Applications of Integration Test Answer Key with Scoring – AP Calculus AB

Question 1 Average value (2 points)

∫₁³ 3x² dx = 27 − 1 = 26. Divide by the width 2: 13. Answer (A).

Scoring: 2 points for the correct choice; 1 point for the integral value 26.

Question 2 Area between curves (2 points)

The curves meet at x = 0 and x = 2, with the line on top. Area = ∫₀² (2x − x²) dx = 4 − 8/3 = 4/3. Answer (A).

Scoring: 2 points for the correct choice; 1 point for finding the limits.

Question 3 Disc method (3 points)

(a) The disc radius is x³, so V = π∫₁² (x³)² dx = π∫₁² x⁶ dx.

(b) V = π[x⁷/7] from 1 to 2 = π(128 − 1)/7 = 127π/7.

Scoring: Part (a): 1 point for π and the squared radius, 1 point for the limits. Part (b): 1 point.

Question 4 Washer method (3 points)

(a) 5x = x² gives x = 0 and 5.

(b) The line is the outer curve: V = π∫₀⁵ (25x² − x⁴) dx.

(c) π[25x³/3 − x⁵/5] from 0 to 5 = π(3125/3 − 625) = 1250π/3.

Scoring: 1 point per part.

Question 5 Cross sections (3 points)

(a) The side is √x, so A = (√3/4)(x) = (√3/4)x.

(b) V = ∫₀⁴ (√3/4)x dx.

(c) (√3/4)(16/2) = 2√3 ≈ 3.46.

Scoring: 1 point per part.

Question 6 Displacement and distance (3 points)

(a) ∫₀⁶ v dt = [t³/3 − 3t² + 5t] from 0 to 6 = 72 − 108 + 30 = −6.

(b) v = (t − 1)(t − 5) changes sign at 1 and 5. Positions: s(1) = 7/3, s(5) = −25/3, s(6) = −6. Distance = 7/3 + 32/3 + 7/3 = 46/3 ≈ 15.3.

Scoring: Part (a): 1 point. Part (b): 1 point for splitting at the sign changes, 1 point for the total.

Question 7 Rates in and out (2 points)

(a) 50 + ∫₀⁸ (3t − 9) dt = 50 + (96 − 72) = 74 gallons.

(b) At t = 2, the inflow is 6 and the outflow is 9, so the net rate is −3 < 0 and the amount is decreasing.

Scoring: 1 point per part.

Question 8 Area with respect to y (2 points)

The curves meet where y² = y + 2, so y = −1 and y = 2. The line is on the right. Area = ∫₋₁² (y + 2 − y²) dy = [y²/2 + 2y − y³/3] from −1 to 2 = 10/3 − (−7/6) = 9/2.

Scoring: 1 point for the integral, 1 point for the value.