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Practice Exercises - Antidifferentiation - AP Calculus Premium 2024

Practice Exercises

mtDdZXG8p5sMbQXvxCvHBLwlk8mzkDyKrj3zghkr

A1. ∫(3x2 – 2x + 3) dx =

(A) x3x2 + C
(B) 3x3x2 + 3x + C
(C) x
3x2 + 3x + C
(D) Ttrifp6a3OPArpLU2ccClCEH42NiUM8S5evelid8

A2PHisAkqcF11SzhsFOmqWGFCe_AeSlWgRd6S4wNM3

JswIomg1QvkXLRf9q9OxWu6OvSTrAhMrBFTjnfpe

A3OySvlqPyy7LC2Oj6ps6pCV-ZQBa3fGpIu6boU3Fv

ufy-nhMVQkuKNFPTjP8n0Qkrzukn32xkaE3VbMbR

A4. ∫(2 – 3x)5dx =

f8SWxChgPqrm7MlKZ1N_xPJo15RrODVBrkTPfEeX

A50OdCiNEWXs0u2YvwRGJzqaZho56bSj5qPQHGMW-7

1lJTDh4B0JWDuMR0GvdJD5_fZ26GQGpmyrl1nmzy

A6-ZgZyJtBoe9o2Hm_g-QfUGLVoNZ8phabR252fYhZ

1_ifnbqVV3x11Za4y-0Olw430D_3nXJOYoHOKAS5

A7tFXxdmG8FtQ13-eIyj2TmHQGIigf2nNd9RMOgHh3

mFUbxKmgTBHkl4iVA6GVhhK3C6m3uAN-AY-4PfjA

A8akZ0k6pi1-2lWsmjOJ2sZDVZ1xphHayDdyffdJU4

OLXvgr490wDpfFVPVT61nvtHaL6Ej7rHW6_6mkVb

A9. ∫ cos 3x dx=

(A) 3 sin 3x + C
(B) –sin 3x + C
(C) 
_7L1GgottdOAqG3NK_-pGJwR1Lp3OWmWCZ_QhVjh
(D) hB7RzsZcOYgQsWVJgRYGVKtmCanItvc9kF5V7zlz

A10TXXaNlldicLXcQ33OC0_iCSlAPzOga2GXEqtEljn

eoWJ6qTeGjJiHY-xIiXGFngOygvf9Fyhd7ZK5WhE

A11Je3qORyZYcfh40z5XkCFPHpWmELQqH2y2KLswbaF

HzNmCLBW0Wc-W8fbRac5mBNZc-xsvP9BjdNqU5b7

A12WRR4XD2Qw5n6n2MNSnELdE2rn5UanEr9xGwXDYMK

StD9lr4sWPZRAi72V-sBMz1xS_9Z7Lce2F1RQTob

A13vm8zelY0_9mw7RGoD8lV8OEfKB9PBquCcwpmOddj

TjR7BeDnqdKsS0S2Pd3yOM7OyEItPnzuoEtNw-IS

A14SKk5dj3fgMHYQm8hOSIZM4W77G41FOQwc_LkBuL6

1DwNvxBZLzBtsBYRzOQAs4r6C6WiiJJmYnNfihDk

A15E8jjNuC9uUtHgI47NBoX7qaB2LDRCEMzoMjpAhwV

Usz8A_MvyKQqhdh_xLi5Gf4WkJzdJYF4vInYQXoj

A16gxx_qP-93fZEXMewfeudM2T52yQjUF5g-B3SJcgi

(A) AZc5gnf2Y_U9VIhhjhhr4OVI4VVMRxp71yQkJbk6
(B) x + 2 ln |x| + C
(C) x + ln |2x| + C
(D) pDWolgvmVDTmhyz4KRLHOr6oJfkraitM6647Vyz8

A17SBu5v78FjT1pH7hOtHZcaBlvu_AavLeHJazuQ1RY

v5L5KN88g1NZnSzdQbG-yHVX7461UXSxzj98aF_L

A189dN4dVQXpE_aFCryusTRtdrsInTPF0wxzfkp7Fdm

kDn7ll18l4kzs0J2nFzqd0wGMU4D8CJMfJ5Tqbpj

A19e_y1XwDqbiQpVnusdT_5IcVCnEz3mCSkbJxOiAZV

(A) 3x4/3 – 2x5/2 – 2x1/2 + C
(B) 3x4/3 – 2x5/2 + 2x1/2 + C
(C) vTAx9b5dylzI4TsoPx37-ci5MEFSc8p6PlukK10P
(D) g6N-YllczkrLugd8gUzALzK6soAlc4Y3aPdqbU2q

