Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question
Evaluate the following integral using integration by parts.
Si²e=-221 at
Use the integration by parts formula so that the new integral is simpler than the original one. Choose the correct answer below.
OA - 17171²²-²22²-1 (1-12 2²0 221) 01
dt
22
OB.
O C.
O D.
1
- 2 2 ² e 22 - S ( - 77 te-221) ot
dt
22
11
21/2²2 221 + 1 (171712²27)
- 12/2²³²e²22² + √(221²e-²
Evaluate the integral
dt
dt
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Transcribed Image Text:Evaluate the following integral using integration by parts. Si²e=-221 at Use the integration by parts formula so that the new integral is simpler than the original one. Choose the correct answer below. OA - 17171²²-²22²-1 (1-12 2²0 221) 01 dt 22 OB. O C. O D. 1 - 2 2 ² e 22 - S ( - 77 te-221) ot dt 22 11 21/2²2 221 + 1 (171712²27) - 12/2²³²e²22² + √(221²e-² Evaluate the integral dt dt
Evaluate the following integral using integration by parts.
See
√ 1²₁=-221dt
OA. -7₁ te ²²²-1|-22te cad
té
11
dt
O B.
- 12/2²6 ²2-S (-771²22) ₁1
22t
1
e
dt
11
O D.
-
1
0 C. 72/27²2²22²1 + 1 ( 77 16²21) 0
O
dt
- 12/2² e 221 + √(221²e-221) dt
Evaluate the integral.
√²²₁=²²₁αdt=
22t dt =
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Transcribed Image Text:Evaluate the following integral using integration by parts. See √ 1²₁=-221dt OA. -7₁ te ²²²-1|-22te cad té 11 dt O B. - 12/2²6 ²2-S (-771²22) ₁1 22t 1 e dt 11 O D. - 1 0 C. 72/27²2²22²1 + 1 ( 77 16²21) 0 O dt - 12/2² e 221 + √(221²e-221) dt Evaluate the integral. √²²₁=²²₁αdt= 22t dt =
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