Difficulty: Intermediate to Advanced | Prerequisites: All previous MAT 210 notes (Limits, Derivatives, Applications)
Integration is the second major operation of calculus and, in a sense, the reverse of differentiation. Where derivatives measure rates of change, integrals measure accumulation: total distance from a velocity function, total cost from a marginal cost function, or total area under a curve. This material covers roughly Topics 16 through 24 and appears on the final exam, which is worth 35% of the course grade.
An antiderivative reverses differentiation: if the derivative of F(x) is f(x), then F is an antiderivative of f. The definite integral adds up the accumulated value of a function over an interval, which geometrically corresponds to the (signed) area under its curve. The Fundamental Theorem of Calculus ties these two ideas together, saying that you can evaluate a definite integral by finding an antiderivative and plugging in the endpoints.
Antiderivative
A function F(x) whose derivative is f(x). For example, an antiderivative of 2x is x², because d/dx(x²) = 2x. Think of it as "undoing" the derivative.
Indefinite integral
The notation ∫ f(x) dx represents the family of all antiderivatives of f(x). The result always includes + C (the constant of integration) because any constant vanishes under differentiation.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It accounts for the fact that infinitely many functions share the same derivative (x² + 3, x² - 7, x² + 100 all have derivative 2x).
Definite integral
The integral ∫ from a to b of f(x) dx, which produces a single number. Geometrically, it gives the net signed area between the curve y = f(x) and the x-axis over the interval [a, b].
Fundamental Theorem of Calculus (FTC)
If F is any antiderivative of f on [a, b], then ∫ from a to b of f(x) dx = F(b) - F(a). This is the bridge between antiderivatives and definite integrals. In simple terms, find the antiderivative, plug in the top number, plug in the bottom number, subtract.
Net change
The integral of a rate of change gives the total (net) change. If f'(x) is the rate, then ∫ from a to b of f'(x) dx = f(b) - f(a).
Area under a curve
When f(x) ≥ 0 on [a, b], the definite integral equals the area between the curve and the x-axis. When f(x) is negative on part of the interval, the integral subtracts that portion.
Riemann sum
An approximation of the definite integral using rectangles. You divide [a, b] into n subintervals, pick a sample point in each, and sum the rectangle areas. The definite integral is the limit of Riemann sums as n → ∞.
u-substitution
An integration technique that reverses the chain rule. You set u = g(x), find du = g'(x) dx, rewrite the integral in terms of u, integrate, and substitute back.
Power rule for integration: ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, provided n ≠ -1.
When n = -1: ∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C.
Exponential: ∫ eˣ dx = eˣ + C. More generally, ∫ e^(kx) dx = (1/k)e^(kx) + C.
General exponential: ∫ aˣ dx = aˣ / ln(a) + C.
Constant multiple: ∫ k · f(x) dx = k · ∫ f(x) dx.
Sum/difference: ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx.
An indefinite integral ∫ f(x) dx is a family of functions (includes + C).
A definite integral ∫ from a to b of f(x) dx is a number (no + C).
The definite integral uses the indefinite integral as an intermediate step: find the antiderivative, then evaluate at the bounds.
Part 1 (evaluation): ∫ from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
Part 2 (connection): if F(x) = ∫ from a to x of f(t) dt, then F'(x) = f(x). The derivative of an accumulation function is the original function.
Example: ∫ from 1 to 3 of (2x + 1) dx. Antiderivative is x² + x. Evaluate: (9 + 3) - (1 + 1) = 12 - 2 = 10.
Identify an inner function u = g(x) and check that g'(x) dx (or a constant multiple of it) appears in the integrand.
Replace g(x) with u and g'(x) dx with du.
Integrate in terms of u, then substitute back.
Example: ∫ 2x · e^(x²) dx. Let u = x², du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = e^(x²) + C.
Example: ∫ from 0 to 1 of x(x² + 1)³ dx. Let u = x² + 1, du = 2x dx, so x dx = du/2. When x = 0, u = 1; when x = 1, u = 2. Integral becomes (1/2) ∫ from 1 to 2 of u³ du = (1/2)[u⁴/4] from 1 to 2 = (1/2)(16/4 - 1/4) = (1/2)(15/4) = 15/8.
When f(x) ≥ 0 on [a, b], the definite integral equals the area between the curve and the x-axis.
When f(x) takes negative values, the integral gives the net signed area: regions below the x-axis contribute negative values.
To find the total area (unsigned), split the interval where f changes sign and take the absolute value of each piece.
Net change interpretation: ∫ from a to b of R'(x) dx = R(b) - R(a), the total change in R.
Example: if the marginal cost is C'(x) = 10 + 0.02x and you want the total cost of increasing production from 100 to 200 units, compute ∫ from 100 to 200 of (10 + 0.02x) dx = [10x + 0.01x²] from 100 to 200 = (2000 + 400) - (1000 + 100) = 1300.
