Difficulty: Introductory to Intermediate | Prerequisites: All derivative rules from Chapters 3 and 4
Big Picture
Integration is differentiation in reverse. If you know the rate of change (the derivative), integration recovers the original function. This chapter introduces the indefinite integral, the notation, and the basic rules. It is the foundation for everything in Ch. 5.5 (definite integrals and applications). If you can differentiate confidently, you can learn to integrate; the rules mirror each other.
An indefinite integral (antiderivative) reverses differentiation. The power rule for integrals adds 1 to the exponent and divides by the new exponent. Always add + C. To find a particular antiderivative, use an initial condition to solve for C.
Antiderivative
A function F(x) such that F'(x) = f(x). If F is one antiderivative, then F + C (for any constant C) is also an antiderivative.
Think of it as: the function whose derivative brings you back to what you started with.
Indefinite integral
The notation for the general antiderivative: the integral sign, the integrand f(x), and dx. Written as integral of f(x) dx = F(x) + C.
In simple terms, this means: the family of all functions whose derivative is f(x).
Constant of integration (C)
The arbitrary constant added to every indefinite integral, because the derivative of a constant is zero and therefore lost during differentiation.
Think of it as: a reminder that infinitely many functions share the same derivative, shifted up or down.
Particular antiderivative
The specific antiderivative found by using an initial condition (such as f(0) = -4) to solve for C.
Differential equation
An equation involving a derivative, such as dy/dx = 3x^2 + 4x + 2. Solving it means finding y as a function of x.
Basic Integration Rules
Constant rule: integral of k dx = kx + C
Power rule: integral of x^n dx = x^(n+1) / (n+1) + C, provided n is not -1
Reciprocal rule: integral of (1/x) dx = ln|x| + C
Exponential rule: integral of e^x dx = e^x + C
Constant multiple: integral of k f(x) dx = k integral of f(x) dx
Sum/difference: integral of [f(x) +/- g(x)] dx = integral of f(x) dx +/- integral of g(x) dx
Rewriting Before Integrating
Many problems require algebraic manipulation before you can apply the rules.
Roots become fractional exponents: sqrt(x) = x^(1/2), cube root of x = x^(1/3), fourth root of x^5 = x^(5/4).
Fractions with x in the denominator: 4/t = 4 t^(-1); 7/x^2 = 7 x^(-2); 1/x^3 = x^(-3).
Multiply out products before integrating: (3x - 1)(2x^2 + 5) = 6x^3 - 2x^2 + 15x - 5.
Split fractions: (x^4 - x) / x^2 = x^2 - x^(-1).
Worked Examples from the Review
integral of 5 dx = 5x + C
integral of 3w dw = (3/2)w^2 + C
integral of (2 - 4z + z^2) dz = 2z - 2z^2 + (1/3)z^3 + C
integral of (4/t) dt = 4 ln|t| + C
integral of (-2e^x) dx = -2e^x + C
integral of (7/x^2 - sqrt(x) + 3) dx = -7/x - (2/3) sqrt(x^3) + 3x + C
integral of (fourth root of x^5 - x^3 + 5e^x) dx = (4/9) fourth root of x^9 - (1/4)x^4 + 5e^x + C
integral of (3x - 1)(2x^2 + 5) dx. Expand first to 6x^3 - 2x^2 + 15x - 5. Integrate: (3/2)x^4 - (2/3)x^3 + (15/2)x^2 - 5x + C
integral of (x^4 - x) / x^2 dx. Simplify to x^2 - 1/x. Integrate: (1/3)x^3 - ln|x| + C
Finding a Particular Antiderivative
When given f'(x) and a condition like f(0) = -4:
Integrate f'(x) to get f(x) + C.
Substitute the given x-value and f(x)-value to solve for C.
Problem 27: f'(x) = 5x^4 - x^3 + 2x^2 - e^x, f(0) = -4.
Integrate: f(x) = x^5 - (1/4)x^4 + (2/3)x^3 - e^x + C.
