An indefinite integral reverses differentiation: given a function, you find the family of functions whose derivative is the original. Every indefinite integral includes a constant of integration (+C) because derivatives destroy constants. When an initial condition is given, you can pin down the exact antiderivative.
Difficulty: Beginner to Intermediate
Prerequisites: Derivative rules (power, product, chain), familiarity with exponential and logarithmic functions.
Integration is the second pillar of calculus (differentiation was the first). Where derivatives answer "how fast is this changing?", integrals answer "how much has accumulated?" In a business calculus course, you need integrals to go from a marginal cost function (the derivative of total cost) back to the total cost function itself. This topic also sets you up for definite integrals and area calculations. If you are comfortable with basic derivatives, integration is mostly about running those rules in reverse.
Antiderivative
A function F(x) whose derivative is f(x). In simple terms, it is the function you get when you "undo" a derivative.
Indefinite integral
The notation ∫ f(x) dx represents the entire family of antiderivatives of f(x). The result is F(x) + C.
Constant of integration (C)
The arbitrary constant added to every indefinite integral. It exists because any constant vanishes when you differentiate, so there are infinitely many antiderivatives that differ only by a constant.
Particular antiderivative
The specific antiderivative that satisfies a given initial condition, such as f(0) = -4. You find it by plugging the condition into the general antiderivative and solving for C.
Initial condition
A known value of the function (or its derivative) at a specific point, used to determine the constant of integration. Think of it as the one extra piece of information that pins down the exact curve.
Power rule for integration
The reverse of the power rule for derivatives. For n ≠ -1: ∫ x^n dx = x^(n+1) / (n+1) + C.
Integration reverses differentiation, term by term. The key rules:
Constant rule: ∫ k dx = kx + C
Power rule: ∫ x^n dx = x^(n+1)/(n+1) + C, provided n ≠ -1
Sum/difference rule: integrate each term separately
Constant multiple rule: ∫ k·f(x) dx = k · ∫ f(x) dx
Worked examples from the source:
∫ 5 dx = 5x + C. A constant integrates to that constant times x.
∫ 3w dw = (3/2)w^2 + C. Apply the power rule with n = 1, then multiply by 3.
∫ (2 - 4z + z^3) dz = 2z - 2z^2 + (1/4)z^4 + C. Integrate each term separately: the constant 2 becomes 2z, the -4z term uses the power rule to give -2z^2, and z^3 gives (1/4)z^4.
When a problem gives you f'(x) and an initial condition like f(0) = -4, the procedure is:
Integrate f'(x) to get f(x) = [general antiderivative] + C.
Substitute the initial condition: plug in the given x-value and the given f(x)-value.
Solve for C.
Write the final answer with the specific value of C.
Example: Suppose f'(x) = 6x + 2 and f(0) = -4.
Integrate: f(x) = 3x^2 + 2x + C.
Apply f(0) = -4: 0 + 0 + C = -4, so C = -4.
Answer: f(x) = 3x^2 + 2x - 4.
The same method works with any initial condition, such as f(4) = 2. You simply plug in x = 4 instead of x = 0.
\int k \, dx = kx + C\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 [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dxIn business, the marginal cost function C'(x) tells you the cost of producing one additional unit. Integrating C'(x) recovers the total cost function C(x), and the initial condition (e.g., fixed costs when x = 0) pins down the constant. The same logic applies to marginal revenue and marginal profit.
Forgetting +C. Every indefinite integral needs a constant of integration. Leaving it off is the single most common mark lost on integration questions.
Misapplying the power rule to 1/x. The power rule does not work when n = -1. ∫ x^(-1) dx = ln|x| + C, not x^0/0.
Adding the exponent before dividing. The correct order is: raise the exponent by 1, then divide by the new exponent. Students sometimes divide by the old exponent.
Treating the integral of a product as the product of integrals. ∫ f(x)·g(x) dx ≠ [∫ f(x) dx]·[∫ g(x) dx]. There is no simple product rule for integration.
⚠️ "Find the antiderivative" and "evaluate the integral" mean the same thing for indefinite integrals. Do not overthink the phrasing.
⚠️ Particular antiderivative questions are a near-certainty on the exam. Practise applying initial conditions until the process is automatic.
⚠️ Look out for functions that require rewriting before integrating, e.g., 1/x^2 = x^(-2). Professors test whether you can spot the power rule in disguise.
Fill in the blank: ∫ 7 dx = ______. (7x + C)
True or false: ∫ x^(-1) dx = x^0/0 + C. (False; the answer is ln|x| + C)
Fill in the blank: The constant of integration exists because derivatives eliminate ______. (constants)
True or false: If f'(x) = 4x and f(2) = 10, then C = 2. (True: f(x) = 2x^2 + C, f(2) = 8 + C = 10, C = 2)
Fill in the blank: ∫ x^3 dx = ______. (x^4/4 + C)
Q: Evaluate ∫ (2 - 4z + z^3) dz.
A: Integrate term by term: 2z - 2z^2 + (1/4)z^4 + C.
Q: Find f(x) given that f'(x) = 3/√x and f(4) = 2.
A: Rewrite as f'(x) = 3x^(-1/2). Integrate: f(x) = 3 · x^(1/2)/(1/2) + C = 6√x + C. Apply f(4) = 2: 6(2) + C = 2, so C = -10. Answer: f(x) = 6√x - 10.
Q: What is ∫ 5e^x dx?
A: 5e^x + C. The exponential function is its own antiderivative, so just carry the constant multiplier.
Q: A marginal cost function is C'(x) = 8x + 50, and fixed costs are £2,000. Find C(x).
A: Integrate: C(x) = 4x^2 + 50x + C. Fixed costs mean C(0) = 2000, so C = 2000. Answer: C(x) = 4x^2 + 50x + 2000.
Integration feeds directly into definite integrals and area under the curve, which you will use to compute total revenue or total cost over a range of production levels. It also connects back to critical points: knowing the antiderivative lets you reconstruct the original function and then apply the extrema-finding techniques from the previous topic. In economics applications, integration turns marginal functions into total functions.
Indefinite integral, antiderivative, integration, power rule for integrals, constant of integration, +C, particular antiderivative, initial condition, marginal cost integration, reverse differentiation, MATH 201, business calculus integrals, FIU calculus, finding f(x) from f'(x)