When x = 0, we haveu = 0 and when x = 8, we have u = 64 Step 5 Therefore, 8. 64 6. x sin(x2) dx | sin(u) du %3D 64 1- cos (x) 0. 3(1- cos (64)) %D
When x = 0, we haveu = 0 and when x = 8, we have u = 64 Step 5 Therefore, 8. 64 6. x sin(x2) dx | sin(u) du %3D 64 1- cos (x) 0. 3(1- cos (64)) %D
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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![Step 4
When x = 0, we have u = 0
0, and when x = 8, we have u = 642
Step 5
Therefore,
64
sin(u) du
2 Jo
8.
9.
x sin(x2) dx = 6. 1
Jo
%3D
164
3 1- cos(x)
3(1– cos(64))
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F27d79205-0ce3-49e4-b539-77503c7c758e%2F4e6b1324-51b9-44d6-9cf7-2387cb66c49a%2F7lc06fc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 4
When x = 0, we have u = 0
0, and when x = 8, we have u = 642
Step 5
Therefore,
64
sin(u) du
2 Jo
8.
9.
x sin(x2) dx = 6. 1
Jo
%3D
164
3 1- cos(x)
3(1– cos(64))
%3D
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