Difference between revisions of "009B Sample Midterm 3, Problem 3"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|  
+
| u substitution
 
|}
 
|}
  
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!Step 1:    
 
!Step 1:    
 
|-
 
|-
|
+
|We proceed using u substitution. Let <math>u=x^3</math>. Then, <math>du=3x^2dx</math>.
 +
|-
 +
|Therefore, we have
 
|-
 
|-
|
+
|<math>\int x^2\sin (x^3) dx=\int \frac{\sin(u)}{3}du</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|
+
|We integrate to get
 
|-
 
|-
|
+
|<math>\int x^2\sin (x^3) dx=\frac{-1}{3}\cos(u)+C=\frac{-1}{3}\cos(x^3)+C</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)'''
+
|'''(a)''' <math>\frac{-1}{3}\cos(x^3)+C</math>
 
|-
 
|-
 
|'''(b)'''  
 
|'''(b)'''  
 
|}
 
|}
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:38, 31 January 2016

Compute the following integrals:

a)
b)


Foundations:  
u substitution

Solution:

(a)

Step 1:  
We proceed using u substitution. Let . Then, .
Therefore, we have
Step 2:  
We integrate to get

(b)

Step 1:  
Step 2:  
Final Answer:  
(a)
(b)

Return to Sample Exam