Practice — Integration and Accumulation of Change

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ftc-part1 chain-rule-with-variable-bounds accumulation-functions Calculator not allowed Difficulty 3/5

If \(g(x) = \int_{1}^{x^{2}} \sqrt{t}\,dt\), then \(g'(x) =\)

A. x

B. 2x^{2}

C. 2x|x|

D. \sqrt{x^{2}}

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Answer: C

Apply FTC Part 1 with the chain rule. The upper bound is \(u(x) = x^2\), so \(u'(x) = 2x\). \[g'(x) = \sqrt{x^2} \cdot 2x = |x| \cdot 2x = 2x|x|\] **Why the distractors fail:** - **(A) \(x\):** Ignores both the chain rule and the integrand evaluation — likely a guess or confusion with \(\frac{d}{dx}[x^2] = 2x\) combined with a wrong simplification. - **(B) \(2x^2\):** Applies the chain rule factor \(2x\) but evaluates \(\sqrt{t}\) at \(t = x\) instead of \(t = x^2\), giving \(\sqrt{x} \cdot 2x = 2x\sqrt{x} = 2x^{3/2}\). Choice B is likely a sign/exponent error variant. - **(D) \(\sqrt{x^2}\):** Correctly evaluates the integrand at \(x^2\) (giving \(\sqrt{x^2} = |x|\)) but **omits the chain-rule factor \(u'(x) = 2x\)** — the most common error on FTC Part 1 with composite bounds.

ftc-part2 definite-integral-evaluation antiderivatives Calculator not allowed Difficulty 1/5

\(\displaystyle\int_{1}^{3} \left(6x^{2} - 4\right)\,dx =\)

A. 40

B. 42

C. 44

D. 52

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Answer: C

Apply FTC Part 2. An antiderivative of \(6x^2 - 4\) is \(F(x) = 2x^3 - 4x\). \[\int_{1}^{3}\left(6x^2 - 4\right)\,dx = F(3) - F(1) = (54 - 12) - (2 - 4) = 42 - (-2) = 44\] **Why the distractors fail:** - **(A) \(40\):** Sign error in the evaluation step — computes \(F(3) + F(1) = 42 + (-2) = 40\) instead of \(F(3) - F(1)\). Subtracting the lower-bound value, not adding it, is the rule. - **(B) \(42\):** Evaluates only at the upper bound, \(F(3) = 42\), forgetting to subtract \(F(1)\) entirely. - **(D) \(52\):** Fails to antidifferentiate the constant term: uses \(F(x) = 2x^3 - 4\) (leaving \(-4\) unintegrated instead of \(-4x\)), giving \((54-4)-(2-4) = 52\).

ftc-part1 accumulation-functions ftc1-vs-ftc2 Calculator not allowed Difficulty 2/5

If \(g(x) = \displaystyle\int_{2}^{x} \left(t^{3} + 1\right)\,dt\), then \(g'(x) =\)

A. \frac{x^{4}}{4} + x - 6

B. x^{3} + 1

C. x^{3} - 8

D. \frac{x^{4}}{4} + x

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Answer: B

By FTC Part 1, the derivative of an accumulation function with a constant lower bound and an upper bound of \(x\) is simply the integrand evaluated at \(x\): \[g'(x) = \frac{d}{dx}\int_{2}^{x}\left(t^3 + 1\right)\,dt = x^3 + 1\] **Why the distractors fail:** - **(A) \(\frac{x^4}{4} + x - 6\):** Confuses FTC Part 1 with FTC Part 2 — first computes the definite integral \(g(x) = \frac{x^4}{4} + x - 6\) and then forgets to differentiate, leaving the accumulation function itself. - **(C) \(x^3 - 8\):** Mistakenly subtracts the integrand evaluated at the lower bound: \((x^3+1) - (2^3+1) = x^3 - 8\). FTC Part 1 never involves the lower bound when it is constant. - **(D) \(\frac{x^4}{4} + x\):** Computes the antiderivative but drops the constant from the lower-limit evaluation — another FTC Part 1 / Part 2 confusion that also mishandles the bounds.

u-substitution definite-integral-evaluation changing-limits-of-integration Calculator not allowed Difficulty 3/5

\(\displaystyle\int_{0}^{2} x\sqrt{x^{2}+1}\,dx =\)

A. \frac{2}{3}\left(5\sqrt{5}-1\right)

B. \frac{1}{3}\left(5\sqrt{5}-1\right)

C. \frac{1}{2}\left(5\sqrt{5}-1\right)

D. \frac{2\sqrt{2}}{3}

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Answer: B

Use the substitution \(u = x^2 + 1\), so \(du = 2x\,dx\) and \(x\,dx = \tfrac{1}{2}\,du\). Change the limits: when \(x = 0\), \(u = 1\); when \(x = 2\), \(u = 5\). \[\int_{0}^{2} x\sqrt{x^2+1}\,dx = \frac{1}{2}\int_{1}^{5} \sqrt{u}\,du = \frac{1}{2}\cdot \frac{2}{3}u^{3/2}\Big|_{1}^{5} = \frac{1}{3}\left(5^{3/2} - 1\right) = \frac{1}{3}\left(5\sqrt{5} - 1\right)\] **Why the distractors fail:** - **(A) \(\frac{2}{3}(5\sqrt{5}-1)\):** Forgets the \(\tfrac{1}{2}\) factor from \(x\,dx = \tfrac{1}{2}\,du\), doubling the answer. - **(C) \(\frac{1}{2}(5\sqrt{5}-1)\):** Power-rule slip — integrates \(\sqrt{u}\) as \(u^{3/2}\) without dividing by \(\tfrac{3}{2}\), so the \(\tfrac{1}{2}\) survives but the \(\tfrac{2}{3}\) is lost. - **(D) \(\frac{2\sqrt{2}}{3}\):** Forgets to change the limits of integration — keeps the antiderivative \(\tfrac{1}{3}u^{3/2}\) but evaluates it at the original \(x\)-limits \(0\) and \(2\) instead of \(u\)-limits \(1\) and \(5\).

