Lesson 26: Antiderivatives & Indefinite Integrals

(a) Opening Question

You already know that the derivative of x² is 2x. Now run the machine backward.

Find a function whose derivative is 2x. Then find another one. How many are there?

Take a minute before reading on.

The obvious answer is x², since d/dx[x²] = 2x. But x² + 1 also works, because the derivative of the constant 1 is 0. So does x² − 7, and x² + 1000. In fact, every function of the form x² + C, where C is any constant, has derivative 2x — and it turns out these are the only functions that do.

This is the central idea of this lesson. Differentiation sends a whole family of functions (all the x² + C) to a single derivative (2x). Going backward, then, we never recover just one function — we recover the entire family. That little + C is not optional decoration; it is the honest answer to the question "which function?" When you run differentiation in reverse, you are doing antidifferentiation, and the family you get is the set of antiderivatives.

(b) Core Concepts

The antiderivative

A function F is an antiderivative of f on an interval if F'(x) = f(x) for every x in that interval. So finding an antiderivative means answering: what function has f as its derivative?

Differentiation and antidifferentiation are inverse processes:

              differentiate
    F(x)  ───────────────────▶  f(x)
          ◀───────────────────
              antidifferentiate

Because the derivative of a constant is zero, if F is one antiderivative of f, then F(x) + C is also an antiderivative for any constant C. A theorem from Unit 5 (a consequence of the Mean Value Theorem) guarantees there are no others: if two functions have the same derivative on an interval, they differ by a constant. So the complete answer to "what are the antiderivatives of f?" is a one-parameter family.

The constant C is called the constant of integration. It is there because antidifferentiation loses information — the derivative throws away constants, and we have no way to recover the specific one without more information (an initial condition, coming later).

Indefinite integral notation

We write the family of all antiderivatives of f using the indefinite integral:

∫ f(x) dx = F(x) + C,   where F'(x) = f(x)

Read this as "the integral of f of x, dee x." The pieces:

• ∫ is the integral sign.
• f(x) is the integrand — the function being antidifferentiated.
• dx tells you the variable of integration (here, x). Do not drop it; it is part of the notation and it tells you what is "varying."
• + C is the constant of integration, included on every indefinite integral.

The word indefinite signals that the answer is a family of functions, not a single number. (Definite integrals, which produce numbers, come in Lesson 28.)

The basic antiderivative rules

Every rule below is just a differentiation rule read in reverse. Verify any of them by differentiating the right side.

The power rule for integrals is the workhorse:

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C,   for n ≠ −1

The mechanics: add one to the exponent, then divide by the new exponent. This works for negative and fractional exponents too, as long as you first rewrite radicals and quotients as powers: √x = x^(1/2), 1/x³ = x⁻³.

The n = −1 special case. The power rule fails when n = −1, because adding one gives exponent 0 and you would divide by 0. That single case is patched by the logarithm:

∫ x⁻¹ dx = ∫ (1/x) dx = ln|x| + C

The absolute value matters: 1/x is defined for negative x too, and d/dx[ln|x|] = 1/x on both sides of zero, so ln|x| (not ln x) is the correct antiderivative.

Constant-multiple and sum/difference rules

Antidifferentiation is linear, exactly like differentiation:

∫ k·f(x) dx = k ∫ f(x) dx
∫ [f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ g(x) dx

Constants pull out front, and you integrate sums term by term. (You only need one + C at the very end — combining all the individual constants into a single constant.) There is no product rule or quotient rule for integrals. If you face a product or a quotient, you must rewrite first.

Rewrite before integrating

The integrand must be a sum of power/trig/exponential terms before the rules apply. Common rewrites:

• Split a fraction over its denominator: (x³ − 4x + 1)/x² = x − 4·x⁻¹ + x⁻².
• Expand a product: (x² + 1)² = x⁴ + 2x² + 1.
• Rewrite radicals as powers: 1/√x = x^(−1/2).

Verify by differentiating

This is the most reliable check in all of calculus: after you integrate, differentiate your answer. If you get the integrand back, you are right. If not, you made an error. The AP exam rewards this discipline — and so does your future self.

Particular antiderivatives and initial conditions

A particular antiderivative is one specific member of the family, pinned down by an initial condition — a known value F(a) = b. The procedure:

1. Antidifferentiate to get the general family F(x) + C.
2. Substitute the initial condition and solve for C.
3. Write the final answer with the specific C (no longer a free constant).

Mini-example. If f'(x) = 6x² − 4x + 1 and f(1) = 2, then f(x) = 2x³ − 2x² + x + C. Substituting: f(1) = 2 − 2 + 1 + C = 1 + C = 2, so C = 1, giving f(x) = 2x³ − 2x² + x + 1. (Check: f'(x) = 6x² − 4x + 1. ✓)

Rectilinear motion: a → v → s

Antiderivatives recover motion. If a particle moves along a line with acceleration a(t), then antidifferentiating once gives velocity and again gives position, each time using an initial condition to fix the constant:

v(t) = ∫ a(t) dt   (use v(0) to find C)
s(t) = ∫ v(t) dt   (use s(0) to find C)

Each integration introduces one constant, so you need one initial condition per integration.

