Integration and Accumulation of Change
Unit 6 of AP Calculus AB, worth 17–20% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Riemann sums and the definite integral, accumulation functions, the Fundamental Theorem of Calculus, antiderivatives, integration by substitution, integrating with long division and completing the square.
How this unit is tested
Start by identifying what kind of integration question you're facing: a numerical approximation from a table or graph (Riemann sum, trapezoidal sum), a symbolic evaluation (antiderivative, FTC), or a derivative-of-an-integral question (FTC Part 1). Numerical questions almost always give you unevenly or evenly spaced data — read the table carefully, note the subinterval width, and be exact about which x-values a left, right, midpoint, or trapezoidal sum actually uses.
For symbolic integrals, first check whether the integrand is already in a recognizable antiderivative form (power rule, trig, exponential). If not, look for a composite structure that signals u-substitution: an inner function and its derivative sitting nearby (up to a constant multiple). If the integrand is a rational function, check the degrees — if the numerator's degree is greater than or equal to the denominator's, do polynomial long division before anything else. If instead the denominator is an irreducible quadratic that doesn't factor nicely, complete the square to reveal an arctan or log form.
For FTC Part 1 problems (derivative of an accumulation function), always ask whether the upper (or lower) limit is a function of x other than x itself — if so, you need the chain rule, and if the variable appears as the lower limit, remember the sign flip. For FTC Part 2, find any antiderivative, evaluate it at the upper bound minus the lower bound, and don't bother with +C since it cancels.
Work example problems by hand, not just by pattern-matching; the AP exam rewards students who can justify each step (which rule, why it applies) rather than just producing a final numeric answer.
What you have to know
Definite integral as a limit of Riemann sums
∫ from a to b of f(x) dx = lim as n→∞ of Σ_{i=1}^n f(x_i*) Δx, where Δx=(b−a)/n and x_i* is any sample point in the i-th subinterval.
Trapezoidal Rule
∫ from a to b of f(x) dx ≈ (Δx/2)[f(x0) + 2f(x1) + 2f(x2) + ... + 2f(x_{n-1}) + f(x_n)], with n equal subintervals of width Δx.
Fundamental Theorem of Calculus, Part 1
If g(x) = ∫ from a to u(x) of f(t) dt, then g'(x) = f(u(x))·u'(x). If the variable is the lower limit instead, an extra factor of −1 appears.
Fundamental Theorem of Calculus, Part 2
If F is any antiderivative of f on [a,b], then ∫ from a to b of f(x) dx = F(b) − F(a).
u-Substitution
If u=g(x) and du=g'(x) dx, then ∫ f(g(x))g'(x) dx = ∫ f(u) du. For a definite integral, convert the bounds to u-values (or convert the antiderivative back to x before evaluating).
Net Change Theorem
The definite integral of a rate of change gives the total (net) change in the quantity: ∫ from a to b of F'(x) dx = F(b) − F(a).
14 practice questions
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A table gives f(0)=3, f(2)=5, f(4)=8, f(6)=12, f(8)=15. Use a left Riemann sum with 4 equal subintervals to approximate ∫ from 0 to 8 of f(x) dx.
- 56
- 68
- 80
- 28
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Answer. 56
Each subinterval has width Δx=2, and a left sum uses f at x=0,2,4,6: 2(3+5+8+12)=2(28)=56. -
Using the same table (f(0)=3, f(2)=5, f(4)=8, f(6)=12, f(8)=15), use a right Riemann sum with 4 equal subintervals to approximate the integral.
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Answer. 80
A right sum uses f at x=2,4,6,8 with Δx=2: 2(5+8+12+15)=2(40)=80. -
A function has values f(0)=2, f(1)=3, f(2)=5, f(3)=7, f(4)=10, f(5)=13, f(6)=16, f(7)=19, f(8)=22. Approximate ∫ from 0 to 8 of f(x) dx using a midpoint sum with 4 subintervals of width 2.
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Answer. 84
The midpoints of the four subintervals [0,2],[2,4],[4,6],[6,8] are x=1,3,5,7. The sum is 2(f(1)+f(3)+f(5)+f(7))=2(3+7+13+19)=2(42)=84. -
Using the table f(0)=3, f(2)=5, f(4)=8, f(6)=12, f(8)=15, approximate ∫ from 0 to 8 of f(x) dx with a trapezoidal sum using 4 subintervals.
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Answer. 68
Trapezoidal sum: (Δx/2)[f(0)+2f(2)+2f(4)+2f(6)+f(8)] = (2/2)[3+10+16+24+15] = 68. -
A function f is increasing on [a,b]. Which Riemann sum approximation is guaranteed to underestimate ∫ from a to b of f(x) dx, regardless of concavity?
