Difficulty: Intermediate | Prerequisites: All differentiation rules, limits
Integration is the reverse of differentiation. Where differentiation finds rates of change, integration recovers total quantities from rates: total cost from marginal cost, total distance from velocity, accumulated profit over a production range. This chapter rounds out the course by connecting the two halves of calculus through the Fundamental Theorem. If you know your derivative rules well, learning to integrate is largely a matter of running them backwards.
An antiderivative undoes a derivative. Indefinite integrals give a family of functions (always add +C). Definite integrals give a number, representing net accumulation between two bounds. The Fundamental Theorem of Calculus links the two: evaluate the antiderivative at the upper bound, subtract its value at the lower bound.
Antiderivative
A function F(x) whose derivative is f(x). In simple terms, it is the function you would have needed to differentiate to get f(x).
Indefinite integral
The general antiderivative of f(x), written ∫f(x) dx = F(x) + C. The "+C" accounts for the fact that infinitely many functions share the same derivative (they differ by a constant).
Definite integral
The integral of f(x) from a to b, written ∫(a to b) f(x) dx = F(b) − F(a). It gives a specific number: the net signed area under the curve between x = a and x = b.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It disappears in definite integrals because it cancels when you subtract F(a) from F(b).
Fundamental Theorem of Calculus
If F is an antiderivative of f on [a,b], then ∫(a to b) f(x) dx = F(b) − F(a). Think of it as the bridge between antiderivatives and definite integrals.
Average value of a function
The mean height of f(x) over [a,b]: (1/(b−a)) ∫(a to b) f(x) dx. In simple terms, if the curve were flattened into a rectangle of the same width, this would be the rectangle's height.
Differential equation
An equation involving a derivative (e.g. dy/dx = 5x⁴ − 3e^x − 2x). Solving it means finding the function y by integrating and using an initial condition to pin down C.
Initial value problem (IVP)
A differential equation plus a condition like y = 5 when x = 1. The condition determines the specific value of C.
Power rule (reversed): ∫x^n dx = x^(n+1)/(n+1) + C, provided n ≠ −1.
Special case n = −1: ∫(1/x) dx = ln|x| + C.
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.
Always rewrite the integrand into power-rule form before integrating.
Example: ∫(5 − 16/x³) dx = ∫(5 − 16x^(−3)) dx = 5x + 16/(2x²) + C = 5x + 8/x² + C.
Example: ∫(x⁷e^x − x⁶)/x⁷ dx = ∫(e^x − 1/x) dx = e^x − ln|x| + C.
Example: ∫(2/√x) dx = ∫2x^(−1/2) dx = 2 · 2x^(1/2) + C = 4√x + C.
Example: ∫(x+3)(4x−1) dx. Expand first: ∫(4x² + 11x − 3) dx = (4/3)x³ + (11/2)x² − 3x + C.
Evaluate the antiderivative at the upper limit and subtract its value at the lower limit.
Example: ∫(2 to 4) (3x − 2) dx. Antiderivative: (3/2)x² − 2x. Evaluate: [(3/2)(16) − 8] − [(3/2)(4) − 4] = [24 − 8] − [6 − 4] = 16 − 2 = 14.
Key property: ∫(a to a) f(x) dx = 0. If the upper and lower limits are the same, the integral is zero regardless of the integrand.
Example: ∫(12 to 12) (e^π − 6x⁷ − ln x)^34 dx = 0.
∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx (sum/difference rule). This is true.
∫(a to b) f(x) dx = F(b) − F(a), not F(b) + F(a). The Fundamental Theorem uses subtraction.
∫(1/x) dx = ln|x| + C. This is true.
Average value = (1/(b−a)) ∫(a to b) f(x) dx.
Example: f(x) = 2e^x + 9x² on [−2, 5]. Average = (1/7) ∫(−2 to 5) (2e^x + 9x²) dx = (1/7)[2e^x + 3x³] from −2 to 5 = (1/7)[(2e⁵ + 375) − (2e^(−2) − 24)] = (1/7)(2e⁵ − 2/e² + 399) ≈ 99.4.
Integrate the marginal cost function and use a known condition (usually the fixed cost at x = 0) to find C.
Example: C'(x) = 3x³ − 4x, C(0) = 12000. Integrate: C(x) = (3/4)x⁴ − 2x² + C. Use C(0) = 12000: C = 12000. So C(x) = (3/4)x⁴ − 2x² + 12000. C(12) = (3/4)(20736) − 2(144) + 12000 = 15552 − 288 + 12000 = $27,264.
Example: P'(x) = 130/x + 3. Change in profit from 40 to 150 units: ∫(40 to 150) (130/x + 3) dx = [130 ln|x| + 3x] from 40 to 150 = [130 ln 150 + 450] − [130 ln 40 + 120] = 130(ln 150 − ln 40) + 330 = 130 ln(150/40) + 330 = 130 ln(3.75) + 330 ≈ 171.83 + 330 = $501.83.
Integrate both sides, then use the initial condition to find C.
Example: dy/dx = 5x⁴ − 3e^x − 2x, with y = 5 when x = 1. Integrate: y = x⁵ − 3e^x − x² + C. Substitute x = 1, y = 5: 5 = 1 − 3e − 1 + C, so C = 5 + 3e. Final answer: y = x⁵ − 3e^x − x² + 5 + 3e.
