Definite integrals tie together nearly everything else in AP Calculus AB: they're evaluated using antiderivatives via the Fundamental Theorem of Calculus, approximated numerically with Riemann sums and trapezoids when no closed-form antiderivative is available (or convenient), and used to compute quantities like average value and area between bounds. This unit is where the algebraic techniques of antidifferentiation meet the geometric and numerical side of calculus, and the AP exam draws from all of it — often within a single multi-part question.
A few ideas are worth having rock-solid. The Fundamental Theorem of Calculus comes in two forms that get tested constantly: the first lets you evaluate ∫abf(x)dx as F(b)−F(a) for any antiderivative F, while the second tells you that d/dx ∫axf(t)dt = f(x) — and when the upper limit is itself a function of x rather than just x, the chain rule tags along. Riemann sums (left, right, midpoint) and the trapezoidal rule approximate area using rectangles or trapezoids built from function values at specific points, and recognizing the limit of a Riemann sum as a definite integral (or vice versa) is its own tested skill. Finally, average value is just a definite integral divided by the width of the interval — the same formula whether the function is a polynomial, a semicircle, or piecewise-defined.
Below are 30 AP Calculus AB-style practice problems covering definite integral evaluation, Riemann sums and their limits, the Fundamental Theorem of Calculus (including chain-rule variations), trapezoidal and rectangular approximation, u-substitution with changing limits, and average value. 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.
∫01 (e−x+2)/e−x dx =
Split the fraction first: (e−x+2)/e−x = 1 + 2ex (dividing each term in the numerator by e−x).
Integrate: ∫(1+2ex)dx = x + 2ex.
Evaluate from 0 to 1: (1+2e) − (0+2) = 2e − 1. The correct answer is A.
∫01 ex/(ex+2) dx =
Let u = ex+2, so du = exdx. The integral becomes ∫du/u = ln|u|.
Substitute back and evaluate: [ln(ex+2)]01 = ln(e+2) − ln(0+2) = ln(e+2) − ln 3.
Combine using log rules: ln((e+2)/3). The correct answer is A.
If we let x = tanθ, then ∫01 √(1+x2) dx is equivalent to
With x=tanθ, dx = sec2θ dθ, and √(1+x2) = √(1+tan2θ) = secθ (using the identity 1+tan² = sec²).
Convert the bounds: when x=0, θ=0; when x=1, θ=π/4.
The integral becomes ∫0π/4 secθ · sec2θ dθ = ∫0π/4sec3θ dθ. The correct answer is B.
Keep the momentum going.
Explore 1,500+ Free AP Calculus AB ProblemsIf the substitution u = √(x+4) is used, then ∫05 dx / [x√(x+4)] is equivalent to
With u = √(x+4), we have u2 = x+4 → x = u2−4, and dx = 2u du.
Convert the bounds: when x=0, u=2; when x=5, u=3.
Substitute everything in: ∫ (2u du) / [(u2−4)u] = ∫ 2 du/(u2−4) (the u's cancel).
With the new bounds: ∫23 2 du/(u2−4). The correct answer is A.
The table below shows some values of a continuous function f and its first derivative. Evaluate ∫80 f′(x)dx.
| x | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| f(x) | 10 | 13 | 17 | 11 | 4 |
| f′(x) | 2 | 3 | −1 | −4 | 0 |
By the Fundamental Theorem of Calculus, ∫ab f′(x)dx = f(b) − f(a), no matter what the formula for f actually is.
Here a=8 and b=0: ∫80 f′(x)dx = f(0) − f(8) = 10 − 4.
= 6. The correct answer is A. The f′ column is a distractor — only the f(0) and f(8) values from the table are needed.
Using a midpoint Riemann Sum with 3 equal-width subintervals, find the approximate area under y = 8x−x2 on [0,6].
Three equal subintervals over [0,6] each have width 2: [0,2], [2,4], [4,6], with midpoints x=1, 3, 5.
Evaluate the function at each midpoint: f(1)=8−1=7, f(3)=24−9=15, f(5)=40−25=15.