- Left Riemann sum
- Right Riemann sum
- Midpoint sum
- Trapezoidal sum
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Answer. Left Riemann sum
For an increasing function, each left-endpoint rectangle sits entirely below the curve on its subinterval (since the left endpoint gives the smallest function value there), so the left sum always underestimates. The right sum always overestimates for increasing functions; midpoint and trapezoidal results depend on concavity. -
Write the exact right Riemann sum limit definition used to define ∫ from 1 to 3 of x^2 dx.
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Answer. lim as n→∞ of Σ_{i=1}^n (1 + 2i/n)^2 · (2/n)
With Δx=(3−1)/n=2/n and right endpoints x_i = 1 + iΔx = 1+2i/n, the definite integral is the limit of the sum f(x_i)Δx = (1+2i/n)^2(2/n) as n→∞. -
A function f is piecewise linear: on [0,2] it rises from 0 to 4, and on [2,5] it stays constant at 4. Let F(x)=∫ from 0 to x f(t) dt. Find F(5).
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Answer. 16
The region from 0 to 2 is a triangle with area (1/2)(2)(4)=4. The region from 2 to 5 is a rectangle with area 3(4)=12. Adding gives F(5)=4+12=16. -
Let g(x) = ∫ from 3 to x^3 of cos(t) dt. Find g'(x).
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Answer. g'(x) = 3x^2·cos(x^3)
By FTC Part 1 with chain rule, differentiating an accumulation function with upper limit u(x)=x^3 gives f(u(x))·u'(x) = cos(x^3)·3x^2. -
Evaluate ∫ from 0 to 2 of (3x^2 − 4x + 1) dx.
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Answer. 2
An antiderivative is x^3−2x^2+x. Evaluating: at x=2, 8−8+2=2; at x=0, 0. By FTC Part 2, the integral equals 2−0=2. -
Find the general antiderivative of f(x) = 4x^3 + sin(x) − 5.
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Answer. x^4 − cos(x) − 5x + C
Apply the power rule to 4x^3 to get x^4, the standard antiderivative of sin(x) is −cos(x), and the antiderivative of the constant −5 is −5x. Include +C since this is an indefinite integral. -
Evaluate ∫ x√(x^2+5) dx using u-substitution.
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Answer. (1/3)(x^2+5)^{3/2} + C
Let u=x^2+5, so du=2x dx, giving x dx = du/2. The integral becomes (1/2)∫√u du = (1/2)(2/3)u^{3/2}+C = (1/3)u^{3/2}+C, and substituting back gives (1/3)(x^2+5)^{3/2}+C. -
Evaluate ∫ from 0 to 1 of 2x·e^{x^2} dx using u-substitution.
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Answer. e − 1
Let u=x^2, du=2x dx. When x=0, u=0; when x=1, u=1. The integral becomes ∫ from 0 to 1 of e^u du = e^1 − e^0 = e − 1. -
Evaluate ∫ (x^2+1)/(x−2) dx.
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Answer. x^2/2 + 2x + 5 ln|x−2| + C
Since the numerator's degree equals the denominator's would-be quotient case, perform long division: (x^2+1)÷(x−2) = x+2 remainder 5, so the integrand equals x+2+5/(x−2). Integrating term by term gives x^2/2+2x+5 ln|x−2|+C. -
Evaluate ∫ dx/(x^2+6x+13).
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Answer. (1/2) arctan((x+3)/2) + C
Complete the square in the denominator: x^2+6x+13=(x+3)^2+4. This matches the arctan form ∫ du/(u^2+a^2) = (1/a) arctan(u/a)+C with u=x+3 and a=2, giving (1/2) arctan((x+3)/2)+C.
What people get wrong
- Forgetting the chain-rule factor in FTC Part 1 when the bound is a function of x other than x itself — always multiply by the derivative of that bound, not just plug it in.
- Forgetting to flip the sign when the variable is the lower limit of an accumulation function — rewrite ∫ from x to a as −∫ from a to x before differentiating.
- Dropping the +C on indefinite integral answers — it costs points on free-response even though it cancels out in definite integrals.
- Using u-substitution on a definite integral but evaluating the u-antiderivative at the original x-bounds — convert the limits to u-values, or convert the antiderivative back to x before plugging in the original bounds.
- Trying to integrate a rational function directly when the numerator's degree is greater than or equal to the denominator's — perform polynomial long division first to get a polynomial plus a proper fraction.
- Assuming a left Riemann sum always underestimates or a right sum always overestimates — this only holds when the function is increasing; for decreasing functions the roles reverse.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Calculus AB. Free, and no account needed to start.