\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{(Fundamental Theorem)}\text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dxIntegrating a marginal cost function recovers the total cost function, which tells a business its full production expenses. Definite integrals of profit rates give the total change in profit over a production range. Average value calculations appear in economics (average price over time) and science (average concentration of a drug).
Students forget the +C on indefinite integrals. Every indefinite integral needs it. Definite integrals do not, because C cancels in F(b) − F(a).
Students write ∫(a to b) f(x) dx = F(b) + F(a). The Fundamental Theorem uses subtraction: F(b) − F(a).
When integrating 1/x, students sometimes write ln(x) instead of ln|x|. The absolute value matters because ln is only defined for positive arguments, but 1/x has domain on both sides of zero.
In initial value problems, students integrate correctly but forget to solve for C using the given condition. Without C, the answer is incomplete.
⚠️ Always rewrite roots and fractions as power-rule form before integrating. ∫(2/√x) dx is much easier as ∫2x^(−1/2) dx.
⚠️ The property ∫(a to a) f(x) dx = 0 is a quick freebie if it appears. Do not waste time computing the antiderivative.
⚠️ Expand products before integrating. ∫(x+3)(4x−1) dx requires you to multiply out first.
⚠️ Differential equation problems give an initial condition. Use it. The answer is not complete without a specific value of C.
Fill in the blank: ∫x⁴ dx = ______. (x⁵/5 + C.)
True or false: ∫(5 to 5) sin(x³ + e^x) dx = 0. (True. Same upper and lower limits.)
Fill in the blank: ∫(1/x) dx = ______. (ln|x| + C.)
True or false: ∫(a to b) f(x) dx = F(b) + F(a). (False. It is F(b) − F(a).)
Fill in the blank: The average value of f on [a,b] is (1/______) times the definite integral. (b − a.)
Q: Evaluate ∫(5 − 16/x³) dx.
A: Rewrite: ∫(5 − 16x^(−3)) dx = 5x + 16/(2x²) + C = 5x + 8/x² + C.
Q: Evaluate ∫(x⁷e^x − x⁶)/x⁷ dx.
A: Simplify: ∫(e^x − 1/x) dx = e^x − ln|x| + C.
Q: Evaluate ∫(2/√x) dx.
A: Rewrite: ∫2x^(−1/2) dx = 2 · 2x^(1/2) + C = 4√x + C.
Q: Evaluate ∫(x+3)(4x−1) dx.
A: Expand: ∫(4x² + 11x − 3) dx = (4/3)x³ + (11/2)x² − 3x + C.
Q: Evaluate ∫(2 to 4) (3x − 2) dx.
A: Antiderivative: (3/2)x² − 2x. Evaluate: [(3/2)(16) − 8] − [(3/2)(4) − 4] = 16 − 2 = 14.
Q: Evaluate ∫(12 to 12) (e^π − 6x⁷ − ln x)^34 dx.
A: The limits are identical, so the integral is 0.
Q: Which are true? I. ∫(a to b) f(x) dx = F(b) + F(a). II. ∫[f ± g] dx = ∫f dx ± ∫g dx. III. ∫(1/x) dx = ln|x| + C.
A: I is false (should be F(b) − F(a)). II and III are true. Answer: II and III.
Q: Find the average value of f(x) = 2e^x + 9x² on [−2, 5].
A: (1/7)[∫(−2 to 5) (2e^x + 9x²) dx] = (1/7)[2e^x + 3x³] from −2 to 5 = (1/7)(2e⁵ + 375 − 2e^(−2) + 24) = (1/7)(2e⁵ − 2/e² + 399) ≈ 99.4.
Q: C'(x) = 3x³ − 4x with C(0) = 12000. Find C(x) and C(12).
A: C(x) = (3/4)x⁴ − 2x² + C. C(0) = 12000 gives C = 12000. C(x) = (3/4)x⁴ − 2x² + 12000. C(12) = 15552 − 288 + 12000 = $27,264.
Q: P'(x) = 130/x + 3. Find the change in profit from 40 to 150 units.
A: ∫(40 to 150) (130/x + 3) dx = [130 ln x + 3x] from 40 to 150 = (130 ln 150 + 450) − (130 ln 40 + 120) = 130 ln(3.75) + 330 ≈ $501.83.
Q: Solve dy/dx = 5x⁴ − 3e^x − 2x with y = 5 when x = 1.
A: Integrate: y = x⁵ − 3e^x − x² + C. Substitute: 5 = 1 − 3e − 1 + C, so C = 5 + 3e. Answer: y = x⁵ − 3e^x − x² + 5 + 3e.
Integration directly reverses differentiation: every derivative rule you learnt has a corresponding antiderivative rule. The Fundamental Theorem of Calculus formally connects the two halves of the course. Business applications here (finding total cost from marginal cost, total profit change from marginal profit) build on the derivative-based marginal analysis from the Applications of Derivatives notes. Initial value problems combine integration with the algebra of solving for constants, completing the circle back to the kind of equation-solving you started the course with.
integration, antiderivative, indefinite integral, definite integral, Fundamental Theorem of Calculus, constant of integration, +C, power rule for integrals, average value, differential equation, initial value problem, IVP, marginal cost, total cost, change in profit, net signed area, MAC 2233, calculus for business, FIU, exponential integral, logarithmic integral, rewrite before integrating