AP Calc AB Integration and Accumulation of Change — Questions and Answers
Question 1: What is the value of ∫ from 0 to 2 of (3x² − 2x) dx?
Correct answer: 4
An antiderivative is x³ − x². Evaluating from 0 to 2 gives (8 − 4) − 0 = 4.
The Fundamental Theorem of Calculus (Part 2) says ∫ from a to b of f(x) dx = F(b) − F(a), where F is any antiderivative of f. Using the power rule, the antiderivative of 3x² is x³ and the antiderivative of −2x is −x². So F(x) = x³ − x². Then F(2) = 8 − 4 = 4 and F(0) = 0, so the integral equals 4. Choosing 8 comes from forgetting the −x² term.
Question 2: If G(x) = ∫ from 1 to x² of sin(t) dt, what is G′(x)?
- sin(x²)
- 2x·cos(x²)
- 2x·sin(x²) (Correct answer)
- sin(x²) − sin(1)
Correct answer: 2x·sin(x²)
By the Fundamental Theorem of Calculus with the chain rule, G′(x) = sin(x²)·d/dx(x²) = 2x·sin(x²).
Part 1 of the Fundamental Theorem says d/dx ∫ from a to u(x) of f(t) dt = f(u(x))·u′(x). Here f(t) = sin t and u(x) = x², so you substitute x² into sine and multiply by u′(x) = 2x. The result is 2x·sin(x²). Forgetting the chain-rule factor gives sin(x²), and differentiating the integrand itself gives the cosine distractor.
Question 3: Which of the following is equal to ∫ 2x·cos(x²) dx?
- sin(x²) + C (Correct answer)
- 2·sin(x²) + C
- −sin(x²) + C
- x²·sin(x²) + C
Correct answer: sin(x²) + C
Let u = x², so du = 2x dx. The integral becomes ∫ cos u du = sin u + C = sin(x²) + C.
This is a standard u-substitution. Set u = x², which gives du = 2x dx, and the integrand 2x·cos(x²) dx becomes cos u du exactly. Its antiderivative is sin u + C, and substituting back gives sin(x²) + C. You can check by differentiating: d/dx sin(x²) = cos(x²)·2x, which matches the original integrand.
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 the value of the sum?
- 30
- 21
- 20
- 14 (Correct answer)
Correct answer: 14
Each width is 1 and the left endpoints are 0, 1, 2, 3. The sum is 1·(0 + 1 + 4 + 9) = 14.
With [0, 4] split into 4 equal pieces, Δx = 1 and the subintervals are [0,1], [1,2], [2,3], [3,4]. A left Riemann sum uses f at the left endpoints 0, 1, 2, 3, giving f values 0, 1, 4, 9. Multiplying the total 14 by Δx = 1 gives 14. The right sum would be 30 and the midpoint sum 21, while the exact integral is 64/3 ≈ 21.3. Since x² is increasing on this interval, the left sum underestimates.
Question 5: What is the value of ∫ from 0 to 1 of x·e^(x²) dx?
- (e − 1)/2 (Correct answer)
- e − 1
- e/2
- (e + 1)/2
Correct answer: (e − 1)/2
Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫ from 0 to 1 of e^u du = (e − 1)/2.
Using u = x², du = 2x dx, so x dx = (1/2) du. When x = 0, u = 0, and when x = 1, u = 1. The integral becomes (1/2)∫ from 0 to 1 of e^u du = (1/2)(e¹ − e⁰) = (e − 1)/2. Omitting the factor 1/2 gives the common wrong answer e − 1.
Question 6: What is the average value of f(x) = x² on the interval [0, 3]?
- 9
- 3 (Correct answer)
- 9/2
- 3/2
Correct answer: 3
The average value is (1/(3−0))·∫ from 0 to 3 of x² dx = (1/3)(9) = 3.
The average value of a continuous function f on [a, b] is (1/(b − a))·∫ from a to b of f(x) dx. Here ∫ from 0 to 3 of x² dx = 3³/3 = 9. Dividing by the interval length 3 gives 3. The value 9 is the integral itself, which is the area rather than the average height.
Question 7: Water flows into a tank at a rate of r(t) = 2t + 1 gallons per minute. If the tank holds 5 gallons at t = 0, how many gallons are in the tank at t = 3 minutes?
- 12
- 14
- 20
- 17 (Correct answer)
Correct answer: 17
The amount is 5 + ∫ from 0 to 3 of (2t + 1) dt = 5 + [t² + t] from 0 to 3 = 5 + 12 = 17 gallons.
The definite integral of a rate of change gives the net change in the quantity. Here ∫ from 0 to 3 of (2t + 1) dt = (9 + 3) − 0 = 12 gallons entered the tank. Because the tank started with 5 gallons, the total is 5 + 12 = 17. Answering 12 forgets the initial amount.
Question 8: If ∫ from 2 to 6 of f(x) dx = 10 and ∫ from 2 to 4 of f(x) dx = 3, what is ∫ from 6 to 4 of f(x) dx?
- 7
- −13
- −7 (Correct answer)
- 13
Correct answer: −7
By additivity, ∫ from 4 to 6 of f = 10 − 3 = 7. Reversing the limits changes the sign, so ∫ from 6 to 4 of f = −7.
Integrals over adjacent intervals add: ∫ from 2 to 6 = ∫ from 2 to 4 + ∫ from 4 to 6, so ∫ from 4 to 6 of f = 10 − 3 = 7. The question asks about the integral from 6 down to 4, and swapping the limits of integration negates the value. Therefore ∫ from 6 to 4 of f(x) dx = −7.
Question 9: What is the value of ∫ from 0 to 1 of x/(x² + 1) dx?
- ln 2
- (1/2)·ln 2 (Correct answer)
- 2·ln 2
- 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 (1/u) du = (1/2)·ln 2.
Set u = x² + 1, so du = 2x dx and x dx = du/2. The limits change from x = 0, 1 to u = 1, 2. The integral becomes (1/2)∫ from 1 to 2 of du/u = (1/2)[ln|u|] from 1 to 2 = (1/2)(ln 2 − ln 1) = (1/2)·ln 2. Leaving out the factor 1/2 would give ln 2.
Question 10: Let F(x) = ∫ from 0 to x of (t² − 4t + 3) dt. At which value of x does F have a local maximum?
- x = 0
- x = 3
- x = 1 (Correct answer)
- x = 2
Correct answer: x = 1
F′(x) = x² − 4x + 3 = (x − 1)(x − 3). F′ changes from positive to negative at x = 1, so F has a local maximum there.
By the Fundamental Theorem of Calculus, F′(x) = x² − 4x + 3 = (x − 1)(x − 3), which is zero at x = 1 and x = 3. For x < 1, F′ is positive, between 1 and 3 it is negative, and after 3 it is positive again. A sign change from positive to negative at x = 1 gives a local maximum, while the change from negative to positive at x = 3 gives a local minimum. The endpoint x = 0 is not a critical point because F′(0) = 3.
