Antiderivatives and indefinite integrals (basic rules) — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Definition of antiderivatives, indefinite integral notation, the constant of integration, basic integration rules including power rule, constant multiple, sum/difference, and basic exponential and trigonometric antiderivative formulas.

You should already know: Derivative rules for polynomials, exponentials, and trigonometric functions; basic algebraic manipulation of polynomial and rational terms; the chain rule for derivatives.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Antiderivatives and indefinite integrals (basic rules)?

This topic is the foundational building block of all integration in AP Calculus AB, and the unit (Integration and Accumulation of Change) accounts for ~17-20% of the total AP exam score, per the official CED. Content from this topic appears in both MCQ and FRQ sections, often as a component of larger problems involving motion, area, or differential equations. An antiderivative of a function $f(x)$ on an interval is any differentiable function $F(x)$ such that $F^{\prime }(x)=f(x)$ for all $x$ in the interval. The indefinite integral of $f(x)$ is the general representation of all possible antiderivatives of $f(x)$, written with the notation $\int f(x)dx$. The notation includes the differential $dx$ to specify that $x$ is the variable of integration, and an arbitrary constant of integration $+C$ to capture that any vertical shift of an antiderivative is also an antiderivative. Unlike definite integrals (which output a numerical value for accumulated change), indefinite integrals output a general function family. This topic introduces the core basic rules that are required for every more complex integration application later in the AB course.

2. Antiderivatives, Indefinite Integrals, and the Constant of Integration

An antiderivative reverses the differentiation process: if the derivative of $F(x)$ is $f(x)$, then $F(x)$ is an antiderivative of $f(x)$. Because the derivative of any constant is zero, if $F(x)$ is an antiderivative of $f(x)$, then $F(x)+C$ (for any real constant $C$) is also an antiderivative of $f(x)$. This means there are infinitely many antiderivatives for any function, differing only by the constant $C$. The indefinite integral $\int f(x)dx$ is the general form of all antiderivatives, so it always includes $+C$ by convention. The $\int$ symbol (an elongated S) comes from the sum notation used for definite integrals, and $dx$ explicitly identifies the variable of integration, which is critical for problems with multiple variables or implicit functions. If you are given an initial condition (a point that the antiderivative must pass through), you can substitute into the general form to solve for a specific value of $C$, resulting in a particular antiderivative.

Worked Example

Problem: Given $f(x)=6x^2$, (1) write the general indefinite integral of $f(x)$, and (2) find the particular antiderivative that satisfies $F(1)=10$.

Solution steps:

1. Recall that the derivative of $x^3$ is $3x^2$, so the antiderivative of $x^2$ is $\frac{x^3}{3}+C$.
2. Apply the constant multiple rule: $\int 6x^{2}dx=6\cdot \frac{x^3}{3}+C=2x^3+C$. This is the general indefinite integral.
3. Substitute the initial condition $F(1)=10$ into the general form: $F(1)=2(1{)}^3+C=2+C=10$.
4. Solve for $C$: $C=10-2=8$.
5. The particular antiderivative is $F(x)=2x^3+8$.

Exam tip: Always check your antiderivative by differentiating it! If the derivative of your result equals the original integrand, you know your work is correct. This catches 90% of common sign and arithmetic errors on the exam.

3. Algebraic Basic Integration Rules

Algebraic integration rules directly mirror differentiation rules, reversed for integration. The constant multiple rule states that $\int k\cdot f(x)dx=k\int f(x)dx$ for any constant $k$, which follows directly from the constant multiple rule for derivatives: $\frac{d}{dx}[kF(x)]=k{F}^{\prime }(x)=kf(x)$. The sum/difference rule states that $\int [f(x)\pm g(x)]dx=\int f(x)dx\pm \int g(x)dx$, again reversing the sum rule for derivatives. The most widely used algebraic rule is the power rule for integration, which reverses the power rule for derivatives. For derivatives, $\frac{d}{dx}[x^n]=n{x}^{n-1}$, so reversing this gives the integration power rule: $$\int x^{n}dx=\frac{x^{n+1}}{n+1}+Cn\ne -1$$ The key exception is when $n=-1$ (i.e., $f(x)=\frac{1}{x}$), which we cover with the logarithmic antiderivative rule. To use the power rule, you must always rewrite terms with roots or negative exponents into the standard $x^n$ form first. For example, $\frac{1}{x^3}=x^{-3}$ and $\sqrt{x}=x^{1/2}$.

Worked Example

Problem: Evaluate the indefinite integral $\int \left(4x^3-\frac{2}{\sqrt{x}}+5\right)dx$.

