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 09:27, 5 February 2017

The rate of reaction to a drug is given by:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r'(t)=2t^2e^{-t}}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t} is the number of hours since the drug was administered.

Find the total reaction to the drug from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=6.}


Foundations:  

Solution:

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

Return to Sample Exam

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