Integration Accumulation Test Answer Key with Scoring – AP Calculus AB

Question 1 Definite integral (2 points)

An antiderivative is x³ − x². At 2: 8 − 4 = 4; at 0: 0. The value is 4. Answer (A).

Scoring: 2 points for the correct choice; 1 point for a correct antiderivative.

Question 2 Accumulation derivative (2 points)

By FTC Part 1, H′(x) = √(x⁴ + 9). At x = 2: √25 = 5. Answer (A).

Scoring: 2 points for the correct choice; 1 point for using FTC Part 1.

Question 3 Riemann sums from a table (3 points)

(a) 8(3) + 10(2) + 7(4) = 24 + 20 + 28 = 72.

(b) 10(3) + 7(2) + 2(4) = 30 + 14 + 8 = 52.

(c) (8 + 10)/2·3 + (10 + 7)/2·2 + (7 + 2)/2·4 = 27 + 17 + 18 = 62 gallons.

Scoring: 1 point per part; units required in (c).

Question 4 Substitution (3 points)

(a) Let u = x³ + 2, so du = 3x² dx. Then (1/3)∫ u⁴ du = u⁵/15 + C = (x³ + 2)⁵/15 + C.

(b) [(x³ + 2)⁵/15] from 0 to 1 = (3⁵ − 2⁵)/15 = (243 − 32)/15 = 211/15.

Scoring: Part (a): 1 point for the substitution, 1 point for the antiderivative. Part (b): 1 point.

Question 5 Accumulation function from a graph (3 points)

(a) g(2) = 2·2 = 4. From 2 to 4 the net signed area is (2 + (−2))/2·2 = 0, so g(4) = 4.

(b) From 4 to 6 the area is (−2)(2) = −4, so g(6) = 0.

(c) g′ = f changes from positive to negative at x = 3 (the line from (2, 2) to (4, −2) crosses zero there). g(3) = 4 + (2 + 0)/2·1 = 5. Comparing with g(0) = 0 and g(6) = 0, the maximum is 5.

Scoring: 1 point per part.

Question 6 Long division and completing the square (3 points)

(a) Divide: x² − 3x + 5 = (x − 1)(x − 2) + 3, so the integrand is x − 2 + 3/(x − 1). The antiderivative is x²/2 − 2x + 3 ln|x − 1| + C.

(b) x² + 4x + 5 = (x + 2)² + 1, so the antiderivative is arctan(x + 2) + C.

Scoring: Part (a): 1 point for the division, 1 point for the antiderivative. Part (b): 1 point.

Question 7 Properties of integrals (2 points)

(a) ∫₄⁶ = −2 − 7 = −9.

(b) ∫₁⁴ (2f(x) + 3) dx = 2(7) + 3(3) = 23. Reversing the limits gives −23.

Scoring: 1 point per part.

Question 8 Net change in a tank (2 points)

(a) 40 + ∫₀⁶ (12 − 2t) dt = 40 + (72 − 36) = 76 gallons.

(b) The amount increases while R > 0 and decreases when R < 0. R changes from positive to negative at t = 6, so the greatest amount occurs at t = 6.

Scoring: 1 point per part.