A20qeHFjULzuwES_LbD4jAY4G-2qrKZZil5LjQKpqLc

Ua7wI72VPujDww4H2qgfvO-opDN4CBNa1Ple54HD

A21jNAKkFsMgSgio5WMehlZ0vqwAq9Cy_DJlMSPOBaV

go30TQ0xygSzGWJTjj-o0vuGUdw5K13k_LQDM4zm

A22qzbfzdBA0r37WubVxV_KRuco1YS62dvpXtZ7MurA

ds1bhbGOSqpKe1kwTW38wnd2xzwJu8oGlCtOnZs3

A23. ∫ sin2θ cosθ =

drwud05PgqYKi0x0N4P175aZPHq9lDaVrVYpwldJ

A24G3UJ_rXGorPTUzBMEEJsjOqPfiC_WKzWKRyOB8Lv

EvybJl4ZN5GK5xvXOFrfENpuLDlqjYlnFtw4XF2r

A25. ∫ t cos(4t2)dt =

OO8sD83lA9ncASesTanhViZjajLecOu79ZLZMDMk

CHALLENGE A26. ∫ cos2 2x dx =

yxj4tUJpaFXRqywQy6akiNlriTOOY_USi1A43Pq4

A27. ∫ sin 2θ =

(A) Y1VBGG5J-f2QQoW6BO4XfISArKeVdljRtE51quUl
(B) –2 cos 2θ + C
(C) cos2θ + C
(D)9SSl8b77LbwyVck8fbHoNwhfd5hocv_KPSe7W_ao

*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 xx sin x + C

A29ZFLstNF1tf02CTxsdnIrh8P28uMuqUZQPjJOQT0y

X_GAqyonCxMsUX7Oz29zjGDwLG0nIvb8Yrxv8kWI

A30sSN09C8DFkB-vaOgWHMwUACbOZjFnFEj0NN4u1A-

5Xa8mPoa2zg8l4n91jmkTEvwaZrNXJ1TvlMN-4pZ

A31OMQkyJy1W5UzpFSpf6W0LyE-_8hfTsUtsoktO7MT

(A) 2 ln sin |θ – 1| + C
(B) –csc(θ – 1) + C
(C) –cot(θ – 1) + C
(D) csc(θ – 1) + C

CHALLENGE A3211yl1VPrryCQkU1wDn32vfuQp73G4gGzMkxhgUOo

ZjEUHNqNl4eLOz1o5_xihYbeHZHdU0yfj6PdyQXP

A33. ∫ sec3/2x tan x dx =

bRt8UReTostqP1i4HySkYoOlmkjcFCpDn55naBXj

A34. ∫tan θ =

(A) –ln |cos θ| + C
(B) sec2θ + C
(C) ln |sin θ| + C
(D) –ln |sec θ| + C

A35yu-UN30mjTVu5_44gYI_vEob8bVNhaLafrO4oTFy

(A) WwClegXBpbgCVwJcLl7rn-Llkff_xhHBioMZbtLp
(B) U_NcahVaRix-44uK60ms2DMQc9Mt3hpS7OaLEf66
(C) –cot x + C
(D) –csc 2x + C

A36PV-Rnl-SkqnjqOMKJmrMoeqh9Nm36S4IGFCbtxDK

(A) (tan–1y)2 + C
(B) ln (1 + y2) + C
(C) ln (tan–1y) + C
(D) AVSS6LRcUeiWswbjxv4Vx_EC0evbKwWRmcwjZbOE

A37. ∫ 2 sin θ cos2θ =
(A) r-vAj_EC1sYON2JE3xB6Tua8u-yVTv1E2czHbFWf
(B) 17KC2LJQdmuAPe56pw5vP48RPNAzl3qozZj_WR4D
(C) sin2θ cos θ + C
(D) cos3θ + C

A38yq31itMg-Kl47lKhXCfRf4KALWklgDVGkmPZPBcy

0YKbnq1eY2KtK5uwZT-4u3MfkdTy4Eyj4Q5noNdT

A39. ∫ cot 2u du =

(A) ln |sin u| + C
(B) 
vroyETi-4NQ53xvx2BwiJ2VTzPOClngNRt0Xz-JI
(C) r9IEemPlD7e9y7smG4R1tBruLvH-l8Blbc37vZP-
(D) 2 ln |sin 2u| + C