Solution steps:

1. Rewrite all terms to match the $x^n$ form required for the power rule: $\frac{2}{\sqrt{x}}=2x^{-1/2}$, and $5=5x^0$. The integral becomes $\int \left(4x^3-2x^{-1/2}+5x^0\right)dx$.
2. Apply the sum/difference rule and constant multiple rule to split the integral: $4\int x^{3}dx-2\int x^{-1/2}dx+5\int x^{0}dx$.
3. Apply the power rule to each term: $4\cdot \frac{x^4}{4}-2\cdot \frac{x^{1/2}}{1/2}+5\cdot \frac{x^1}{1}+C$
4. Simplify: $x^4-4x^{1/2}+5x+C=x^4-4\sqrt{x}+5x+C$.
5. Verify by differentiation: $\frac{d}{dx}\left(x^4-4\sqrt{x}+5x+C\right)=4x^3-\frac{2}{\sqrt{x}}+5$, which matches the original integrand.

Exam tip: Always rewrite roots and reciprocals as power terms before applying the power rule. Skipping this step is the most common cause of miscalculating the exponent in the power rule on AP exams.

4. Basic Transcendental Antiderivatives (Exponential and Trigonometric)

We reverse the derivative rules for non-algebraic (transcendental) functions to get basic antiderivative rules for exponential and trigonometric functions. For exponentials: the derivative of $e^x$ is $e^x$, so $\int e^{x}dx=e^x+C$. For a general base $a>0,a\ne 1$, the derivative of $a^x$ is $a^x\ln a$, so rearranging gives $\int a^{x}dx=\frac{a^x}{\ln a}+C$. For $f(x)=\frac{1}{x}$ (the $n=-1$ exception to the power rule), the derivative of $\ln |x|$ is $\frac{1}{x}$, so $\int \frac{1}{x}dx=\ln |x|+C$. The absolute value is required because $\ln x$ is only defined for $x>0$, while $\frac{1}{x}$ is defined for all non-zero $x$.

For trigonometric functions, all rules are direct reverses of derivative rules. The basic rules required for AP Calculus AB are:

• $\int \cos xdx=\sin x+C$
• $\int \sin xdx=-\cos x+C$
• $\int {\sec }^{2}xdx=\tan x+C$
• $\int \sec x\tan xdx=\sec x+C$ Pay close attention to negative signs, which are a common source of error.

Worked Example

Problem: Evaluate $\int \left(3e^x+2\sin x-\sec x\tan x+\frac{4}{x}\right)dx$.

Solution steps:

1. Split the integral into separate terms using the sum/difference rule: $3\int e^{x}dx+2\int \sin xdx-\int \sec x\tan xdx+4\int \frac{1}{x}dx$
2. Apply the corresponding antiderivative rule to each term: $3(e^x)+2(-\cos x)-(\sec x)+4(\ln |x|)+C$
3. Simplify: $3e^x-2\cos x-\sec x+4\ln |x|+C$
4. Verify by differentiation: $\frac{d}{dx}\left(3e^x-2\cos x-\sec x+4\ln |x|+C\right)=3e^x+2\sin x-\sec x\tan x+\frac{4}{x}$, which matches the original integrand.

Exam tip: If you are unsure about the sign of a trigonometric antiderivative, take 10 seconds to differentiate your result to confirm it matches the original integrand. FRQ graders deduct points for incorrect signs, so this check is well worth the time.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Writing $\int x^{-1}dx=\frac{x^0}{0}+C$, leaving the answer undefined or incorrectly simplifying $\frac{1}{0}$ to 1. Why: Students blindly apply the power rule without remembering the exception for $n=-1$. Correct move: Always check if $n=-1$ before applying the power rule; if yes, use $\int \frac{1}{x}dx=\ln |x|+C$ instead.
• Wrong move: For a particular antiderivative problem, substituting the initial condition into the original integrand $f(x)$ instead of the general antiderivative $F(x)+C$ to solve for $C$. For example, given $f(x)=6x^2$ and $F(1)=10$, calculating $6(1{)}^2=10+C$ to get $C=-4$. Why: Students confuse the original function with its antiderivative. Correct move: Always find the general antiderivative first, then substitute the given $x$-value into the general antiderivative to solve for $C$.
• Wrong move: Omitting the absolute value in $\ln |x|+C$ and writing $\ln x+C$ instead. Why: Students memorize the rule without remembering the domain of $\frac{1}{x}$ includes negative $x$. Correct move: Always write the absolute value around the logarithm argument for $\int \frac{1}{x}dx$, even if the problem doesn't specify the domain, to get full credit on FRQs.
• Wrong move: Writing $\int (2x+1{)}^{2}dx=\frac{(2x+1{)}^3}{3}+C$, without accounting for the inner coefficient of $x$. Why: Students apply the basic power rule directly to composite functions before learning u-substitution. Correct move: Expand polynomial powers first when using only basic integration rules, or wait for u-substitution to integrate non-basic composite functions.
• Wrong move: Adding a separate constant of integration to every term, e.g., $\int 4x^3-2xdx=x^4+C_1-x^2+C_2$. Why: Students think each term needs its own constant. Correct move: Add only one constant of integration $+C$ at the end of the entire antiderivative, since the sum of multiple arbitrary constants is just one arbitrary constant.
• Wrong move: Writing $\int 2^{x}dx=\frac{2^x}{2}+C$. Why: Students mix up the power rule for polynomials with the exponential integration rule. Correct move: Remember $\int a^{x}dx=\frac{a^x}{\ln a}+C$; the denominator is the natural log of the base, not the base itself.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

