Integration is where AP Calculus AB shifts from finding rates of change to accumulating them, and the exam tests it across a wide range of formats: basic power-rule antiderivatives, definite integrals evaluated with the Fundamental Theorem of Calculus, u-substitution, average value of a function, and integrals built from piecewise or abstract functions. A strong integration toolkit means recognizing quickly which technique a given integrand calls for — direct power-rule integration, a trig identity, a substitution that simplifies a composite function, or a geometric shortcut like recognizing an odd function integrated over a symmetric interval.
Several of the trickiest AP integration questions aren't really about computing an antiderivative at all — they're about applying the right theorem correctly. The Fundamental Theorem of Calculus connects an integral of f directly to any antiderivative F, so a question can hand you F′(x) = f(x) and ask for a definite integral without ever giving you an explicit formula for f. The average value formula turns a definite integral into a single representative height for a function over an interval. And substitution-based integrals, where the limits of integration change along with the variable, test whether you can track that change of variable correctly all the way to the final numeric answer.
Below are 14 AP Calculus AB-style practice problems covering definite and indefinite integrals, u-substitution, average value, the Fundamental Theorem of Calculus, and integrals of piecewise functions. Work through each problem, select your answer, and check whether you got it right — every problem includes a complete, step-by-step explanation. Once you've worked through these, keep building your skills with thousands more official-style questions in our free AP Calculus AB QBank.
∫03 (x+1)1/2 dx =
Apply the power rule for integration: ∫(x+1)1/2dx = (2/3)(x+1)3/2 + C (the inner function x+1 has derivative 1, so no extra chain-rule factor is needed).
Evaluate from 0 to 3: (2/3)(3+1)3/2 − (2/3)(0+1)3/2 = (2/3)(4)3/2 − (2/3)(1)3/2
(4)3/2 = (√4)3 = 23 = 8, so this becomes (2/3)(8) − (2/3)(1) = 16/3 − 2/3 = 14/3.
The correct answer is D.
∫12 x−3 dx =
Apply the power rule: ∫x−3dx = x−2/(−2) + C = −1/(2x2) + C.
Evaluate from 1 to 2: [−1/(2x2)]12 = −1/(2·4) − (−1/(2·1)) = −1/8 + 1/2
= −1/8 + 4/8 = 3/8.
The correct answer is C.
∫01/2 2x / √(1−x2) dx =
Let u = 1−x2, so du = −2x dx, meaning 2x dx = −du. The integral becomes ∫ −du/√u = −2√u + C = −2√(1−x2) + C.
Evaluate from x=0 to x=1/2: [−2√(1−x2)]01/2 = −2√(1−0.25) − (−2√(1−0))
= −2√0.75 + 2√1 = −2·(√3/2) + 2 = −√3 + 2 = 2 − √3.
The correct answer is D.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB ProblemsThe average value of √x over the interval 0 ≤ x ≤ 2 is
The average value formula is (1/(b−a)) ∫ab f(x) dx. Here a=0, b=2, and f(x)=√x=x1/2.
Find the antiderivative: ∫x1/2dx = (2/3)x3/2. Evaluate from 0 to 2: (2/3)(2)3/2 − 0 = (2/3)·2√2 = (4/3)√2.
Multiply by 1/(b−a) = 1/2: (1/2)·(4/3)√2 = (2/3)√2.
The correct answer is C.
If F and f are continuous functions such that F′(x) = f(x) for all x, then ∫ab f(x) dx is
This is exactly the statement of the Fundamental Theorem of Calculus: if F is any antiderivative of f (meaning F′=f), then the definite integral of f from a to b equals F evaluated at the upper limit minus F evaluated at the lower limit.
∫ab f(x) dx = F(b) − F(a)
This holds no matter what specific formula f or F have — the theorem doesn't require knowing them explicitly, only that F′=f.
The correct answer is D. Choices A and B use F′ instead of F, which doesn't match the theorem's statement.
∫0π/4 tan2x dx =
There's no direct antiderivative rule for tan2x, so rewrite it using the Pythagorean identity tan2x = sec2x − 1.
Integrate term by term: ∫(sec2x − 1) dx = tan x − x + C.
Evaluate from 0 to π/4: (tan(π/4) − π/4) − (tan(0) − 0) = (1 − π/4) − 0
= 1 − π/4. The correct answer is B.
If ∫−22 (x7+k) dx = 16, then k =
Split the integral into two parts: ∫−22 x7 dx + ∫−22 k dx.
