AP-CALCBC-6-FRQ

U6 FRQ Practice

Master AP Calculus BC Unit 6 free-response questions: learn to attack multi-part FRQs by selecting integration techniques, showing setup, and earning every point.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

What this lesson covers

Unit 6 FRQs rarely ask you to compute a single integral in isolation. Instead, they weave together accumulation functions, the Fundamental Theorem of Calculus, and antidifferentiation techniques—substitution, parts, partial fractions, and improper integrals—into a multi-part story. The graders reward correct setup, correct notation, and correct answers, often on separate lines.

This guide will not re-teach each technique; it assumes you know them from U6.8 through U6.14. Instead, it shows you how to read a Unit 6 FRQ, decide which tool fits, structure your written response so every earned point is visible, and avoid the small slips that cost partial credit. Treat this as a workshop on turning what you know into the maximum score.

Reading the Prompt and Choosing a Technique

The first skill an FRQ tests is diagnosis: given an integrand, which method applies? Rushing into the wrong technique wastes minutes you do not have. Train yourself to scan the integrand and match it to a signal.
Integrand featureTechniqueExample
Inner function with its derivative presentSubstitution2xex2dx\int 2x\,e^{x^2}\,dx
Product of unlike functionsIntegration by partsxcosxdx\int x\cos x\,dx
Proper rational function, factorable denominatorPartial fractions1x21dx\int \frac{1}{x^2-1}\,dx
Degree of numerator \ge denominatorLong division firstx2x+1dx\int \frac{x^2}{x+1}\,dx
Quadratic denominator, no real rootsComplete the square1x2+2x+5dx\int \frac{1}{x^2+2x+5}\,dx
Infinite limit or discontinuityImproper integral (limit)11x2dx\int_1^\infty \frac{1}{x^2}\,dx
A common misconception is that a hard-looking integrand needs a hard technique. Often a quick substitution collapses it. Always check for the simplest tool first. On multi-part questions, one part may hand you an antiderivative you must reuse in a later part, so keep your intermediate work legible and labeled.

Structuring a Multi-Part Response for Points

AP readers score against a rubric with specific point allocations. Each part typically awards points for setup and for the final answer separately, so writing the definite integral expression before evaluating protects at least partial credit.

Write the integral in full definite-integral notation, including dxdx and limits, before you compute. If a part says "set up but do not evaluate," do exactly that—an evaluated answer with a wrong setup earns nothing, while a correct unevaluated setup earns the setup point. When a later part references "the function gg defined in part (a)," restate what gg is so your work is self-contained.

Show the antiderivative explicitly, then substitute limits using the notation [F(x)]ab=F(b)F(a)\left[F(x)\right]_a^b = F(b)-F(a). This line alone often earns a point even if arithmetic later slips. Keep units and context: if the problem describes a rate in liters per minute, your accumulated answer should be stated in liters with a sentence of interpretation when asked. Interpretation points are among the easiest to earn and the most frequently skipped.

Calculator vs. No-Calculator Sections

Unit 6 material appears in both the calculator-active and no-calculator FRQ sections, and the expected work differs sharply.

In the calculator-active section, you may evaluate a definite integral numerically. Write the exact integral setup, then present the numerical result to three decimal places. You do not need to show an antiderivative—the setup plus the numeric answer earns full credit. Store intermediate values rather than rounding early; rounding a value used in a later part propagates error and can cost accuracy points.

In the no-calculator section, you must produce an exact antiderivative by hand. This is where substitution, parts, partial fractions, and completing the square are truly tested. Show every step: the substitution and its differential, the parts choice of uu and dvdv, or the partial-fraction decomposition with solved constants. Leave answers exact, using ln\ln, ee, π\pi, and fractions rather than decimals.
SectionIntegral evaluationAnswer form
No calculatorAntiderivative by handExact (ln3\ln 3, π4\frac{\pi}{4})
Calculator activeNumeric integration allowedDecimal to 3 places

Improper Integrals and Common FRQ Traps

BC-only FRQ parts frequently involve improper integrals (U6.13), and the rubric almost always requires the limit notation. Writing 1f(x)dx\int_1^\infty f(x)\,dx and jumping straight to an answer forfeits the setup point. Instead write limb1bf(x)dx\lim_{b\to\infty}\int_1^b f(x)\,dx, evaluate the antiderivative, then take the limit explicitly.

Several recurring traps cost points. First, dropping the constant or mishandling the differential in substitution—if u=x2u=x^2, then du=2xdxdu=2x\,dx, so a stray factor of 2 must be accounted for. Second, forgetting to change limits of integration after substitution, or converting back to xx inconsistently; either approach is fine, but be consistent. Third, sign errors in integration by parts: the formula is udv=uvvdu\int u\,dv = uv - \int v\,du, and that minus sign is the most common slip. Fourth, treating a discontinuity inside the interval as if the integral were proper—if the integrand blows up at an interior point, the integral must be split. Finally, failing to state divergence: if a limit is infinite, write "the integral diverges" explicitly rather than leaving a blank.