Sum the areas: (7+15+15) × 2 = 37 × 2 = 74.
The correct answer is C.
6 down, 24 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB ProblemsThe graph of a continuous function f passes through the points (2,4), (5,10), (6,8), and (9,12). Using trapezoids, estimate ∫29 f(x)dx.
Apply the trapezoid area formula, (width) × (average of the two heights), to each of the three consecutive intervals between given points.
[2,5]: width 3, average height (4+10)/2=7: area = 21. [5,6]: width 1, average height (10+8)/2=9: area = 9. [6,9]: width 3, average height (8+12)/2=10: area = 30.
Add the three trapezoid areas: 21+9+30 = 60.
The correct answer is C.
The shaded region under y = 1/x from x=1 to x=4 is equal to ln 4. If we approximate ln 4 using both a left Riemann sum and a right Riemann sum with 3 equal-width subintervals, which inequality follows for just the subinterval from x=3 to x=4?
Since y=1/x is decreasing, the right endpoint value gives an underestimate and the left endpoint value gives an overestimate of the area on any subinterval.
On [3,4] (width 1): the right endpoint gives f(4) = 1/4, and the left endpoint gives f(3) = 1/3.
So the true area on this subinterval is trapped between these: 1/4 < ∫34(1/x)dx < 1/3.
The correct answer is A.
Let A = ∫01 sin x dx. We estimate A using the L, R, and T approximations with n=100 subintervals. Which is true?
On [0,1], sin x is increasing (since it's within the first quarter-period) and concave down (since (sin x)″ = −sin x < 0 for x in (0,π)).
For an increasing function, the left sum underestimates and the right sum overestimates: L < A < R.
The trapezoidal estimate T is always exactly the average of L and R, so T sits between them: L < T < R. And since the curve is concave down, the straight-line trapezoids fall below the curve, so T underestimates the true area: T < A.
Putting it together: L < T < A < R. The correct answer is B.
∫−24 |x| dx =
The graph of |x| forms two right triangles: one from x=−2 to x=0, and one from x=0 to x=4. Each triangle's area is easiest to find geometrically.
From −2 to 0: base 2, height 2 (since |−2|=2), area = (2)(2)/2 = 2. From 0 to 4: base 4, height 4, area = (4)(4)/2 = 8.
Total: 2+8 = 10. The correct answer is C.
10 down, 20 to go.
Explore 1,500+ Free AP Calculus AB Problems∫−43 |x+2| dx =
The expression |x+2| equals zero at x=−2, splitting the integral into two triangular regions there.
From x=−4 to x=−2: base 2, height |−4+2|=2, area = (2)(2)/2 = 2.
From x=−2 to x=3: base 5, height |3+2|=5, area = (5)(5)/2 = 25/2.
Total: 2 + 25/2 = 4/2+25/2 = 29/2. The correct answer is C.
The average value of y = √(36−x2) on its domain is
This function's domain is [−6,6], and its graph is the upper half of a circle of radius 6 (since y = √(36−x2) comes from x2+y2=36, y≥0).
Instead of integrating directly, use the geometric area of a semicircle: ∫−66√(36−x2)dx = (1/2)π(6)2 = 18π.
Average value = (1/(6−(−6))) × 18π = (1/12)(18π) = 1.5π.
The correct answer is B.
The average value of sin x over the interval π/6 ≤ x ≤ π/3 is
The interval width is π/3−π/6 = π/6, so the averaging factor is 1/(π/6) = 6/π.
Find the antiderivative and evaluate: ∫sin x dx = −cos x. [−cos x]π/6π/3 = −cos(π/3)+cos(π/6) = −1/2+√3/2 = (√3−1)/2.
Multiply by the averaging factor: (6/π) × (√3−1)/2 = 3(√3−1)/π.
The correct answer is A.
13 down, 17 to go — keep practicing.
Explore 1,500+ Free AP Calculus AB ProblemsThe average value of sec2x over the interval from x=π/6 to x=π/4 is
The interval width is π/4−π/6 = π/12, giving an averaging factor of 12/π.
