AP Calc BC Integration and Accumulation of Change — Questions and Answers
Question 1: What is the value of ∫ from 0 to 1 of x·eˣ dx?
- e − 1
- e
- 1 (Correct answer)
- 0
Correct answer: 1
Integration by parts with u = x and dv = eˣ dx gives x·eˣ − eˣ. Evaluating from 0 to 1 gives (e − e) − (0 − 1) = 1.
Integration by parts uses ∫u dv = uv − ∫v du. Here u = x, so du = dx, and dv = eˣ dx, so v = eˣ. This gives ∫x·eˣ dx = x·eˣ − ∫eˣ dx = x·eˣ − eˣ. At x = 1 the antiderivative is e − e = 0. At x = 0 it is 0 − 1 = −1. The definite integral is 0 − (−1) = 1.
Question 2: Which expression equals ∫ 1/((x − 1)(x + 2)) dx?
- (1/3)·ln|(x − 1)/(x + 2)| + C (Correct answer)
- (1/3)·ln|(x − 1)(x + 2)| + C
- ln|(x − 1)/(x + 2)| + C
- −(1/3)·ln|(x − 1)/(x + 2)| + C
Correct answer: (1/3)·ln|(x − 1)/(x + 2)| + C
Partial fractions give 1/((x−1)(x+2)) = (1/3)/(x−1) − (1/3)/(x+2). Integrating gives (1/3)ln|x−1| − (1/3)ln|x+2| + C.
Write 1 = A(x + 2) + B(x − 1). Setting x = 1 gives 3A = 1, so A = 1/3. Setting x = −2 gives −3B = 1, so B = −1/3. Each term integrates to a logarithm: (1/3)ln|x−1| − (1/3)ln|x+2| + C. The log difference rule combines this into (1/3)ln|(x−1)/(x+2)| + C. The coefficient 1/3 must stay in front of the logarithm.
Question 3: If F(x) = ∫ from 0 to x² of √(1 + t³) dt, what is F′(x)?
- √(1 + x⁶)
- 2x·√(1 + x³)
- x²·√(1 + x⁶)
- 2x·√(1 + x⁶) (Correct answer)
Correct answer: 2x·√(1 + x⁶)
By the Fundamental Theorem of Calculus with the chain rule, F′(x) = √(1 + (x²)³) · d/dx(x²) = 2x·√(1 + x⁶).
The FTC says d/dx ∫ from a to u(x) of f(t) dt = f(u(x))·u′(x). Here f(t) = √(1 + t³) and u(x) = x². Substitute the upper limit into the integrand to get √(1 + (x²)³) = √(1 + x⁶). Then multiply by u′(x) = 2x. Leaving out the chain-rule factor is the most common error.
Question 4: A left Riemann sum with 4 subintervals of equal width is used to approximate ∫ from 0 to 4 of x² dx. What is its value?
- 30
- 14 (Correct answer)
- 22
- 21
Correct answer: 14
The width is 1 and the left endpoints are 0, 1, 2, 3. The sum is 1·(0 + 1 + 4 + 9) = 14.
With Δx = (4 − 0)/4 = 1, the left endpoints are x = 0, 1, 2, 3. The left sum is Δx·[f(0) + f(1) + f(2) + f(3)] = 0 + 1 + 4 + 9 = 14. The right sum would be 30, the trapezoidal sum 22, and the midpoint sum 21. Since x² is increasing on [0, 4], the left sum underestimates the true value of 64/3 ≈ 21.33.
Question 5: What is the value of ∫ from 0 to ∞ of e^(−2x) dx?
- 2
- 1
- The integral diverges
- 1/2 (Correct answer)
Correct answer: 1/2
The integral is the limit as b→∞ of [−(1/2)e^(−2x)] from 0 to b. That is 0 − (−1/2) = 1/2.