Key terms

Accumulation function.
A function defined as an integral with a variable upper limit, g(x)=axf(t)dtg(x)=\int_a^x f(t)\,dt, whose derivative is f(x)f(x) by the Fundamental Theorem of Calculus.
Integration by parts.
A technique reversing the product rule: udv=uvvdu\int u\,dv = uv - \int v\,du, used for products of unlike functions such as xexx e^x.
Partial fraction decomposition.
Rewriting a proper rational function as a sum of simpler fractions with linear or irreducible quadratic denominators so each can be integrated directly.
Improper integral.
An integral with an infinite limit or an unbounded integrand, evaluated as a limit; it converges to a finite value or diverges.
Setup point.
A rubric point awarded for writing a correct integral expression, independent of whether the evaluation is correct.
u-substitution.
A technique that reverses the chain rule by substituting u=g(x)u=g(x) and du=g(x)dxdu=g'(x)\,dx to simplify an integrand.
Exact form.
An answer left in symbolic terms such as ln2\ln 2, π6\frac{\pi}{6}, or 34\frac{3}{4}, required in the no-calculator section rather than a decimal approximation.

Worked example

A tank is being drained so that water leaves at a rate of R(t)=4tt2+1R(t)=\frac{4t}{t^2+1} liters per minute for 0t30\le t\le 3. (a) Write and evaluate an expression for the total volume of water that leaves the tank during the first 3 minutes. (b) The tank held 20 liters at t=0t=0. How much remains at t=3t=3? (No calculator.)
Part (a): The total volume leaving is the accumulated rate, so set up the definite integral 034tt2+1dt\int_0^3 \frac{4t}{t^2+1}\,dt. Writing this line earns the setup point.

The integrand signals substitution: the denominator t2+1t^2+1 has derivative 2t2t, and the numerator contains a factor of tt. Let u=t2+1u=t^2+1, so du=2tdtdu=2t\,dt, meaning 4tdt=2du4t\,dt = 2\,du.

Change the limits: when t=0t=0, u=1u=1; when t=3t=3, u=10u=10. The integral becomes 1102udu=2lnu110\int_1^{10} \frac{2}{u}\,du = 2\ln|u|\Big|_1^{10}.

Evaluate: 2ln102ln1=2ln100=2ln102\ln 10 - 2\ln 1 = 2\ln 10 - 0 = 2\ln 10 liters. Leave this exact.

Part (b): The remaining volume is the initial amount minus what left: 202ln1020 - 2\ln 10 liters. State it with units. Numerically this is about 15.4 liters, but in the no-calculator section the exact form 202ln1020-2\ln 10 is the required answer.

Practice questions

Which integration technique is most efficient for evaluating 01xe2xdx\int_0^1 x\,e^{2x}\,dx?
  1. u-substitution with u=2xu=2x
  2. Integration by parts with u=xu=x, dv=e2xdxdv=e^{2x}\,dx
  3. Partial fraction decomposition
  4. Completing the square

Answer: Integration by parts with u=xu=x, dv=e2xdxdv=e^{2x}\,dx

The integrand is a product of a polynomial (xx) and an exponential (e2xe^{2x}), the classic signal for integration by parts. Choosing u=xu=x makes du=dxdu=dx simpler, while dv=e2xdxdv=e^{2x}\,dx gives v=12e2xv=\frac{1}{2}e^{2x}. Substitution alone fails because there is no factor matching the derivative of an inner function, and partial fractions and completing the square apply only to rational integrands.
Evaluate the improper integral 11x2dx\int_1^\infty \frac{1}{x^2}\,dx, showing correct limit notation, or state that it diverges.

Answer: The integral converges to 11.

Write it as a limit: limb1bx2dx\lim_{b\to\infty}\int_1^b x^{-2}\,dx. The antiderivative is x1-x^{-1}, so evaluating gives limb[1x]1b=limb(1b+1)\lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^b = \lim_{b\to\infty}\left(-\frac{1}{b}+1\right). As bb\to\infty, 1b0-\frac{1}{b}\to 0, so the value is 11. Omitting the limit notation would forfeit the setup point even though the final number is correct.
A rational function integrand is x2+1x+2\frac{x^2+1}{x+2}. Describe the first step needed before integrating and why.

Answer: Perform polynomial long division first, because the numerator's degree is not less than the denominator's.

Partial fractions require a proper rational function, where the numerator degree is strictly less than the denominator degree. Here the numerator is degree 2 and the denominator is degree 1, so long division is required first. Dividing gives x2+5x+2x-2+\frac{5}{x+2}, which integrates term by term to x222x+5lnx+2+C\frac{x^2}{2}-2x+5\ln|x+2|+C.

FAQ

Do I have to show an antiderivative on calculator-active FRQs?
No. In the calculator-active section you may write the correct definite integral setup and then report the numerical value to three decimal places using your calculator's integration feature. The setup expression plus the numeric answer earns full credit, so you do not need a hand-computed antiderivative there.
How much work do I need to show for full credit?
Enough that a reader can follow your reasoning to the answer. Show the integral setup, the substitution or parts choices, the antiderivative, and the limit evaluation. Bare answers with no supporting work usually earn only the answer point and lose all setup and method points.
What is the most common way students lose points on Unit 6 FRQs?
Skipping the setup line and losing the setup point, dropping the limit notation on improper integrals, sign errors in integration by parts, and forgetting to interpret the answer with correct units when the problem is in context. Writing each step separately protects partial credit.
Will one FRQ combine several integration techniques?
Often, yes. A multi-part FRQ may require substitution in one part, reuse that antiderivative in an accumulation function in another, and ask for interpretation in a third. Read all parts before starting so you recognize how the setup in early parts feeds later ones.

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The Crimsora tutor teaches U6 FRQ Practice live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.