Difference between revisions of "009B Sample Midterm 3, Problem 4"

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<span class="exam">Evaluate the integral:
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<span class="exam"> The rate of reaction to a drug is given by:
  
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::<math>\int \sin (\ln x)~dx.</math>
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::::::<math>r'(t)=2t^2e^{-t}</math>
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<span class="exam">where <math>t</math> is the number of hours since the drug was administered.
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<span class="exam">Find the total reaction to the drug from <math>t=1</math> to <math>t=6.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int vdu.</math>
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|'''2.''' How could we break up <math style="vertical-align: -5px">\sin (\ln x)~dx</math> into <math style="vertical-align: -1px">u</math> and <math style="vertical-align: -1px">dv?</math>
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::Notice that <math style="vertical-align: -5px">\sin (\ln x)</math> is one term. So, we need to let <math style="vertical-align: -5px">u=\sin (\ln x)</math> and <math style="vertical-align: -1px">dv=dx.</math>
 
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We proceed using integration by parts.
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|Let <math style="vertical-align: -6px">u=\sin(\ln x)</math> and <math style="vertical-align: -1px">dv=dx.</math> Then, <math style="vertical-align: -14px">du=\cos(\ln x)\frac{1}{x}dx</math> and <math style="vertical-align: -1px">v=x.</math>
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|Therefore, we get
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::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.</math>
 
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we need to use integration by parts again.
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|Let <math style="vertical-align: -6px">u=\cos(\ln x)</math> and <math style="vertical-align: -1px">dv=dx.</math> Then, <math style="vertical-align: -14px">du=-\sin(\ln x)\frac{1}{x}dx</math> and <math style="vertical-align: -1px">v=x.</math>
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|Therfore, we get
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::<math>\begin{array}{rcl}
 
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\displaystyle{\int \sin (\ln x)~dx} & = & \displaystyle{x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx.}\\
 
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\end{array}</math>
 
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
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|So, if we add the integral on the right to the other side of the equation, we get
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::<math>2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x).</math>
 
 
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|Now, we divide both sides by 2 to get
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::<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.</math>
 
 
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|Thus, the final answer is
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::<math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C.</math>
 
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp;&nbsp; <math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
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|&nbsp;&nbsp;
 
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[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:27, 5 February 2017

The rate of reaction to a drug is given by:

where is the number of hours since the drug was administered.

Find the total reaction to the drug from to


Foundations:  

Solution:

Step 1:  
Step 2:  
Step 3:  
Final Answer:  
  

Return to Sample Exam

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