Since ∫sec2x dx = tan x: [tan x]π/6π/4 = tan(π/4)−tan(π/6) = 1−1/√3 = (3−√3)/3.
Multiply by the averaging factor: (12/π) × (3−√3)/3 = 4(3−√3)/π = (12−4√3)/π.
The correct answer is A.
Choose the Riemann Sum whose limit is the integral ∫26 x2 dx.
For ∫abf(x)dx, the standard Riemann sum uses Δx = (b−a)/n and right endpoints xk = a+kΔx.
Here a=2, b=6, so Δx = 4/n and xk = 2+4k/n.
The sum is Σ f(xk)Δx = Σ(2+4k/n)2 · (4/n). The correct answer is A.
Choose the Riemann Sum whose limit is the integral ∫0π cos(2x) dx.
Here a=0, b=π, so Δx = π/n and xk = 0+k(π/n) = πk/n.
Substitute into the function: f(xk) = cos(2xk) = cos(2πk/n).
The sum is Σcos(2πk/n) · (π/n). The correct answer is B.
Choose the integral that is the limit of the Riemann Sum limn→∞ Σk=1n sin(2+4k/n) · (4/n).
Compare the sum to the general form Σf(a+kΔx)·Δx. Here the width factor is Δx = 4/n, and the argument of sin is exactly 2+4k/n = a+kΔx with a=2.
Since Δx = (b−a)/n = 4/n and a=2, the upper bound is b = a+4 = 6.
The function being sampled is simply f(x)=sin(x), evaluated directly at each xk=2+4k/n — so this is the Riemann sum for ∫26sin(x)dx.
The correct answer is A.
∫0π/4 sin3α cosα dα is equal to
Let u = sinα, so du = cosα dα. When α=0, u=0; when α=π/4, u=√2/2.
Substitute: ∫0√2/2 u3du = [u4/4]0√2/2.
Compute (√2/2)4 = (2/4)2 = 1/4, so [u4/4] = (1/4)/4 = 1/16 at the upper limit, and 0 at the lower limit.
The result is 1/16 − 0 = 1/16. The correct answer is A.
18 down, 12 to go — over halfway there.
Explore 1,500+ Free AP Calculus AB ProblemsA function f rises in a straight line from (0,0) to (3,12), then stays constant at 12 from x=3 to x=6. Find the average value of f on the interval [0,6].
Find the area under f using geometry. From 0 to 3, f forms a triangle: (3)(12)/2 = 18. From 3 to 6, f forms a rectangle: (3)(12) = 36.
Total area (i.e. ∫06f(x)dx): 18+36 = 54.
Average value: 54/6 = 9. The correct answer is B.
The integral ∫−55 √(25−x2) dx gives the area of
The function y = √(25−x2) comes from solving x2+y2=25 for the non-negative branch of y — this is the upper half of a circle of radius 5.
The bounds −5 to 5 cover the entire diameter of that circle, not just a quarter of it.
So the integral represents the full upper half — a semicircle of radius 5. The correct answer is B.
20 down, 10 to go.
Explore 1,500+ Free AP Calculus AB ProblemsIf ∫06 f′(x)dx = 9, find ∫02 f′(3x)dx.
Let u = 3x, so du = 3 dx → dx = du/3. When x=0, u=0; when x=2, u=6.
Substitute: ∫02f′(3x)dx = ∫06f′(u)·(du/3) = (1/3)∫06f′(u)du.
Substitute the given value: (1/3)(9) = 3. The correct answer is C.
Let g(x) = ∫02x f(t)dt. If f(6) = 5, then g′(3) =
By the Fundamental Theorem of Calculus combined with the chain rule (since the upper limit is 2x, not just x): g′(x) = f(2x) · 2.
Evaluate at x=3: g′(3) = 2f(6).
Substitute the given value: 2(5) = 10. The correct answer is C.
Let h(x) = x2−f(x). If ∫06 f(x)dx = 20, find ∫06 h(x)dx.
