a. One way to integrate xx+8 b.Another way to evaluate the integral in part a is by using the substitution u =x 8. Do so to find the antiderivative c. Compare the results from the two methods. If they do not look the same, explain how this can happen. dx is to use integration by parts. Do so to find the antiderivative JxVx+8 dx? a. Which of the following is the result of using integration by parts on 3 2 OA. 3*(x+8)2 4 3 16 (x + 8) 5 2 Ов. 5X(X + 8) (X+ 8 15 C + C 5 4 5 Ос. 8 16 (x+8) 15 C O D. 15X(X+8) +C 3 x+8) 3 2 5 5 2 (X8) C E. 75X(X8) 15 16 C O F. 3x(X+ 8) 3(X8)
a. One way to integrate xx+8 b.Another way to evaluate the integral in part a is by using the substitution u =x 8. Do so to find the antiderivative c. Compare the results from the two methods. If they do not look the same, explain how this can happen. dx is to use integration by parts. Do so to find the antiderivative JxVx+8 dx? a. Which of the following is the result of using integration by parts on 3 2 OA. 3*(x+8)2 4 3 16 (x + 8) 5 2 Ов. 5X(X + 8) (X+ 8 15 C + C 5 4 5 Ос. 8 16 (x+8) 15 C O D. 15X(X+8) +C 3 x+8) 3 2 5 5 2 (X8) C E. 75X(X8) 15 16 C O F. 3x(X+ 8) 3(X8)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Please solve for a, b and c
![a. One way to integrate xx+8
b.Another way to evaluate the integral in part a is by using the substitution u =x 8. Do so to find the antiderivative
c. Compare the results from the two methods. If they do not look the same, explain how this can happen.
dx is to use integration by parts. Do so to find the antiderivative
JxVx+8 dx?
a. Which of the following is the result of using integration by parts on
3
2
OA.
3*(x+8)2
4
3
16
(x + 8)
5
2
Ов.
5X(X + 8)
(X+ 8
15
C
+ C
5
4
5
Ос.
8
16
(x+8)
15
C
O D.
15X(X+8)
+C
3 x+8)
3
2
5
5
2
(X8) C
E.
75X(X8)
15
16
C
O F.
3x(X+ 8)
3(X8)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e2c5a88-a31f-4e9d-bcbd-75e956d26fb0%2F6a8d60a9-9b53-4d02-9dae-35f1b69fb234%2F785emh.png&w=3840&q=75)
Transcribed Image Text:a. One way to integrate xx+8
b.Another way to evaluate the integral in part a is by using the substitution u =x 8. Do so to find the antiderivative
c. Compare the results from the two methods. If they do not look the same, explain how this can happen.
dx is to use integration by parts. Do so to find the antiderivative
JxVx+8 dx?
a. Which of the following is the result of using integration by parts on
3
2
OA.
3*(x+8)2
4
3
16
(x + 8)
5
2
Ов.
5X(X + 8)
(X+ 8
15
C
+ C
5
4
5
Ос.
8
16
(x+8)
15
C
O D.
15X(X+8)
+C
3 x+8)
3
2
5
5
2
(X8) C
E.
75X(X8)
15
16
C
O F.
3x(X+ 8)
3(X8)
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