Source: Calc AB Lecture Notes, UCF Advanced Calculus
Difficulty: Intermediate | Prerequisites: Differentiation rules (power rule, basic derivatives)
Tags: antiderivative, indefinite integral, reverse power rule, differential equation, general solution, particular solution, constant of integration, integration
This is where calculus flips direction. After spending the previous units learning how to take derivatives, you now learn how to undo them. Antidifferentiation (integration) asks: "Given a rate of change, what was the original function?" This skill is foundational for everything that follows in the course, including area calculations, the Fundamental Theorem of Calculus, and applied problems involving motion and accumulation. If you are comfortable differentiating power functions, you are ready for this material.
An antiderivative reverses differentiation: you recover the original function from its derivative. The reverse power rule is the main tool, and you must always include a constant of integration (+C) for indefinite integrals. Differential equations use initial conditions to pin down that constant and find a specific function.
Antiderivative
The function F(x) whose derivative equals f(x). In other words, if F'(x) = f(x), then F(x) is an antiderivative of f(x). Think of it as: "working backwards from the derivative to find the original function."
Indefinite integral
The notation ∫ f(x) dx, which represents the family of all antiderivatives of f(x). The result always includes + C. In simple terms, this means: "find all possible original functions that could have produced this derivative."
Constant of integration (C)
Because the derivative of any constant is zero, antidifferentiation can never recover that lost constant on its own. The +C accounts for every possible vertical shift of the antiderivative. Think of it as: "we lost a number when we differentiated, and +C is the placeholder for whatever it was."
Reverse power rule
The core antidifferentiation formula: ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C, valid for all n ≠ −1. You increase the exponent by one and divide by the new exponent. In simple terms: "do the opposite of the power rule for derivatives."
Differential equation
An equation that contains a derivative (e.g. y' = 8t³ + 5). Solving it means finding the function y that satisfies the equation. Think of it as: "a puzzle where you know the rate of change and need to find the function itself."
General solution
The antiderivative of a differential equation, expressed with +C. It represents a whole family of curves (one for each value of C).
Particular solution
A single specific antiderivative found by using an initial condition (a known point on the curve) to solve for C.
The formula: ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C
Add 1 to the exponent, then divide by the new exponent.
Always attach + C at the end. Losing it is the single most common mark lost on exams.
Worked example (positive exponent):
∫ x dx = x¹⁺¹ / (1+1) = x² / 2 + C
Worked example (negative exponent):
∫ x⁻⁵ dx = x⁻⁴ / (−4) = −x⁻⁴ / 4 + C
Worked example (polynomial):
∫ (2x − 3x²) dx = 2x² / 2 − 3x³ / 3 = x² − x³ + C
Finding a general solution:
Integrate both sides of the equation.
The result includes + C and represents a family of functions.
Finding a particular solution:
Start by finding the general solution (integrate).
Substitute the given initial condition to solve for C.
Write the final answer with C replaced by its numeric value.
Worked example (particular solution):
Given: y' = 8t³ + 5, and y = −4 when t = 1.
Integrate: y = ∫ (8t³ + 5) dt = 2t⁴ + 5t + C
Apply the condition: −4 = 2(1)⁴ + 5(1) + C → −4 = 7 + C → C = −11
Final answer: y = 2t⁴ + 5t − 11
Rule | Formula |
|---|---|
Reverse power rule | ∫ xⁿ dx = xⁿ⁺¹ / (n+1) + C, where n ≠ −1 |
Constant multiple rule | ∫ k·f(x) dx = k · ∫ f(x) dx |
Sum/difference rule | ∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx |
Antiderivatives are how you recover position from velocity, or velocity from acceleration. Every time a physics problem gives you an acceleration function and asks for the position, you are solving a differential equation with initial conditions, exactly as shown above.
Forgetting + C on indefinite integrals. This is not a stylistic choice. Without + C, your answer represents one curve instead of the entire family. Exams penalise this consistently.
Dividing by the old exponent instead of the new one. The reverse power rule divides by (n + 1), not by n. Double-check by differentiating your answer.
Applying the reverse power rule when n = −1. The formula ∫ xⁿ dx = xⁿ⁺¹/(n+1) breaks when n = −1 because you would divide by zero. The integral of x⁻¹ is ln|x| + C (covered separately).
Sign errors with negative exponents. When n is negative, (n + 1) may also be negative. Write it out fully rather than doing it in your head.
⚠️ The +C is tested relentlessly. Expect at least one question where omitting it costs marks.
⚠️ Particular solution problems (initial value problems) are a staple of free-response sections. You need to show every step: integrate, substitute, solve for C, write the final function.
⚠️ Know the difference between "general solution" and "particular solution" as vocabulary. A question asking for the general solution should not have C solved out.
True or false: ∫ x³ dx = x⁴ + C.
Fill in the blank: The constant of integration accounts for the fact that the derivative of any ______ is zero.
True or false: A particular solution to a differential equation still contains + C.
Fill in the blank: ∫ x⁻⁵ dx = ______.
True or false: ∫ (2x + 3x²) dx = x² + x³ + C.
Answers: 1. False (should be x⁴/4 + C). 2. Constant. 3. False (C is solved for using the initial condition). 4. −x⁻⁴/4 + C. 5. True.
Q: Find the general solution of ∫ (6x² − 4x + 1) dx.
A: 6x³/3 − 4x²/2 + x + C = 2x³ − 2x² + x + C.
Q: Given that dy/dx = 3x² + 2 and y(0) = 5, find the particular solution.
A: y = x³ + 2x + C. Using y(0) = 5: 5 = 0 + 0 + C, so C = 5. Particular solution: y = x³ + 2x + 5.
Q: A student writes ∫ x⁻¹ dx = x⁰/0 + C. What went wrong?
A: The reverse power rule does not apply when n = −1, because it produces division by zero. The correct integral is ln|x| + C.
Q: Explain why indefinite integrals always include + C.
A: Differentiation destroys constant terms (their derivative is zero), so when you reverse the process, you cannot recover the original constant. The + C represents all possible values that constant could have been.
This material connects directly to the Fundamental Theorem of Calculus (next set of notes), which links antiderivatives to definite integrals and area. It also underpins motion problems in later units, where position, velocity, and acceleration are connected through integration and differentiation.
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