Definite integrals on the AP Calculus BC exam combine every technique from antidifferentiation with the extra bookkeeping that comes from having actual bounds: converting limits of integration when you substitute, tracking which piece of a piecewise function applies over which part of the interval, and applying the Fundamental Theorem of Calculus — including its chain-rule variant, when the bounds themselves are functions of x rather than just x.
Whenever a definite integral calls for u-substitution or a trigonometric substitution (like x=a sinθ), the bounds must be converted into the new variable at the same time the integrand is — forgetting this is one of the most common ways to get an otherwise-correct integral wrong. Average value problems are just a definite integral divided by the length of the interval, but they get harder when the function is piecewise, since the integral itself needs to be split at the breakpoint first. And whenever the upper (or lower) limit of an integral is a function of x rather than x itself, differentiating that integral requires the chain rule on top of the ordinary Fundamental Theorem of Calculus.
Below are 16 AP Calculus BC-style practice problems covering substitution with changing limits, parametric and trigonometric substitution, average value (including piecewise functions), left Riemann sum approximation from a table, and the Fundamental Theorem of Calculus. 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 BC QBank.
If x = 5cosθ and y = 2sinθ, then ∫2.55 xy dx is equivalent to
Convert the bounds: x=5cosθ=2.5 → cosθ=1/2 → θ=π/3. x=5cosθ=5 → cosθ=1 → θ=0.
Find dx: dx = −5sinθ dθ. Also xy = (5cosθ)(2sinθ) = 10sinθcosθ.
Substitute everything: ∫π/30 10sinθcosθ · (−5sinθ)dθ = ∫π/30 −50sin2θcosθ dθ.
Flip the bounds to remove the negative sign: 50∫0π/3 sin2θcosθ dθ. The correct answer is D.
A continuous function f takes on the values shown in the table below. Estimate ∫030 f(x)dx using a left rectangular approximation with five subintervals.
| x | 0 | 4 | 10 | 16 | 22 | 30 |
|---|---|---|---|---|---|---|
| f(x) | 10 | 14 | 8 | 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, 6, and 8.
[0,4]: width 4, left height 10, area 40. [4,10]: width 6, left height 14, area 84. [10,16]: width 6, left height 8, area 48. [16,22]: width 6, left height 4, area 24. [22,30]: width 8, left height 2, area 16.
Sum all five areas: 40+84+48+24+16 = 212.
The correct answer is C.
Find the value of x at which the function y = x2 reaches its average value on the interval [0,8].
Find the average value: (1/8)∫08x2dx = (1/8)[x3/3]08 = (1/8)(512/3) = 64/3.
Set the function equal to this average value and solve for x: x2 = 64/3 → x = √(64/3).
√(64/3) = 8/√3 ≈ 4.619.
The correct answer is B.
Keep the momentum going.
Explore 1,500+ Free AP Calculus BC ProblemsThe average value of f(x) = x2 for x<3, and f(x) = 5x for x≥3, on the interval 0≤x≤4 is
Split the integral at x=3 to match each piece of f: ∫04f(x)dx = ∫03x2dx + ∫345x dx.
First piece: [x3/3]03 = 27/3 = 9. Second piece: [2.5x2]34 = 2.5(16)−2.5(9) = 40−22.5 = 17.5.
Total: 9+17.5 = 26.5. Average value: 26.5/4 = 6.625.
The correct answer is B.
∫1e2 (ln x)/x dx =
Let u = ln x, so du = dx/x. Convert the bounds: at x=1, u=0; at x=e2, u=2.
The integral becomes ∫02u du = [u2/2]02.
= 4/2 − 0 = 2. The correct answer is B.
∫02 xex dx =
Use integration by parts: let u=x, dv=exdx, so du=dx and v=ex.
∫xexdx = xex − ∫exdx = xex − ex. Evaluate from 0 to 2: [xex−ex]02.
At x=2: 2e2−e2 = e2. At x=0: 0−1 = −1.