Since integration distributes over subtraction: ∫06h(x)dx = ∫06x2dx − ∫06f(x)dx.
Compute the first piece directly: ∫06x2dx = [x3/3]06 = 216/3 = 72.
Subtract the given value: 72 − 20 = 52. The correct answer is B.
d/dt ∫0t √(x2+4) dx =
This is the first part of the Fundamental Theorem of Calculus in its most direct form: d/dt ∫atf(x)dx = f(t), since the upper limit is simply t (no chain rule needed here).
Just replace x with t in the integrand: √(t2+4).
The correct answer is A.
24 down, 6 to go — almost there.
Explore 1,500+ Free AP Calculus AB ProblemsIf F(u) = ∫1u (3−x2)2 dx, then F′(u) is equal to
By the Fundamental Theorem of Calculus, differentiating an integral with respect to its own (simple) upper limit just substitutes that variable into the integrand.
Here the upper limit is u itself (not a function of u composed with something else), so F′(u) = (3−u2)2 — no chain rule is needed.
The correct answer is B.
d/dx ∫π/2x3 √(cos t) dt =
Here the upper limit is x3, a function of x, so the Fundamental Theorem of Calculus must be combined with the chain rule: d/dx ∫ag(x)f(t)dt = f(g(x)) · g′(x).
Here f(t) = √(cos t) and g(x) = x3, so g′(x) = 3x2.
Substitute: √(cos(x3)) · 3x2 = 3x2√(cos(x3)).
The correct answer is B. A common mistake is squaring the exponent inside the cosine (getting x6 instead of x3), which confuses the chain-rule factor with the integrand's own argument.
If x = 4cosθ and y = 5sinθ, then ∫24 xy dx is equivalent to
Convert the bounds: x=4cosθ=2 → cosθ=1/2 → θ=π/3. x=4cosθ=4 → cosθ=1 → θ=0.
Find dx: dx = −4sinθ dθ. Also xy = (4cosθ)(5sinθ) = 20sinθcosθ.
Substitute everything: ∫π/30 20sinθcosθ · (−4sinθ)dθ = ∫π/30 −80sin2θcosθ dθ.
Flip the bounds to remove the negative sign: 80∫0π/3 sin2θcosθ dθ. The correct answer is B.
A continuous function f takes on the values shown in the table below. Estimate ∫025 f(x)dx using a left rectangular approximation with five subintervals.
| x | 0 | 4 | 10 | 16 | 20 | 25 |
|---|---|---|---|---|---|---|
| f(x) | 8 | 12 | 6 | 4 | 2 | 3 |
A left rectangular approximation multiplies each subinterval's width by the function value at its left endpoint. The five subintervals here have widths 4, 6, 6, 4, and 5.
[0,4]: width 4, left height 8, area 32. [4,10]: width 6, left height 12, area 72. [10,16]: width 6, left height 6, area 36. [16,20]: width 4, left height 4, area 16. [20,25]: width 5, left height 2, area 10.
Sum all five areas: 32+72+36+16+10 = 166.
The correct answer is C.
28 down, 2 to go.
Explore 1,500+ Free AP Calculus AB ProblemsFind the value of x at which the function y = x2 reaches its average value on the interval [0,12].
Find the average value: (1/12)∫012x2dx = (1/12)[x3/3]012 = (1/12)(1728/3) = (1/12)(576) = 48.
Set the function equal to this average value and solve for x: x2 = 48 → x = √48.
√48 = √(16·3) = 4√3 ≈ 6.928.
The correct answer is B.
The average value of f(x) = x2 for x<3, and f(x) = 3x for x≥3, on the interval 0≤x≤6 is
Split the integral at x=3 to match each piece of f: ∫06f(x)dx = ∫03x2dx + ∫363x dx.
First piece: [x3/3]03 = 27/3 = 9. Second piece: [1.5x2]36 = 1.5(36)−1.5(9) = 54−13.5 = 40.5.
Total: 9+40.5 = 49.5. Average value: 49.5/6 = 8.25.
The correct answer is B.
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