AP Calculus AB: Antidifferentiation
Antidifferentiation reverses the process you've been building fluency with all along: given a derivative, find the function it came from. This free, complete lesson covers basic antiderivative formulas, u-substitution, logarithmic and arcsin/arctan forms, and applications to initial-value problems and motion along a line. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.
1. Basic Antiderivative Formulas
The antiderivative (or indefinite integral) of f(x) is a function F(x) whose derivative is f(x), written ∫f(x) dx = F(x) + C. A constant factor pulls out of the integral, and the integral of a sum is the sum of the integrals:
| Rule | Formula |
|---|---|
| Power Rule | ∫un du = un+1n + 1 + C (n ≠ −1) |
| Cosine | ∫cos u du = sin u + C |
| Sine | ∫sin u du = −cos u + C |
| Secant squared | ∫sec² u du = tan u + C |
| Exponential | ∫eu du = eu + C |
Worked Example: Find ∫(3x⁴ − 2x + 7) dx.
3x⁵5 − 2x²2 + 7x + C = 35x⁵ − x² + 7x + C
1. Find ∫7x³ dx.
Explanation: By the Power Rule: 7 · x⁴4 + C = 74x⁴ + C.
2. Find ∫(4 − 6x + 3x²) dx.
Explanation: Integrating term by term: 4x − 6(x²/2) + 3(x³/3) + C = 4x − 3x² + x³ + C.
3. Find ∫cos(3x) dx.
Explanation: The inside function 3x has derivative 3, so dividing by that factor: 13sin(3x) + C.
4. Find ∫sec²(5x) dx.
Explanation: Dividing by the inside function's derivative (5): 15tan(5x) + C.
5. Find ∫e4x dx.
Explanation: Dividing by the inside function's derivative (4): 14e4x + C.
2. U-Substitution
When an integral contains a function and (a multiple of) its own derivative, let u equal the "inside" function, compute du, then rewrite the whole integral in terms of u before applying a basic formula.
Worked Example: Find ∫3(2 − 5x)⁴ dx.
Let u = 2 − 5x, so du = −5 dx.
3∫u⁴ · du−5 = −35 · u⁵5 + C = −325(2 − 5x)⁵ + C
1. Find ∫4(1 − 3x)³ dx.
Explanation: u = 1 − 3x, du = −3 dx. 4∫u³(du/−3) = −43 · u⁴4 + C = −13(1 − 3x)⁴ + C.
2. Find ∫(3x² − 1)⁴ · x dx.
Explanation: u = 3x² − 1, du = 6x dx, so x dx = du/6. ∫u⁴(du/6) = 16 · u⁵5 + C = 130(3x² − 1)⁵ + C.
3. Find ∫x√(4 − x²) dx.
Explanation: u = 4 − x², du = −2x dx, so x dx = du/−2. ∫u1/2(du/−2) = −13u3/2 + C = −13(4 − x²)3/2 + C.
4. Find ∫sin²x · cos x dx.
Explanation: u = sin x, du = cos x dx. ∫u² du = u³3 + C = 13sin³x + C.
5. Find ∫x²ex³ dx.
Explanation: u = x³, du = 3x² dx, so x² dx = du/3. ∫eu(du/3) = 13eu + C = 13ex³ + C.
3. Logarithmic and Arcsin/Arctan Forms
∫duu = ln|u| + C ∫du√(a² − u²) = sin⁻¹ua + C ∫dua² + u² = 1atan⁻¹ua + C
Worked Example: Find ∫xx² + 9 dx.
u = x² + 9, du = 2x dx, so x dx = du/2.
12∫duu = 12ln|x² + 9| + C
1. Find ∫2xx² − 4 dx.
Explanation: u = x² − 4, du = 2x dx, matching the numerator exactly: ∫du/u = ln|x² − 4| + C.
2. Find ∫dxx² + 16.
Explanation: Here a² = 16, so a = 4: 14tan⁻¹x4 + C.
