Difference between revisions of "009B Sample Midterm 1, Problem 1"

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!Foundations:    
 
!Foundations:    
 
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|Answers:
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'''Solution:'''
 
'''Solution:'''
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'''(a)'''
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!Step 1:  
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|We need to use substitution. Let <math>u=1+x^3</math>. Then, <math>du=3x^2dx</math> and <math>\frac{du}{3}=x^2dx</math>.
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|Therefore, the integral becomes <math>\frac{1}{3}\int \sqrt{u}du</math>.
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!Step 2: &nbsp;
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|We now have:
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|<math>\int x^2\sqrt{1+x^3}dx=\frac{1}{3}\int \sqrt{u}du=\frac{2}{9}u^{\frac{3}{2}}+C=\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>.
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|(a) <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
 
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|(b) <math></math>
 
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[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:56, 26 January 2016

Evaluate the indefinite and definite integrals.

a)
b)


Foundations:  

Solution:

(a)

Step 1:  
We need to use substitution. Let . Then, and .
Therefore, the integral becomes .
Step 2:  
We now have:
.

(b)

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

Return to Sample Exam