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

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!Final Answer:    
 
!Final Answer:    
 
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|'''(a)''' &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
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|&nbsp;&nbsp; '''(a)''' &nbsp; <math>\frac{2}{9}(1+x^3)^{\frac{3}{2}}+C</math>
 
|-
 
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|'''(b)''' &nbsp; <math>-1+\sqrt{2}</math>
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|&nbsp;&nbsp; '''(b)''' &nbsp; <math>-1+\sqrt{2}</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 15:06, 18 April 2016

Evaluate the indefinite and definite integrals.

a) Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{2}{\sqrt {1+x^{3}}}~dx}
b)


Foundations:  
How would you integrate
You could use -substitution. Let Then, Thus,

Solution:

(a)

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

(b)

Step 1:  
Again, we need to use -substitution. Let Then, Also, we need to change the bounds of integration.
Plugging in our values into the equation we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u_{1}=\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}} and
Therefore, the integral becomes
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du.}
Step 2:  
We now have:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx}&=&\displaystyle {\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {\left.{\frac {-1}{u}}\right|_{\frac {\sqrt {2}}{2}}^{1}}\\&&\\&=&\displaystyle {-{\frac {1}{1}}-{\frac {-1}{\frac {\sqrt {2}}{2}}}}\\&&\\&=&\displaystyle {-1+{\sqrt {2}}.}\\\end{array}}}
Final Answer:  
   (a)  
   (b)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -1+{\sqrt {2}}}

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