3. Find ∫dx√(25 − x²).
Explanation: Here a² = 25, so a = 5: sin⁻¹x5 + C.
4. Find ∫tan(2x) dx.
Explanation: ∫tan u du = ln|sec u| + C with u = 2x; dividing by the inside derivative (2): 12ln|sec(2x)| + C.
5. Find ∫dxx + 7.
Explanation: u = x + 7, du = dx, so ∫du/u = ln|x + 7| + C.
4. Applications: Initial-Value Problems and Motion
Given f'(x) and a single known point (a, f(a)), antidifferentiate to find the general form (with an unknown constant C), then substitute the known point to solve for C. This same idea extends directly to motion: antidifferentiate acceleration to get velocity, and velocity to get position, using an initial condition at each stage.
Worked Example: Find f(x) if f'(x) = 4x³ and f(1) = 6.
f(x) = ∫4x³ dx = x⁴ + C.
f(1) = 1 + C = 6, so C = 5. Thus f(x) = x⁴ + 5.
1. Find f(x) if f'(x) = 6x² and f(2) = 10.
Explanation: f(x) = 2x³ + C. f(2) = 16 + C = 10, so C = −6. f(x) = 2x³ − 6.
2. Find a curve whose slope at each point (x, y) equals 1x², if the curve passes through (1, 3).
Explanation: dy/dx = x⁻², so y = ∫x⁻² dx = −x⁻¹ + C = −1/x + C. At (1, 3): 3 = −1 + C, so C = 4. y = −1/x + 4.
3. A particle moves along a line with velocity v(t) = 6t² − 4t. If x(0) = 5, find the position function x(t).
Explanation: x(t) = ∫(6t² − 4t) dt = 2t³ − 2t² + C. x(0) = 0 + C = 5, so C = 5.
4. Using x(t) = 2t³ − 2t² + 5 from the previous problem, find the particle's position at t = 2.
Explanation: x(2) = 2(8) − 2(4) + 5 = 16 − 8 + 5 = 13.
5. A particle's acceleration is a(t) = 6t − 4, and v(0) = 3. Find the velocity function v(t).
Explanation: v(t) = ∫(6t − 4) dt = 3t² − 4t + C. v(0) = 0 + C = 3, so C = 3.
Common Mistakes to Avoid
- Forgetting the constant of integration, C. Every indefinite integral needs it, dropping it is one of the most common and easily avoidable point losses.
- Forgetting to divide by the "extra" constant factor when the inside function's derivative isn't exactly 1. If u = 3x, du = 3 dx, not just dx, always account for that factor before applying the basic formula.
- Choosing u incorrectly for substitution. Look for a function whose derivative (up to a constant multiple) also appears in the integral, that's the signal for what to let u equal.
- Mixing up the arcsin and arctan forms. The arcsin form has a square root and a minus sign under it (√(a² − u²)); the arctan form has no square root and a plus sign (a² + u²). Confusing these leads to entirely the wrong antiderivative.
- Substituting the initial condition before finishing the general antiderivative. Always integrate completely first, including any leftover constant multiples, before plugging in the known point to solve for C.
Frequently Asked Questions
How can I tell when a problem needs u-substitution instead of a direct formula?
If the integrand isn't a basic function of x alone, but instead involves a composite function (like (2 − 5x)⁴ or ex³) together with something resembling its derivative, that's the signal to try substitution.
Why does the antiderivative of 1/u involve an absolute value, but the derivative of ln u doesn't need one?
ln u is only defined for positive u, but 1/u is defined for any nonzero u, including negative values. Using ln|u| extends the antiderivative to work correctly on both sides of 0.
What's the difference between finding f(x) from f'(x) and solving a motion problem?
They use exactly the same technique, antidifferentiate, then use a given condition to solve for C. A motion problem is simply this same process applied twice in a row: once from acceleration to velocity, and again from velocity to position, each stage getting its own initial condition.
Where can I practice more problems like these?
The Antidifferentiation quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.