A40Mpne57I1gfVJxkQj1jfK7mGMbcNOWnm-CpBBZkbZ

(A) x + ln |ex – 1| + C
(B) xex + C
(C) 
84hONOt7ifaA_i3Ubeg7OdPInxI91nXIfePkiQfJ
(D) ln |ex – 1| + C

*A41yVgRG49RH8EEAqYOKM0FvQ0m9TmAQwtekKLqhnmW

IbIJQJAmaCy9tFjv8ksF6z9i3m7YFU9lDz4SVgXM

A422ofyT5Co8v1h0kQoNEfh7KvOkTrKBFU22xje-8r4

w2AFBt19cb2O1PsBuaRGEbtEdTYmlrdWh706ze82

A43. ∫ cos θ esin θ =

(A) esin θ + 1 + C
(B) esin θ + C
(C) –esin θ + C
(D) ecos θ + C

A44. ∫ e2θ sin e2θ =

(A) cos e2θ + C
(B) 2e4θ (cos e2θ + sin e2θ) + C
(C) kYURDKr-qRx4xd9Gl4gR-cg5MperhusFKjErxPZJ
(D) –2 cos e2θ+ C

A45Rn63HTPRXbnxN97MaBYGp52tUXIY-BIEnG0DoGfP

--wGHeiZ6oZEF1VNwZEYBDQj11m4xM58eYWNO3Zh

*A46. ∫ xe–x dx =

(A) e–x (1 – x) + C
(B) 
8Z8PBoLpgLFa7J-4xS_3_9QlQ9pEkLkQerwc1p-Z
(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

A481dR0HIMCwoazFunPyAEzhP9dlx3-SrBGHEV0jG8H

(A) x – ln |exex| + C
(B) x + 2 ln |ex – e–x| + C
(C) ln |ex – e–x| + C
(D) ln (ex + e–x) + C

A49Cl3u0jOUz-hDBlzGP1dUa6nJaC1x3MkjHwyKeopc

(A) tan–1ex + C
(B) hWhVlerf5_vMGTdW8Oc-Tlw9i7QBsCV8MYI--RYo
(C) ln (1 + e2x) + C
(D) RvjeVMkU-z9zqfuAO5kObjtMpivXH-N4KYIv-eKa

A503MykixR6LAHXaOta13hH0x8AxDuqveGifTBR2ziO

i0vz-CXt4YC4NB3UV2rE38HripFG8TQw2h_Mn0fE

A51L1fklcz1ZlLVMAi2qY0znnibYBt0ys6hDGKn42WG

_Vae5-vfWyypjSR-vfXnH2omhavjrza0AixSKIQG

*A52. ∫ x3 ln x dx 

pm_JotKFR3W9MPbftsV_xtqjFJyYW3GsGlOj8d2j

*A53 ln η =
(A) GAeC9DaF64k6n--gNQmvZLqeuD7EJBpO6y9KTL_2
(B) η (ln η – 1) + C
(C) pD-SOHhGSeMWP8PyAEXSjW8XGwmghGVX01FOHMvy
(D) η ln η + η + C

*A54. ∫ ln x3dx =

(A) AvfSSj-hSk4XFynP1HbfzIIDyRF4YvQGY4If4k8e
(B) 3x (ln x – 1) + C
(C) 3 ln x (x – 1) + C
(D) 1J4HC_KxtlkPTuPGJgOhzIfqnSus1xJOt580Dl9q

*A55uiaImdSkel_xyIEK7iyjHMDPLwgxcp9dPPwTsg5b

co8GHR0e3ru_OEkvuFXKapWtsbCvoTIw2-pGsKZK

A56TacPtZngLZ62frlsIgdLF5139FvgXLe7bVXAfCZ6

XP3HeHOIe2e2-CRpBCD8JbRkwzuUMJVDL0IudIQo

A571y9ADSFYmmIwNLSjzkTrOkvfkAgFW7G4xuwo2m8G

(A) y – 2 ln |y + 1| + C
(B) C8cyA942-2ni6CsvKkZuxXFMd8pUTrOm6S_VCxSV
(C) BLdfxJHzik7He3dKBwbBOTx8jeki8sZwj7hPXoHM
(D) 1 – 2 ln |y + 1| + C