An improper integral with an infinite limit is defined as lim b→∞ of ∫ from 0 to b. The antiderivative of e^(−2x) is −(1/2)e^(−2x). At b it equals −(1/2)e^(−2b), which tends to 0 as b→∞. At 0 it equals −1/2. The result is 0 − (−1/2) = 1/2, so the integral converges. A common error is forgetting the 1/2 from the chain-rule factor.
Question 6: What is the value of ∫ from 0 to 1 of x/(x² + 1) dx?
- ln 2
- (1/2)·ln 2 (Correct answer)
- π/4
- 1/2
Correct answer: (1/2)·ln 2
Let u = x² + 1, so du = 2x dx. The integral becomes (1/2)∫ from 1 to 2 of du/u = (1/2)ln 2.
Substitute u = x² + 1, so x dx = du/2. The limits change from x = 0 and x = 1 to u = 1 and u = 2. The integral becomes (1/2)∫ from 1 to 2 of (1/u) du = (1/2)[ln u] from 1 to 2 = (1/2)(ln 2 − ln 1) = (1/2)ln 2. The result is not arctan, because the numerator is x, not 1.
Question 7: What is the average value of f(x) = 3x² on the interval [1, 3]?
Correct answer: 13
The average value is (1/(3 − 1))·∫ from 1 to 3 of 3x² dx = (1/2)(27 − 1) = 13.
The average value of f on [a, b] is (1/(b − a))·∫ from a to b of f(x) dx. Here ∫ from 1 to 3 of 3x² dx = [x³] from 1 to 3 = 27 − 1 = 26. The interval length is 2, so the average is 26/2 = 13. The value 26 is the total accumulation, not the average.
Question 8: A tank holds 10 gallons of water at time t = 0. Water flows in at a rate of R(t) = 12 − 3t gallons per minute for 0 ≤ t ≤ 6, with t in minutes. How many gallons are in the tank at t = 6?
- 18
- 38
- 28 (Correct answer)
- 10
Correct answer: 28
The net change is ∫ from 0 to 6 of (12 − 3t) dt = 72 − 54 = 18 gallons. Adding the initial 10 gives 28 gallons.
The accumulation form of the FTC says Q(6) = Q(0) + ∫ from 0 to 6 of R(t) dt. The antiderivative of 12 − 3t is 12t − (3/2)t². At t = 6 it equals 72 − 54 = 18, and at t = 0 it equals 0. The net inflow is therefore 18 gallons. Add the starting 10 gallons to get 28. The integral alone, 18, gives only the change.
Question 9: What is the value of ∫ from 1 to e of x·ln x dx?
- (e² + 1)/4 (Correct answer)
- (e² − 1)/4
- e²/4
- (e² + 1)/2
Correct answer: (e² + 1)/4
By parts with u = ln x and dv = x dx, the antiderivative is (x²/2)ln x − x²/4. Evaluating from 1 to e gives (e²/2 − e²/4) − (−1/4) = (e² + 1)/4.
Set u = ln x and dv = x dx, so du = dx/x and v = x²/2. Then ∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4. At x = e, ln e = 1, so the value is e²/2 − e²/4 = e²/4. At x = 1, ln 1 = 0, so the value is −1/4. The difference is e²/4 + 1/4 = (e² + 1)/4.
Question 10: What is the value of ∫ from 0 to 3 of 1/(x² + 9) dx?
- π/4
- π/36
- π/12 (Correct answer)
- 1/3
Correct answer: π/12
∫ 1/(x² + 9) dx = (1/3)arctan(x/3) + C. Evaluating from 0 to 3 gives (1/3)(π/4 − 0) = π/12.
The standard form is ∫ 1/(x² + a²) dx = (1/a)arctan(x/a) + C. With a = 3 the antiderivative is (1/3)arctan(x/3). At x = 3 the value is (1/3)arctan(1) = (1/3)(π/4) = π/12. At x = 0 the value is 0. The 1/a factor is easy to forget, and omitting it gives π/4.