Use f(0) = -4: 0 - 0 + 0 - 1 + C = -4, so C = -3.
f(x) = x^5 - (1/4)x^4 + (2/3)x^3 - e^x - 3.
Problem 28: f'(x) = (x + 1)(3x + 2), f(4) = 2.
Expand: f'(x) = 3x^2 + 5x + 2. Integrate: f(x) = x^3 + (5/2)x^2 + 2x + C.
Use f(4) = 2: 64 + 40 + 8 + C = 2, so C = -110.
f(x) = x^3 + (5/2)x^2 + 2x - 110.
Solving Differential Equations
Problem 29: dy/dx = 3x^2 + 4x + 2. General solution: y = x^3 + 2x^2 + 2x + C.
Problem 30: dy/dx = 4x^3 + 3x^(-1) + e^x - 1, y(1) = 0. Integrate: y = x^4 + 3 ln|x| + e^x - x + C. Use y(1) = 0: 1 + 0 + e - 1 + C = 0, so C = -e. Solution: y = x^4 + 3 ln|x| + e^x - x - e.
integral of k dx = kx + C
integral of x^n dx = x^(n+1) / (n+1) + C (n is not -1)
integral of (1/x) dx = ln|x| + C
integral of e^x dx = e^x + C
integral of k f(x) dx = k integral of f(x) dx
integral of [f +/- g] dx = integral of f dx +/- integral of g dx
Forgetting + C. Every indefinite integral needs it. Points are lost on exams for this consistently.
Applying the power rule when n = -1. The integral of x^(-1) dx is ln|x| + C, not x^0 / 0 (which is undefined). This is the single most common rule-application error.
Not rewriting before integrating. You cannot integrate (3x - 1)(2x^2 + 5) directly with the power rule. Expand it first, then integrate term by term.
Confusing integration with differentiation. The power rule for integrals adds 1 to the exponent and divides. The power rule for derivatives subtracts 1 and multiplies. Mixing these up is easy under exam pressure.
Expect several "find the indefinite integral" problems on the exam, testing your fluency with the basic rules.
At least one problem will require you to find a particular antiderivative using an initial condition. Do not forget to solve for C.
Rewriting expressions (splitting fractions, converting roots to exponents, expanding products) before integrating is tested heavily.
Differential equation problems (dy/dx = ...) are integration problems in disguise. Integrate and apply the initial condition.
Fill in the blank: The integral of x^n dx = ______ (state the formula). (x^(n+1) / (n+1) + C, n not -1)
True or False: The integral of (1/x) dx = x^0 / 0 + C. (False, it is ln|x| + C)
Fill in the blank: To find a particular antiderivative, you use an ______ to solve for C. (initial condition)
True or False: You can integrate a product of two functions term by term without expanding first. (False, you must expand first)
Fill in the blank: Every indefinite integral must include ______ at the end. (+ C)
Q: Find the indefinite integral of (7/x^2 - sqrt(x) + 3) dx.
A: Rewrite as 7x^(-2) - x^(1/2) + 3. Integrate: -7/x - (2/3)x^(3/2) + 3x + C.
Q: Find the indefinite integral of (x^4 - x) / x^2 dx.
A: Simplify to x^2 - x^(-1). Integrate: (1/3)x^3 - ln|x| + C.
Q: Given f'(x) = (x + 1)(3x + 2) and f(4) = 2, find f(x).
A: Expand f'(x) = 3x^2 + 5x + 2. Integrate: f(x) = x^3 + (5/2)x^2 + 2x + C. Use f(4) = 2: 64 + 40 + 8 + C = 2, so C = -110. f(x) = x^3 + (5/2)x^2 + 2x - 110.
Q: Find the general solution to dy/dx = 3x^2 + 4x + 2.
A: Integrate: y = x^3 + 2x^2 + 2x + C.
Indefinite integrals are the direct reverse of derivatives from Chapters 3 and 4. The particular-antiderivative technique (integrating a marginal cost function to recover the cost function) appears in the applied problems of Ch. 5.5. Definite integrals in Ch. 5.5 use the same antiderivative rules but then evaluate at bounds, so mastering these rules here is essential for that chapter.
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