ftc-part1 chain-rule-with-variable-bounds accumulation-functions Calculator not allowed Difficulty 3/5

If \(h(x) = \displaystyle\int_{1}^{x^{3}} \frac{1}{t}\,dt\) for \(x > 0\), then \(h'(x) =\)

A. \frac{1}{x^{3}}

B. \frac{3}{x}

C. 3x

D. 3\ln x

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Answer: B

Apply FTC Part 1 with the chain rule. The upper bound is \(u(x) = x^3\), so \(u'(x) = 3x^2\), and the integrand is \(f(t) = \tfrac{1}{t}\). \[h'(x) = f\big(u(x)\big)\cdot u'(x) = \frac{1}{x^3}\cdot 3x^2 = \frac{3}{x}\] **Why the distractors fail:** - **(A) \(\frac{1}{x^3}\):** Evaluates the integrand at the upper bound but omits the chain-rule factor \(u'(x) = 3x^2\) — the most common FTC Part 1 error with a composite bound. - **(C) \(3x\):** Includes the chain-rule factor \(3x^2\) but evaluates \(\tfrac{1}{t}\) at \(t = x\) instead of \(t = x^3\), giving \(\tfrac{1}{x}\cdot 3x^2 = 3x\). - **(D) \(3\ln x\):** Confuses FTC Part 1 with FTC Part 2 — computes the integral \(h(x) = \ln(x^3) = 3\ln x\) and forgets to differentiate.

riemann-sums left-riemann-sum definite-integral-approximation table-of-values Calculator not allowed Difficulty 4/5

Selected values of a continuous function \(f\) are given in the table below. | \(x\) | \(0\) | \(2\) | \(4\) | \(6\) | \(8\) | |---|---|---|---|---|---| | \(f(x)\) | \(3\) | \(5\) | \(6\) | \(10\) | \(12\) | Using a left Riemann sum with the four subintervals indicated by the table, the approximation of \(\displaystyle\int_{0}^{8} f(x)\,dx\) is

A. 24

B. 48

C. 57

D. 66

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Answer: B

Each subinterval has width \(\Delta x = 2\). A left Riemann sum uses the function value at the **left** endpoint of each subinterval, i.e. at \(x = 0, 2, 4, 6\): \[\sum f(x_{\text{left}})\,\Delta x = 2\big(f(0) + f(2) + f(4) + f(6)\big) = 2(3 + 5 + 6 + 10) = 2(24) = 48\] **Why the distractors fail:** - **(A) \(24\):** Sums the left-endpoint heights \(3+5+6+10 = 24\) but forgets to multiply by the width \(\Delta x = 2\). - **(C) \(57\):** Applies the **trapezoidal** rule, \(\tfrac{\Delta x}{2}\big(f_0 + 2f_1 + 2f_2 + 2f_3 + f_4\big) = 1(3 + 10 + 12 + 20 + 12) = 57\), instead of the requested left sum. - **(D) \(66\):** Uses **right** endpoints (\(x = 2,4,6,8\)): \(2(5 + 6 + 10 + 12) = 66\) — a left/right endpoint mix-up.

accumulation-functions definite-integral-as-signed-area properties-of-definite-integrals net-change Calculator not allowed Difficulty 5/5

Let \(F(x) = \displaystyle\int_{0}^{x} f(t)\,dt\), where the graph of \(f\) consists of three pieces. Between \(x=0\) and \(x=2\) the region between the graph and the \(x\)-axis lies **above** the axis with area \(5\); between \(x=2\) and \(x=5\) the region lies **below** the axis with area \(8\); between \(x=5\) and \(x=6\) the region lies **above** the axis with area \(3\). What is the value of \(F(6)\)?

A. -6

B. -3

C. 0

D. 16

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Answer: C

\(F(6) = \displaystyle\int_{0}^{6} f(t)\,dt\) is the **net signed area**. Area above the axis counts as positive; area below counts as negative: \[F(6) = (+5) + (-8) + (+3) = 0\] **Why the distractors fail:** - **(A) \(-6\):** An extra sign error — treats the final region as negative too, computing \(5 - 8 - 3 = -6\) instead of adding the \(+3\) above-axis piece. - **(B) \(-3\):** Stops early and omits the last region, computing only \(5 - 8 = -3\) over \([0,5]\) rather than continuing to \(x = 6\). - **(D) \(16\):** Ignores the sign of the below-axis region and adds all areas as positive: \(5 + 8 + 3 = 16\). A definite integral uses signed area, so area below the axis must be subtracted.

Practice: Integration and Accumulation of Change