Practice Exercises - Antidifferentiation - AP Calculus Premium 2024
Practice Exercises
A1. ∫(3x2 – 2x + 3) dx =
(A) x3 – x2 + C
(B) 3x3 – x2 + 3x + C
(C) x3 – x2 + 3x + C
(D)
A2.
A3.
A4. ∫(2 – 3x)5dx =
A5.
A6.
A7.
A8.
A9. ∫ cos 3x dx=
(A) 3 sin 3x + C
(B) –sin 3x + C
(C)
(D)
A10.
A11.
A12.
A13.
A14.
A15.
A16.
(A)
(B) x + 2 ln |x| + C
(C) x + ln |2x| + C
(D)
A17.
A18.
A19.
(A) 3x4/3 – 2x5/2 – 2x1/2 + C
(B) 3x4/3 – 2x5/2 + 2x1/2 + C
(C)
(D)
A20.
A21.
A22.
A23. ∫ sin2θ cosθ dθ =
A24.
A25. ∫ t cos(4t2)dt =
CHALLENGE A26. ∫ cos2 2x dx =
A27. ∫ sin 2θ dθ =
(A)
(B) –2 cos 2θ + C
(C) cos2θ + C
(D)
*A28. ∫ x cos x dx =
(A) x sin x + C
(B) x sin x + cos x + C
(C) x sin x – cos x + C
(D) cos x – x sin x + C
A29.
A30.
A31.
(A) 2 ln sin |θ – 1| + C
(B) –csc(θ – 1) + C
(C) –cot(θ – 1) + C
(D) csc(θ – 1) + C
CHALLENGE A32.
A33. ∫ sec3/2x tan x dx =
A34. ∫tan θ dθ =
(A) –ln |cos θ| + C
(B) sec2θ + C
(C) ln |sin θ| + C
(D) –ln |sec θ| + C
A35.
(A)
(B)
(C) –cot x + C
(D) –csc 2x + C
A36.
(A) (tan–1y)2 + C
(B) ln (1 + y2) + C
(C) ln (tan–1y) + C
(D)
A37. ∫ 2 sin θ cos2θ dθ =
(A)
(B)
(C) sin2θ cos θ + C
(D) cos3θ + C
A38.
A39. ∫ cot 2u du =
(A) ln |sin u| + C
(B)
(C)
(D) 2 ln |sin 2u| + C
A40.
(A) x + ln |ex – 1| + C
(B) x – ex + C
(C)
(D) ln |ex – 1| + C
*A41.
A42.
A43. ∫ cos θ esin θdθ =
(A) esin θ + 1 + C
(B) esin θ + C
(C) –esin θ + C
(D) ecos θ + C
A44. ∫ e2θ sin e2θdθ =
(A) cos e2θ + C
(B) 2e4θ (cos e2θ + sin e2θ) + C
(C)
(D) –2 cos e2θ+ C
A45.
*A46. ∫ xe–x dx =
(A) e–x (1 – x) + C
(B)
(C) –e–x(x + 1) + C
(D) e–x (x + 1) + C
*A47. ∫ x2ex dx =
(A) ex(x2 + 2x) + C
(B) ex(x2 – 2x – 2) + C
(C) ex (x2 – 2x + 2) + C
(D) ex(x – 1)2 + C
A48.
(A) x – ln |ex – e–x| + C
(B) x + 2 ln |ex – e–x| + C
(C) ln |ex – e–x| + C
(D) ln (ex + e–x) + C
A49.
(A) tan–1ex + C
(B)
(C) ln (1 + e2x) + C
(D)
A50.
A51.
*A52. ∫ x3 ln x dx
*A53. ∫ ln η dη =
(A)
(B) η (ln η – 1) + C
(C)
(D) η ln η + η + C
*A54. ∫ ln x3dx =
(A)
(B) 3x (ln x – 1) + C
(C) 3 ln x (x – 1) + C
(D)
*A55.
A56.
A57.
(A) y – 2 ln |y + 1| + C
(B)
(C)
(D) 1 – 2 ln |y + 1| + C
A58.
(A) ln(x2 + 2x + 2) + C
(B) ln |x + 1| + C
(C) arctan(x + 1) + C
(D)
A59.
*A60. ∫ eθ cos θ dθ =
(A) eθ(cos θ – sin θ) + C
(B)
(C) 2eθ(sin θ + cos θ) + C
(D)
A61.
*A62. ∫ u sec2u du =
(A) u tan u + ln |cos u| + C
(B)
(C)
(D) u tan u – ln |sin u| + C
CHALLENGE A63.
CHALLENGE A64.
CHALLENGE A65.
CHALLENGE A66.
(A) tan–1ex + C
(B) ex– ln (1 + ex) + C
(C) ex – x + ln |1 + ex| + C
(D)
A67.
(A) sec θ tan θ + C
(B) ln (1 + sin2θ) + C
(C) tan–1 (sin θ) + C
(D)
*A68. ∫ arctan x dx =
(A) x arctan x – ln (1 + x2) + C
(B) x arctan x + ln (1 + x2) + C
(C)
(D)
CHALLENGE A69.
(A) –ln |1 – ex| + C
(B) x – ln |l – ex| + C
(C)
(D) e–x ln |l + ex| + C
A70.
A71. ∫ e2 ln udu =
A72.
CHALLENGE A73. ∫ (tan θ – 1)2dθ =
(A) sec θ + θ + 2 ln |cos θ| + C
(B) tan θ + 2 ln |cos θ| + C
(C) tan θ – 2 sec2θ + C
(D) tan θ – 2 ln |cos θ| + C
CHALLENGE A74.
(A) sec θ – tan θ + C
(B) ln (1 + sin θ) + C
(C) ln |sec θ + tan θ| + C
(D) tan θ – sec θ + C
A75. A particle starting at rest at t = 0 moves along a line so that its acceleration at time t is 12t ft/sec2. How much distance does the particle cover during the first 3 seconds?
(A) 32 feet
(B) 48 feet
(C) 54 feet
(D) 108 feet
A76. The equation of the curve whose slope at point (x,y) is x2 – 2 and which contains the point (1,–3) is
A77. A particle moves along a line with acceleration 2 + 6t at time t. When t = 0, its velocity equals 3 and it is at position s = 2. When t = 1, it is at position s =
(A) 5
(B) 6
(C) 7
(D) 8
A78. Find the acceleration (in ft/sec2) needed to bring a particle moving with a velocity of 75 ft/sec to a stop in 5 seconds.
(A) –3
(B) –6
(C) –15
(D) –25
*A79.