AP Calculus AB Integration and Accumulation of Change — Worked Answer Explanations

Unit 6 · 18% of the AP exam · 8 questions explained

Below is a complete answer key for our AP Calculus AB Integration and Accumulation of Change practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Integration and Accumulation of Change practice test and come back here to review, or head back to the Integration and Accumulation of Change unit overview.

In-content ad

Question 1 · Easy
Evaluate $\int (3 x^{2} + 4 x - 5) d x$ .
- A
  $x^{3} + 2 x^{2} - 5 x + C$ Correct
- B
  $3 x^{3} + 4 x^{2} - 5 x + C$
  Why not B: Forgot to divide by the new exponent.
- C
  $x^{3} + 2 x^{2} - 5 x$
  Why not C: Forgot the $+ C$ .
- D
  $6 x + 4 + C$
  Why not D: Differentiated instead of integrated.
Explanation
Power rule for integration: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ . Term-by-term: $x^{3} + 2 x^{2} - 5 x + C$ .

Key takeaway
Integration power rule: raise the exponent by 1, divide by the new exponent.
Question 2 · Easy
Evaluate $\int \frac{1}{x} d x$ for $x > 0$ .
- A
  $ln ∣ x ∣ + C$ Correct
- B
  $- \frac{1}{x ^{2}} + C$
  Why not B: Differentiated rather than integrated.
- C
  $x ln x + C$
  Why not C: Confused with $\int ln x d x$ .
- D
  $\frac{x ^{0}}{0} + C$
  Why not D: Power rule fails at $n = - 1$ .
Explanation
Special case of integration: $\int x^{- 1} d x = ln ∣ x ∣ + C$ (the power rule fails because we'd divide by zero).

Key takeaway
$\int (1/x)\,dx = \ln|x| + C$ — exception to the power rule.
Question 3 · Easy
Evaluate $\int_{0}^{2} (x^{2} + 1) d x$ .
- A
  $\frac{8}{3}$
  Why not A: Forgot the $+ 1$ term integrated.
- B
  $\frac{14}{3}$ Correct
- C
  $5$
  Why not C: Computed integrand at upper limit only.
- D
  $10$
  Why not D: Confused with multiplying width by max value.
Explanation
$\int_{0}^{2} (x^{2} + 1) d x = [\frac{x ^{3}}{3} + x]_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}$ .

Key takeaway
FTC: evaluate antiderivative at upper limit minus lower limit.
Question 4 · Easy
If $F (x) = \int_{0}^{x} t^{3} d t$ , what is $F^{'} (x)$ ?
- A
  $3 x^{2}$
  Why not A: Differentiated $t^{3}$ inside the integral.
- B
  $x^{3}$ Correct
- C
  $\frac{x ^{4}}{4}$
  Why not C: Reported $F (x)$ , not $F^{'} (x)$ .
- D
  $0$
  Why not D: Treated as a constant.
Explanation
Fundamental Theorem of Calculus: $\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)$ . So $F^{'} (x) = x^{3}$ .

Key takeaway
FTC Part 1: derivative of an integral function evaluates the integrand at the upper limit.
Question 5 · Medium
Find $\int x cos (x^{2}) d x$ .
- A
  $sin (x^{2}) + C$
  Why not A: Forgot the $\frac{1}{2}$ factor from $u$ -substitution.
- B
  $\frac{1}{2} sin (x^{2}) + C$ Correct
- C
  $x sin (x^{2}) + C$
  Why not C: Did not properly substitute.
- D
  $2 sin (x^{2}) + C$
  Why not D: Inverted the substitution coefficient.
Explanation
Let $u = x^{2}$ , then $d u = 2 x d x$ , so $x d x = \frac{1}{2} d u$ . Integral becomes $\frac{1}{2} \int cos u d u = \frac{1}{2} sin u + C$ .

Key takeaway
$u$-substitution: pick $u$ so that $du$ matches an inner factor.
Question 6 · Medium
Find the average value of $f (x) = x^{2}$ on $[0, 3]$ .
- A
  $3$ Correct
- B
  $4.5$
  Why not B: Computed $f$ at midpoint only.
- C
  $9$
  Why not C: Reported $f (3)$ , not the average.
- D
  $\frac{27}{3}$
  Why not D: Forgot to divide integrating result by 3.
Explanation
Average value: $\frac{1}{b - a} \int_{a}^{b} f (x) d x = \frac{1}{3} \int_{0}^{3} x^{2} d x = \frac{1}{3} \cdot 9 = 3$ .

Key takeaway
Average value of $f$ on $[a,b]$ is $\dfrac{1}{b-a}\int_a^b f\,dx$.
Question 7 · Medium
If a particle has velocity $v (t) = 3 t^{2} - 6 t$ (m/s), what is its displacement from $t = 0$ to $t = 4$ ?
- A
  $8 m$
  Why not A: Used distance traveled, not signed displacement.
- B
  $16 m$ Correct
- C
  $24 m$
  Why not C: Forgot to subtract the term $- 3 t^{2}$ .
- D
  $32 m$
  Why not D: Doubled the answer.
Explanation
Displacement $= \int_{0}^{4} (3 t^{2} - 6 t) d t = [t^{3} - 3 t^{2}]_{0}^{4} = 64 - 48 = 16 m$ .

Key takeaway
Displacement is the (signed) integral of velocity; distance traveled uses $|v|$.
Question 8 · Medium
Use a left Riemann sum with 4 equal subintervals to approximate $\int_{0}^{4} x^{2} d x$ .
- A
  $14$ Correct
- B
  $21 \frac{1}{3}$
  Why not B: Reported the exact value.
- C
  $30$
  Why not C: Used a right Riemann sum.
- D
  $8$
  Why not D: Off by a factor.
Explanation
$Δ x = 1$ . Left endpoints: $0, 1, 2, 3$ . $f$ values: $0, 1, 4, 9$ . Sum: $0 + 1 + 4 + 9 = 14$ .

Key takeaway
Left Riemann sum: heights from left endpoints; underestimates for increasing $f$.