Subtract: e2−(−1) = e2+1. The correct answer is B.
6 down, 10 to go — keep practicing.
Explore 1,500+ Free AP Calculus BC Problems∫0π/6 cosθ/(1+3sinθ) dθ =
Let u = 1+3sinθ, so du = 3cosθ dθ. Convert the bounds: at θ=0, u=1; at θ=π/6, u = 1+3(1/2) = 5/2.
The integral becomes ∫15/2(du/3)/u = (1/3)[ln|u|]15/2.
= (1/3)[ln(5/2)−ln1] = (1/3)ln(5/2).
The correct answer is B.
∫√33 u/(u2−1) du =
Let w = u2−1, so dw = 2u du. Convert the bounds: at u=√3, w=3−1=2; at u=3, w=9−1=8.
The integral becomes ∫28(dw/2)/w = (1/2)[ln|w|]28.
= (1/2)[ln8−ln2] = (1/2)ln(8/2) = (1/2)ln4.
Since (1/2)ln4 = ln41/2 = ln2, this simplifies to ln 2. The correct answer is A.
∫√33 u/(u2−1)2 du =
Let w = u2−1, so dw = 2u du. Convert the bounds: at u=√3, w=2; at u=3, w=8.
The integral becomes ∫28(dw/2)/w2 = (1/2)[−1/w]28.
= (1/2)[−1/8−(−1/2)] = (1/2)(−1/8+4/8) = (1/2)(3/8).
= 3/16. The correct answer is C.
9 down, 7 to go.
Explore 1,500+ Free AP Calculus BC Problems∫π/12π/6 cos(3x)/sin2(3x) dx =
Let w = sin(3x), so dw = 3cos(3x)dx. Convert the bounds: at x=π/12, 3x=π/4 → w=√2/2; at x=π/6, 3x=π/2 → w=1.
The integral becomes ∫√2/21(dw/3)/w2 = (1/3)[−1/w]√2/21.
= (1/3)[−1−(−2/√2)] = (1/3)[−1+√2] = (√2−1)/3.
The correct answer is A.
If we let x = 3sinθ, then ∫1.53 √(9−x2)/x dx is equivalent to
With x=3sinθ, dx=3cosθ dθ, and √(9−x2) = √(9−9sin2θ) = 3cosθ (using 1−sin2θ=cos2θ).
Convert the bounds: at x=1.5, sinθ=1/2 → θ=π/6; at x=3, sinθ=1 → θ=π/2.
Substitute everything: ∫ (3cosθ)/(3sinθ) · 3cosθ dθ = ∫3cos2θ/sinθ dθ.
The integral becomes 3∫π/6π/2cos2θ/sinθ dθ. The correct answer is B.
∫0π/2 sin2θcosθ dθ =
Let u = sinθ, so du = cosθ dθ. Convert the bounds: at θ=0, u=0; at θ=π/2, u=1.
The integral becomes ∫01u2du = [u3/3]01.
= 1/3 − 0 = 1/3. The correct answer is B.
12 down, 4 to go — almost there.
Explore 1,500+ Free AP Calculus BC ProblemsIf g(x) is continuous on [p,q] and c is a constant, then ∫pq c·g(x) dx is equal to
A constant multiple can always be pulled outside a definite integral — this is a basic linearity property that holds for any continuous g, regardless of what g actually is.
Choices A, B, and C all try to guess a specific closed-form value using g's endpoint behavior, but none of these are valid in general (g could be any continuous function, not necessarily one with a simple antiderivative expressible via its endpoint values).
The only statement that's always true is ∫pqc· g(x)dx = c∫pqg(x)dx.
The correct answer is D.
d/ds ∫0s √(x4+1) dx =
This is the first part of the Fundamental Theorem of Calculus in its most direct form: d/ds ∫asf(x)dx = f(s), since the upper limit is simply s (no chain rule needed here).
Just replace x with s in the integrand: √(s4+1).
The correct answer is A.
If 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.
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