Difference between revisions of "022 Exam 2 Sample A, Problem 6"

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(Created page with "<span class="exam">Find the area under the curve of  <math style="vertical-align: -60%">y\,=\,\frac{8}{\sqrt{x}}</math>  between <math style="vertical-align: -5%...")
 
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::<math>
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::<math>\begin{array}{rcl}
\begin{array}{rcl}
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\int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{8 x^{1/2}}{2} \vert_1^4\\
\int_1^4 \frac{8}{\sqrt{x}}dx & = & \frac{x^{1/2}}
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& = & 4x^{1/2} \vert_1^4
 
\end{array}</math>
 
\end{array}</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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| Now we need to substitute back into our original variables using our original substitution <math>u = 3x + 2</math>
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| Now we need to evaluate to get:
 
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| to get <math>\frac{\log(u)}{3} = \frac{\log(3x + 2}{3}</math>
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| <math>4\cdot 4^{1/2} - 4\cdot 1^{1/2} = 8 - 4 = 4</math>  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 4: &nbsp;
 
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| Since this integral is an indefinite integral we have to remember to add ''' "+ C" ''' at the end.
 
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|<math>\int \frac{1}{3x + 2} dx = \frac{\ln(3x + 2)}{3} + C</math>
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|4
 
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[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]
 
[[022_Exam_2_Sample_A|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:37, 15 May 2015

Find the area under the curve of    between and .

Foundations:  
This problem requires two rules of integration. In particular, you need
Integration by substitution (U - sub): If and are differentiable functions, then

    

The Product Rule: If and are differentiable functions, then

    

The Quotient Rule: If and are differentiable functions and  , then

    
Additionally, we will need our power rule for differentiation:
for ,
as well as the derivative of natural log:

 Solution:

Step 1:  
Set up the integral:
Step 2:  
Using the power rule we have:
Step 3:  
Now we need to evaluate to get:
Final Answer:  
4

Return to Sample Exam