xeˣ - eˣ + C.
If f(x) ≥ g(x) on [a, b], the area between the curves is ∫ from a to b of [f(x) - g(x)] dx.
First find the intersection points by solving f(x) = g(x). These give your limits of integration.
If the curves cross, split the integral at the crossing points and take the absolute value of each piece, or integrate |f(x) - g(x)|.
Always sketch the curves to confirm which function is on top in each interval.
Average value of f on [a, b]: f_avg = (1/(b - a)) · ∫ from a to b of f(x) dx.
This gives the constant value that would produce the same total accumulation over the interval.
Moving average: the average value computed over a sliding window. Used in business and economics to smooth out short-term fluctuations in data.
Consumer surplus: ∫ from 0 to q̄ of [D(q) - p̄] dq, where D(q) is the demand curve and p̄ is the equilibrium price at quantity q̄. It measures the total benefit consumers receive above what they paid.
Producer surplus: ∫ from 0 to q̄ of [p̄ - S(q)] dq, where S(q) is the supply curve. It measures the total benefit producers receive above the minimum they would have accepted.
Continuous income stream: If income flows at rate R(t) dollars per year, the total income over T years is ∫ from 0 to T of R(t) dt. The present value of that stream, discounted at rate r, is ∫ from 0 to T of R(t) · e^(-rt) dt.
Future value of a continuous income stream: e^(rT) · ∫ from 0 to T of R(t) · e^(-rt) dt.
Type 1 (infinite interval): ∫ from a to ∞ of f(x) dx = lim (b→∞) ∫ from a to b of f(x) dx. Similarly for -∞ to a.
Type 2 (integrand has a vertical asymptote at c in [a, b]): replace the bad endpoint with a limit.
If the limit exists and is a finite number, the integral converges. If not, it diverges.
Key example: ∫ from 1 to ∞ of (1/xᵖ) dx converges when p > 1 and diverges when p ≤ 1.
Improper integrals model situations where a quantity accumulates forever (perpetuity, infinite time horizon).
Power rule for integrals: ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C (n ≠ -1)
Logarithm: ∫ (1/x) dx = ln |x| + C
Exponential: ∫ eˣ dx = eˣ + C; ∫ e^(kx) dx = (1/k) · e^(kx) + C
General exponential: ∫ aˣ dx = aˣ / ln a + C
Fundamental Theorem of Calculus: ∫ from a to b of f(x) dx = F(b) - F(a)
Integration by parts: ∫ u dv = uv - ∫ v du
Area between curves: ∫ from a to b of [f(x) - g(x)] dx (where f ≥ g)
Average value: (1/(b - a)) · ∫ from a to b of f(x) dx
Consumer surplus: ∫ from 0 to q̄ of [D(q) - p̄] dq
Producer surplus: ∫ from 0 to q̄ of [p̄ - S(q)] dq
Present value of income stream: ∫ from 0 to T of R(t) · e^(-rt) dt
Convergence test: ∫ from 1 to ∞ of (1/xᵖ) dx converges iff p > 1
Integration computes totals from rates. If you know how fast water flows into a tank (rate), integrating gives you how much water is in the tank (total). In economics, consumer and producer surplus quantify the gains from trade in a market. Present value of income streams is how businesses decide whether an investment that pays out over years is worth its upfront cost. Improper integrals model perpetuities in finance, where payments continue indefinitely.
Students forget +C on indefinite integrals. The constant of integration is not optional; it represents the entire family of antiderivatives.
Students apply the power rule to ∫ (1/x) dx and write x⁰/0, which is undefined. The correct answer is ln |x| + C. The power rule does not work when n = -1.
In substitution, students sometimes forget to convert dx to du, leaving a mix of x and u in the integrand. Everything must be in terms of u before integrating.
Students confuse the definite integral (a number) with the indefinite integral (a function + C). The definite integral has no +C.
When computing area between curves, students sometimes take f - g without checking which function is on top, producing a negative area.
⚠️ The FTC is the centrepiece of Exam 3. Be able to evaluate definite integrals by finding an antiderivative and plugging in limits.
⚠️ Substitution problems are heavily tested. Practise choosing u and converting the entire integral.
⚠️ Integration by parts: expect at least one problem. Know the LIATE priority and be ready to apply the formula twice if needed.
⚠️ Consumer and producer surplus are favourite business-calculus exam topics. Know the setup: sketch the curves, find equilibrium, set up the integral.
⚠️ Improper integrals: know how to rewrite with a limit and determine convergence vs. divergence.
Fill in the blank: ∫ x³ dx = ____.
True or false: ∫ from 2 to 2 of f(x) dx = 0.
Fill in the blank: In integration by parts, u is typically the factor that becomes ____ when differentiated.
True or false: ∫ from 1 to ∞ of (1/x) dx converges.
Fill in the blank: Consumer surplus is the area between the ____ curve and the equilibrium price.
Answers: 1. x⁴/4 + C. 2. True. 3. Simpler. 4. False (it diverges, since p = 1). 5. Demand.
Q: Evaluate ∫ (3x² + 2x - 1) dx.
A: x³ + x² - x + C.
Q: Use substitution to find ∫ 2x · e^(x²) dx.
A: Let u = x². Then du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = e^(x²) + C.
Q: Evaluate ∫ from 0 to 3 of (6x - x²) dx.
A: Antiderivative: 3x² - x³/3. At x = 3: 27 - 9 = 18. At x = 0: 0. Result: 18.
Q: Find the area between f(x) = x² and g(x) = x on [0, 1].
A: On [0, 1], x ≥ x² (check: at x = 0.5, 0.5 > 0.25). Area = ∫ from 0 to 1 of (x - x²) dx = [x²/2 - x³/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
Q: Determine whether ∫ from 1 to ∞ of (1/x³) dx converges or diverges. If it converges, find its value.
A: ∫ from 1 to b of x⁻³ dx = [-1/(2x²)] from 1 to b = -1/(2b²) + 1/2. As b → ∞, -1/(2b²) → 0. The integral converges to 1/2.
Integration is the payoff of all the derivative work from the first two thirds of the course. The chain rule (11.4) is the basis for substitution (13.2). The product rule (11.3) is the basis for integration by parts (14.1). Optimisation (12.1, 12.2) reappears when you maximise consumer surplus or present value. Elasticity (12.6) connects to surplus analysis because both deal with how demand responds to price in a market.
antiderivative, indefinite integral, constant of integration, power rule for integrals, substitution, u-substitution, definite integral, Riemann sum, Fundamental Theorem of Calculus, FTC, net area, signed area, integration by parts, LIATE, area between curves, average value, moving average, consumer surplus, producer surplus, demand curve, supply curve, equilibrium, continuous income stream, present value, future value, improper integral, convergent, divergent, p-integral, MAT 210, brief calculus, ASU, Waner Costenoble