Which of the following is the general indefinite integral $\int \left(\frac{3x^2-2\sqrt{x}}{x}\right)dx$? (A) $\frac{3}{2}{x}^2-4\sqrt{x}+C$ (B) $3x^2-2\sqrt{x}+C$ (C) $3x-4\sqrt{x}+C$ (D) $x^3-2x^2+C$

Worked Solution: First, simplify the integrand by dividing each term in the numerator by $x$ to rewrite all terms in $x^n$ form: $\frac{3x^2-2\sqrt{x}}{x}=\frac{3x^2}{x}-\frac{2x^{1/2}}{x}=3x-2x^{-1/2}$. Next, apply the sum rule and power rule for integration: $\int (3x-2x^{-1/2})dx=3\int x^{1}dx-2\int x^{-1/2}dx=3\left(\frac{x^2}{2}\right)-2\left(\frac{x^{1/2}}{1/2}\right)+C=\frac{3}{2}{x}^2-4\sqrt{x}+C$. Common distractors: Forgetting to simplify the integrand first leads to option C, incorrect exponent calculation leads to B, and misapplying the quotient rule for integration leads to D. The correct answer is (A).

Question 2 (Free Response)

Let $f^{\prime \prime }(x)=2\cos x+12x^2$. It is known that $f^{\prime }(0)=2$ and $f(0)=5$. (a) Find the general expression for $f^{\prime }(x)$. (b) Find the particular expression for $f^{\prime }(x)$ using the given initial condition. (c) Find the particular expression for $f(x)$ using the given initial condition.

Worked Solution: (a) Integrate $f^{\prime \prime }(x)$ to get the general form of $f^{\prime }(x)$: $$f^{\prime }(x)=\int (2\cos x+12x^2)dx=2\int \cos xdx+12\int x^{2}dx=2\sin x+12\left(\frac{x^3}{3}\right)+C=2\sin x+4x^3+C$$ General form: $f^{\prime }(x)=2\sin x+4x^3+C$

(b) Substitute $f^{\prime }(0)=2$ to solve for $C$: $f^{\prime }(0)=2\sin (0)+4(0{)}^3+C=0+0+C=2⟹C=2$ Particular $f^{\prime }(x)$: $f^{\prime }(x)=2\sin x+4x^3+2$

(c) Integrate $f^{\prime }(x)$ to get $f(x)$, then solve for the second constant: $$f(x)=\int (2\sin x+4x^3+2)dx=2(-\cos x)+4\left(\frac{x^4}{4}\right)+2x+K=-2\cos x+x^4+2x+K$$ Substitute $f(0)=5$: $f(0)=-2\cos (0)+0+0+K=-2(1)+K=5⟹K=7$ Particular $f(x)$: $f(x)=x^4-2\cos x+2x+7$

Question 3 (Application / Real-World Style)

The velocity of a projectile moving along a straight line at time $t$ (in seconds) is given by $v(t)=3t^2-2t+4$ meters per second. The initial position of the projectile at $t=0$ is 10 meters. Find the position function $s(t)$, and calculate the position of the projectile at $t=3$ seconds.

Worked Solution: Velocity is the derivative of position, so position is the antiderivative of velocity: $$s(t)=\int v(t)dt=\int (3t^2-2t+4)dt$$ Apply the power rule to each term: $s(t)=3\cdot \frac{t^3}{3}-2\cdot \frac{t^2}{2}+4t+C=t^3-t^2+4t+C$. Use the initial position condition $s(0)=10$: substituting $t=0$ gives $C=10$, so the position function is $s(t)=t^3-t^2+4t+10$. Evaluate at $t=3$: $s(3)=27-9+12+10=40$. After 3 seconds, the projectile is 40 meters from the starting origin of the coordinate system.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the basic antiderivative rules in this chapter is a non-negotiable prerequisite for every integration topic that follows in AP Calculus AB. Next, you will learn u-substitution, the core technique for integrating composite functions, which relies on your ability to quickly recall and apply basic antiderivative rules after completing the substitution step. Without automatic mastery of these basic rules, you will not be able to focus on the substitution logic, and will make frequent calculation errors on even routine problems. This topic is also the foundation for solving separable differential equations, calculating accumulated change, finding areas between curves, and solving kinematics motion problems—all high-weight content on the AP exam. The follow-on topics that build directly on this chapter are:

• u-substitution for indefinite integrals
• Indefinite integrals and differential equations
• Definite integrals and the Fundamental Theorem of Calculus