Since x7 is an odd function (symmetric about the origin) and the interval [−2,2] is symmetric, the positive and negative areas cancel exactly: ∫−22 x7 dx = 0.
This leaves ∫−22 k dx = k · (2−(−2)) = 4k. Set this equal to 16: 4k = 16 → k = 4.
The correct answer is D.
7 down, 7 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB Problems∫08 dx / √(1+x) =
Rewrite the integrand as a power: 1/√(1+x) = (1+x)−1/2. Apply the power rule (the inner function 1+x has derivative 1, so no chain-rule adjustment is needed):
∫(1+x)−1/2dx = 2(1+x)1/2 + C = 2√(1+x) + C
Evaluate from 0 to 8: 2√(1+8) − 2√(1+0) = 2√9 − 2√1 = 2(3) − 2(1) = 6 − 2
= 4. The correct answer is D.
Given f(x) = x+1 for x < 0, and f(x) = cosπx for x ≥ 0, find ∫−11 f(x) dx.
Since f is piecewise, split the integral at x=0 to match each piece: ∫−11 f(x) dx = ∫−10(x+1) dx + ∫01 cos(πx) dx.
First piece: antiderivative is x2/2 + x. Evaluate −1 to 0: (0+0) − (1/2 − 1) = 0 − (−1/2) = 1/2.
Second piece: antiderivative is sin(πx)/π. Evaluate 0 to 1: sin(π)/π − sin(0)/π = 0 − 0 = 0.
Add the two results: 1/2 + 0 = 1/2. The correct answer is D.
∫01 √(x2−2x+1) dx is
Recognize that x2−2x+1 = (x−1)2 is a perfect square, so √(x2−2x+1) = √((x−1)2) = |x−1| (a square root always returns the non-negative value, so an absolute value is required, not just x−1 itself).
On the interval [0,1], x−1 is always ≤ 0, so |x−1| = 1−x on this whole interval.
Integrate: ∫01(1−x) dx = [x − x2/2]01 = (1 − 0.5) − 0 = 0.5.
The correct answer is C. Skipping the absolute value and integrating √((x−1)2) as if it were simply (x−1) is a common error that leads to the wrong sign.
What is the average (mean) value of 3t3−t2 over the interval −1 ≤ t ≤ 2?
The average value formula is (1/(b−a)) ∫ab f(t) dt, with a=−1, b=2, giving a factor of 1/3 out front.
Find the antiderivative: ∫(3t3−t2) dt = (3/4)t4 − (1/3)t3.
Evaluate at t=2: (3/4)(16) − (1/3)(8) = 12 − 8/3 = 28/3. Evaluate at t=−1: (3/4)(1) − (1/3)(−1) = 3/4 + 1/3 = 13/12.
Subtract: 28/3 − 13/12 = 112/12 − 13/12 = 99/12 = 33/4. Multiply by the averaging factor 1/3: (1/3)(33/4) = 11/4.
The correct answer is A.
11 down, 3 to go.
Explore 1,500+ Free AP Calculus AB Problems∫ sin(2x+3) dx =
The antiderivative of sin(u) is −cos(u), but since the inner function is 2x+3 (derivative 2, not 1), a correction factor is needed.
Let u = 2x+3, so du = 2 dx, meaning dx = du/2. The integral becomes ∫ sin(u) · (1/2) du = −(1/2)cos(u) + C.
Substitute back: −(1/2)cos(2x+3) + C.
The correct answer is D. Forgetting the 1/2 correction factor (choice C) is the most common mistake on this type of problem.
∫ (x3−3x) dx =
Integrate term by term using the power rule, ∫xndx = xn+1/(n+1).
∫x3dx = x4/4.
∫−3x dx = −3 · x2/2 = −(3/2)x2.
Combining both terms: x4/4 − (3/2)x2 + C. The correct answer is D. Choice A shows the derivative of the original expression rather than its integral — a common mix-up under time pressure.
If ∫12 f(x−c) dx = 5, where c is a constant, then ∫1−c2−c f(x) dx =
Use substitution on the original integral: let u = x−c, so du = dx. When x=1, u = 1−c; when x=2, u = 2−c.
Rewriting the original integral entirely in terms of u: ∫12 f(x−c) dx = ∫1−c2−c f(u) du.
This is exactly the integral the question is asking about (just with u relabeled as x, which doesn't change its value) — it's the same integral as the one already given to equal 5.
The correct answer is B, 5. No further computation involving c is needed, since the substitution shows the two integrals are identical.
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