A5820blI3wOaAkhAzChOeggHYu5hIL6LVzSHrZGEHNm

(A) ln(x2 + 2x + 2) + C
(B) ln |x + 1| + C
(C) arctan(x + 1) + C
(D) fIU2R_FGO4UzMRRFkHb6L6oWPD_j72DWKGe9SdQ0

A59ejADxcC4wyNGy8eA4u_gAgGHiBcHkW3qvPX_qehm

hwGTUJ_PytBUM16QUn9XOzji61mt6kb4dY3w6erk

*A60. ∫ eθ cos θ dθ =

(A) eθ(cos θ – sin θ) + C
(B) _M_Eyga25xkXT3DBeHoIcTQmJG2s6ex1jmik4EHp
(C) 2eθ(sin θ + cos θ) + C
(D) nXYZvNNIMmAKop1Nb3vqRdaDjmsieEHZkoaXsAYX

A61C3Y8zb4Je1Bp8AqyDvG76n7qCt1HD9KEvCqXwppe

urhum-w7WWqrnrXxMXsoY66rI_sD_5Tqt7xAGCF2

*A62. ∫ u sec2u du =

(A) u tan u + ln |cos u| + C
(B) O8oUcI8kQn_-SfcfAjsCUANMh8MfqupEUPhGJj0T
(C) uZQQKNVSRhz9IFDAQGPCAftORut5sasDMVJOzXy_
(D) u tan u – ln |sin u| + C

CHALLENGE A63vVJZ9OM4rd53P8uS_A82oEsU9hyqjhYwj3bcjNsT

BA2SnVWguL86244cfXfk5VnElO6iWX2z25A38agV

CHALLENGE A64iAPTNYLg10hfrijtIx6_bCXLdsBKmU9MZyNML1k-

ExkqhqZrBeSpwn3vK9CrAKarZCxNKzB5hr9VtDLC

CHALLENGE A65P_8gnhWQYMp1IJZT6tVK4D_gaA11Cw2UuKfXLYzx

-f-we0QMWaBx34ByKkS5QyE8GaMh_Gvjj0sJqMOy

CHALLENGE A66ga9p951ZwcQA5mkUMjGGX0d4nJXxZeWZBeJnqpYT

(A) tan–1ex + C
(B) ex– ln (1 + ex) + C
(C) ex x + ln |1 + ex| + C
(D) zQL4K-dzak2tKlR4QUqn52QnVFfGlq81mLGcb2xm

A67ROUtTejzejReKrtp_ThKJdF6fSoMOc-Bf17YJBl5

(A) sec θ tan θ + C
(B) ln (1 + sin2θ) + C
(C) tan–1 (sin θ) + C
(D) w59hFhioH21ExWsUr23J5fVk9ZD8O9bbggGYzuj2

*A68. ∫ arctan x dx =

(A) x arctan x – ln (1 + x2) + C
(B) x arctan x + ln (1 + x2) + C
(C) VyvFs0u0deHzxxoH97FUJ3DN9Op9lZ7EE03OlB-k
(D) 42L5QtG6J92JDQdaJW-IGcgSXJkrSExceKdTS-L7

CHALLENGE A69qxLhm0TJQ78fFwTpTWD6rUV7-YpEv8sdH5a8cfyD

(A) –ln |1 – ex| + C
(B) x – ln |l – ex| + C
(C) -81rjLRKBYzWm5tJoLg7F2uc5nnQ6SRsUqw_lGxD
(D) e–x ln |l + ex| + C

A70eMTbfhSiwBE1--uSBOT9hvZz3UhviTFryVdGUV5P

Sr1htB6Vp5WoAN2fCEVyuzs3c44bP8SE5YxF92xb

A71. ∫ e2 ln udu =

w-qbvdFHQAbg1jakR0GK6KBx8lJ5p66iVDUT4e00

A72kOPTJmNp3Kq4THEcUUy4s_z8UK9SCZmYzc0cGlOX

fSqSFcNlUglh2gIRjkSmSQ5h391N-mGfpmNAWgvD

CHALLENGE A73. ∫ (tan θ – 1)2 =

(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 A740ydIHTZHNhBh5h89Vc6QzN6yXIQ-_k-pNEDDeeUK

(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
Cp96m8VKlUA9JqQhLMHwu4zsHaeTYiVSKPaML9Qe

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

*A794iAnV61Ilz_icFuXwkCKLvDuEj_DerJ-kLOu454B

snYGIxKeJ8zKhBss4Ht0FfVi6jlkK6TMqzkHMsqI

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