Total revenue from selling units a through b: ∫ from a to b of R'(x) dx.
Total quantity from a rate: if water flows into a tank at rate r(t) litres per minute, the total amount added from t = 0 to t = 10 is ∫ from 0 to 10 of r(t) dt.
Average value of f on [a, b]: (1/(b - a)) · ∫ from a to b of f(x) dx.
\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 \qquad \int a^x \, dx = \frac{a^x}{\ln a} + C\int_a^b f(x)\, dx = F(b) - F(a) \quad \text{(FTC)}\text{Net change: } \int_a^b f'(x)\, dx = f(b) - f(a)\text{Average value} = \frac{1}{b-a} \int_a^b f(x)\, dxIntegration is how you go from a rate to a total. A company with a known marginal cost function can integrate it to find the total cost of a production run. A biologist measuring a population growth rate integrates to find the total population change. Economists compute consumer and producer surplus as areas between curves, which are definite integrals.
Forgetting + C on indefinite integrals. Every antiderivative has a family of solutions. On a definite integral, however, the C cancels, so you omit it.
Confusing ∫ xⁿ dx with d/dx [xⁿ]. Integration raises the exponent by 1 and divides; differentiation lowers it by 1 and multiplies. They are inverse operations, not the same operation.
Applying the power rule when n = -1. ∫ (1/x) dx = ln|x| + C, not x⁰/0 (which is undefined). This is a special case that must be memorised separately.
Forgetting to change the limits of integration when using u-substitution on a definite integral. If you substitute u = g(x), the new limits are g(a) and g(b), not a and b.
⚠️ The Fundamental Theorem of Calculus is the most important single result in the course. Expect it on the final.
⚠️ u-substitution appears in multiple problems on the final. Practise identifying the inner function and changing variables until it is automatic.
⚠️ Net change problems often combine integration with marginal analysis: "Given C'(x), find the total cost of producing units 50 through 100." These are definite integrals of marginal functions.
⚠️ Know how to set up and compute the average value of a function. It is a common short-answer or multiple-choice item.
True or false: ∫ from 2 to 2 of f(x) dx = 0. Answer: True. The integral over an interval of zero width is always 0.
Fill in the blank: ∫ e^(4x) dx = ______. Answer: (1/4)e^(4x) + C.
True or false: the antiderivative of 1/x is x⁰. Answer: False. The antiderivative of 1/x is ln|x| + C.
What does the Fundamental Theorem of Calculus say? Answer: ∫ from a to b of f(x) dx = F(b) - F(a), where F is any antiderivative of f.
Fill in the blank: the average value of f on [a, b] is (1/(b - a)) times ______. Answer: ∫ from a to b of f(x) dx.
Q: Evaluate ∫ (3x² - 4x + 7) dx.
A: x³ - 2x² + 7x + C.
Q: Evaluate ∫ from 0 to 2 of (6x² + 1) dx.
A: Antiderivative is 2x³ + x. Evaluate: (2·8 + 2) - (0 + 0) = 16 + 2 = 18.
Q: Use u-substitution to evaluate ∫ 4x · (x² + 3)⁵ dx.
A: Let u = x² + 3, du = 2x dx, so 4x dx = 2 du. Integral becomes 2 ∫ u⁵ du = 2 · u⁶/6 + C = (x² + 3)⁶/3 + C.
Q: The marginal revenue function is R'(x) = 50 - 0.4x. Find the total revenue gained from selling units 10 through 30.
A: ∫ from 10 to 30 of (50 - 0.4x) dx = [50x - 0.2x²] from 10 to 30 = (1500 - 180) - (500 - 20) = 1320 - 480 = 840.
Q: Find the average value of f(x) = x² on [0, 3].
A: (1/3) · ∫ from 0 to 3 of x² dx = (1/3) · [x³/3] from 0 to 3 = (1/3)(27/3) = (1/3)(9) = 3.
The definite integral ties back to limits (it is defined as a limit of Riemann sums) and to derivatives (the FTC says differentiation and integration are inverse processes). Accumulation problems are the mirror image of marginal analysis: marginal cost is the derivative of total cost, and integrating marginal cost recovers total cost.
The average value formula also connects to the mean value theorem for integrals, which guarantees that a continuous function achieves its average value at least once on the interval.
antiderivative, indefinite integral, definite integral, constant of integration, Fundamental Theorem of Calculus, FTC, net change theorem, Riemann sum, area under a curve, signed area, u-substitution, integration by substitution, power rule for integrals, exponential integral, logarithmic integral, average value of a function, accumulation function, total cost from marginal cost, consumer surplus, producer surplus, MAT 210, Brief Calculus, ASU