(c) Worked Examples

Example 1 — Power rule with radicals and negative powers (NO CALC)

Problem. Find ∫ (3x⁴ − 2/x³ + √x) dx.

Strategy. Rewrite every term as a power of x, then apply the power rule term by term.

Solution. Rewrite: 3x⁴ − 2x⁻³ + x^(1/2). Integrate each term (add one to the exponent, divide by the new exponent):

• ∫ 3x⁴ dx = 3 · x⁵/5 = (3/5)x⁵
• ∫ −2x⁻³ dx = −2 · x⁻²/(−2) = x⁻² = 1/x²
• ∫ x^(1/2) dx = x^(3/2)/(3/2) = (2/3)x^(3/2)

∫ (3x⁴ − 2/x³ + √x) dx = (3/5)x⁵ + 1/x² + (2/3)x^(3/2) + C

Verify. Differentiate: 3x⁴ + (−2)x⁻³ + (2/3)(3/2)x^(1/2) = 3x⁴ − 2/x³ + √x. ✓ Note the −2/x³ term: integrating x⁻³ gives x⁻²/(−2), and the leading −2 cancels the −2 in the denominator to give +1/x².

Example 2 — Rewrite a fraction, then integrate (NO CALC)

Problem. Find ∫ (x³ − 4x + 1)/x² dx.

Strategy. You cannot integrate a quotient directly. Split the single fraction into separate terms over x², simplify each, and watch for the 1/x term.

Solution. Divide each term in the numerator by x²:

(x³ − 4x + 1)/x² = x³/x² − 4x/x² + 1/x² = x − 4·(1/x) + x⁻²

Now integrate term by term. The middle term is the n = −1 case:

• ∫ x dx = x²/2
• ∫ −4·(1/x) dx = −4 ln|x|
• ∫ x⁻² dx = x⁻¹/(−1) = −1/x

∫ (x³ − 4x + 1)/x² dx = x²/2 − 4 ln|x| − 1/x + C

Verify. d/dx[x²/2 − 4 ln|x| − 1/x] = x − 4/x + 1/x² = (x³ − 4x + 1)/x². ✓ The middle term is the trap: −4/x is not a power-rule term — it integrates to a logarithm.

Example 3 — Trig and exponential antiderivatives (NO CALC)

Problem. Find ∫ (4 cos x − 3 sec²x + 2eˣ) dx.

Strategy. Each term matches a basic rule directly. Pull constants out front and watch the sign on the trig pieces.

Solution.

• ∫ 4 cos x dx = 4 sin x
• ∫ −3 sec²x dx = −3 tan x
• ∫ 2eˣ dx = 2eˣ

∫ (4 cos x − 3 sec²x + 2eˣ) dx = 4 sin x − 3 tan x + 2eˣ + C

Verify. d/dx[4 sin x − 3 tan x + 2eˣ] = 4 cos x − 3 sec²x + 2eˣ. ✓ (Recall d/dx[tan x] = sec²x.) Had the first term been sin x instead of cos x, the sign would flip: ∫ sin x dx = −cos x + C.

Example 4 — Rectilinear motion: acceleration to position (NO CALC)

Problem. A particle moves along a line with acceleration a(t) = 6t − 4 (units: m/s²). At t = 0 its velocity is 2 m/s and its position is 5 m. Find the position function s(t).

Strategy. Antidifferentiate a(t) to get v(t), using v(0) = 2 to find the constant. Then antidifferentiate v(t) to get s(t), using s(0) = 5.

Solution. First integration:

v(t) = ∫ (6t − 4) dt = 3t² − 4t + C₁

Apply v(0) = 2: 3(0)² − 4(0) + C₁ = 2, so C₁ = 2, giving v(t) = 3t² − 4t + 2.

Second integration:

s(t) = ∫ (3t² − 4t + 2) dt = t³ − 2t² + 2t + C₂

Apply s(0) = 5: 0 − 0 + 0 + C₂ = 5, so C₂ = 5. Therefore:

s(t) = t³ − 2t² + 2t + 5

Verify. s'(t) = 3t² − 4t + 2 = v(t) ✓ and v'(t) = 6t − 4 = a(t) ✓, with s(0) = 5 and v(0) = 2 matching the initial conditions.

(d) Common Mistakes

Omitting + C (the #1 error). Writing ∫ 2x dx = x² is incomplete. The indefinite integral is a family of functions, so the answer is x² + C. On the AP exam this drops easy points. Fix: make + C the last thing you write on every indefinite integral, automatically.

Using the power rule when n = −1. Students write ∫ (1/x) dx = ∫ x⁻¹ dx = x⁰/0, which is undefined. Fix: memorize the special case ∫ (1/x) dx = ln|x| + C. Whenever you see an exponent of −1 (an x in the denominator to the first power), the answer is a logarithm — and keep the absolute value.

Sign error on ∫ sin x dx. Many write ∫ sin x dx = cos x + C. But d/dx[cos x] = −sin x, the wrong sign. Fix: ∫ sin x dx = −cos x + C. Check by differentiating: d/dx[−cos x] = sin x. ✓ (Memory hook: integrating introduces the minus sign that the cosine derivative removes.)

Not rewriting before integrating. Students try to integrate products and quotients term-as-written — e.g., treating ∫ (x² + 1)/x dx as if the rules apply directly. There is no product or quotient rule for integrals. Fix: always split fractions and expand products into a sum of powers first. Here: (x² + 1)/x = x + 1/x, so the integral is x²/2 + ln|x| + C.

Forgetting to solve for C with the initial condition. After finding F(x) + C, students stop and report the general family even when given F(a) = b. Fix: an initial-condition problem demands a particular answer — substitute, solve for C, and write the specific function with the number plugged in.

(e) Practice Problems

12 (short answer). [NO CALC] Find ∫ (x³ − 4x + 1)/x² dx. Show the rewrite step.

13 (short answer, justification). [NO CALC] A student writes ∫ (1/x) dx = x⁰/0 + C. Identify the error and give the correct antiderivative. Explain why the power rule does not apply here.

14 (short answer, justification/interpretation). [NO CALC] Acceleration of a particle is a(t) = 4 − 2t (m/s²), with v(0) = 1 m/s. Find v(t), and explain the meaning of the constant of integration in this context.

15 (short answer, initial condition). [NO CALC] If f'(x) = 6√x + 1/x² and f(1) = 0, find the particular function f(x).

(f) AP Exam Focus

This is a non-calculator free-response question (AP Section II, Part B style). Budget about 15 minutes. Show all antiderivative work and label units where appropriate.

FRQ. A particle moves along the x-axis. Its acceleration at time t ≥ 0 seconds is given by a(t) = 6t − 4 (in meters per second squared). At time t = 0, the particle has velocity v(0) = 2 meters per second and position s(0) = 5 meters.

(a) Find the velocity function v(t). (2 points)

(b) Show that the particle is never at rest for t ≥ 0, and state whether the particle is moving in the positive or negative direction. Justify your answer. (3 points)

(c) Find the position function s(t). (2 points)

(d) Find the average velocity of the particle over the interval 0 ≤ t ≤ 3, and explain what this value represents about the particle's motion. (2 points)

Total: 9 points

Model Solution

(a) Velocity is the antiderivative of acceleration:

v(t) = ∫ (6t − 4) dt = 3t² − 4t + C₁

Use the initial condition v(0) = 2: 3(0)² − 4(0) + C₁ = 2, so C₁ = 2.

v(t) = 3t² − 4t + 2   (m/s)

(b) The particle is at rest when v(t) = 0. Solving 3t² − 4t + 2 = 0, the discriminant is (−4)² − 4(3)(2) = 16 − 24 = −8 < 0, so there are no real solutions. Therefore v(t) = 0 has no solution, and the particle is never at rest for t ≥ 0. Since v is continuous with no zeros and the leading coefficient 3 > 0 (so the parabola opens upward and never crosses the axis), v(t) > 0 for all t ≥ 0. (As a check, v(0) = 2 > 0.) Because the velocity is always positive, the particle is always moving in the positive direction.

(c) Position is the antiderivative of velocity:

s(t) = ∫ (3t² − 4t + 2) dt = t³ − 2t² + 2t + C₂

Use the initial condition s(0) = 5: 0 − 0 + 0 + C₂ = 5, so C₂ = 5.

s(t) = t³ − 2t² + 2t + 5   (m)

(d) Average velocity over [0, 3] is the change in position divided by the change in time:

average velocity = [s(3) − s(0)] / (3 − 0)

Compute s(3) = 27 − 18 + 6 + 5 = 20 and s(0) = 5. So:

average velocity = (20 − 5)/3 = 15/3 = 5 m/s

Interpretation: Over the first 3 seconds, the particle's net displacement is 15 meters in the positive direction, so on average it moved at 5 meters per second. Because the velocity is always positive (part b), this average reflects steady forward motion with no backtracking.

Scoring Commentary — where students lose points

• (a) 1 point for the correct antiderivative 3t² − 4t + 2; 1 point for using v(0) = 2 to find C₁ = 2. Trap: writing v(t) = 3t² − 4t + C₁ and never solving for C₁ — that loses the second point, and also breaks parts (b)–(d).
• (b) 1 point for setting v(t) = 0; 1 point for the discriminant (or equivalent) showing no real solution; 1 point for justifying the direction via the sign of v. Trap: stating "the particle moves forward" without showing v > 0 — direction claims must be justified by the sign of velocity, not asserted.
• (c) 1 point for the antiderivative; 1 point for C₂ = 5. Trap: integrating a(t) instead of v(t), or forgetting + C₂ and so missing the initial-condition point.
• (d) 1 point for the average-velocity computation 5 m/s; 1 point for a correct interpretation in context (net displacement over elapsed time). Trap: confusing average velocity (displacement ÷ time) with average speed or with the average value of acceleration. On the AP rubric, units and a context-grounded sentence are what earn the interpretation point.