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

From Math Wiki
Jump to navigation Jump to search
(Created page with "<span class="exam">Evaluate the integral: ::<math>\int \sin (\ln x)~dx.</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations:   |- |'...")
 
Line 1: Line 1:
−
<span class="exam">Evaluate the integral:
+
<span class="exam"> The rate of reaction to a drug is given by:
  
−
::<math>\int \sin (\ln x)~dx.</math>
+
::<math>r'(t)=2t^2e^{-t}</math>
 +
 
 +
<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the number of hours since the drug was administered.
 +
 
 +
<span class="exam">Find the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=6.</math>
  
  
Line 7: Line 11:
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int vdu.</math>
+
|If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b r'(t)~dt,</math>&nbsp; what are we calculating?
 
|-
 
|-
−
|'''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>
+
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; We are calculating &nbsp;<math style="vertical-align: -5px">r(b)-r(a).</math>&nbsp; This is the total reaction to the 
 
|-
 
|-
 
|
 
|
−
::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>
+
&nbsp; &nbsp; &nbsp; &nbsp; drug from &nbsp;<math style="vertical-align: 0px">t=a</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=b.</math>  
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
Line 19: Line 25:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|We proceed using integration by parts.
+
|To calculate the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: -4px">t=6,</math>
 
|-
 
|-
−
|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>
+
|we need to calculate
−
|-
 
−
|Therefore, we get
 
 
|-
 
|-
 
|
 
|
−
::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_1^6 r'(t)~dt=\int_1^6 2t^2e^{-t}~dt.</math>
 
|}
 
|}
  
Line 32: Line 36:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we need to use integration by parts again.  
+
|We proceed using integration by parts.  
 +
|-
 +
|Let &nbsp;<math style="vertical-align: 0px">u=2t^2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
 
|-
 
|-
−
|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>  
+
|Then, &nbsp;<math style="vertical-align: -1px">du=4t~dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
 
|-
 
|-
−
|Therfore, we get
+
|Then, we have
 
|-
 
|-
−
|
+
|&nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.</math>
−
::<math>\begin{array}{rcl}
 
−
\displaystyle{\int \sin (\ln x)~dx} & = & \displaystyle{x\sin(\ln x)-\bigg(x\cos(\ln x)+\int \sin(\ln x)~dx\bigg)}\\
 
−
&&\\
 
−
& = & \displaystyle{x\sin(\ln x)-x\cos(\ln x)-\int \sin(\ln x)~dx.}\\
 
−
\end{array}</math>
 
 
|}
 
|}
  
Line 49: Line 50:
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
−
|Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
+
|Now, we need to use integration by parts again.  
 
|-
 
|-
−
|So, if we add the integral on the right to the other side of the equation, we get
+
|Let &nbsp;<math style="vertical-align: 0px">u=4t</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
 
|-
 
|-
−
|
+
|Then, &nbsp;<math style="vertical-align: -1px">du=4dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
−
::<math>2\int \sin(\ln x)~dx=x\sin(\ln x)-x\cos(\ln x).</math>
 
 
|-
 
|-
−
|Now, we divide both sides by 2 to get  
+
|Thus, we get
 
|-
 
|-
 
|
 
|
−
::<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
|-
+
\displaystyle{\int_1^62t^2e^{-t}~dt} & = & \displaystyle{\left. -2t^2e^{-t}-4te^{-t}\right|_1^6+\int_1^6 4e^{-t}}\\
−
|Thus, the final answer is
+
&&\\
−
|-
+
& = & \displaystyle{\left. -2t^2e^{-t}-4te^{-t}-4e^{-t}\right|_1^6}\\
−
|
+
&&\\
−
::<math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C.</math>
+
& = & \displaystyle{-2(6)^2e^{-6}-4(6)e^{-6}-4e^{-6}}-(-2(1)^2e^{-1}-4(1)e^{-1}-4e^{-1}) \\
 +
&&\\
 +
& = & \displaystyle{\frac{-100+10e^5}{e^6}.}
 +
\end{array}</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; <math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
+
|&nbsp; &nbsp; &nbsp;&nbsp; <math>\frac{-100+10e^5}{e^6}</math>
 
|}
 
|}
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 11:20, 9 April 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  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=1}   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:  
If we calculate  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 \int_a^b r'(t)~dt,}   what are we calculating?

        We are calculating  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(b)-r(a).}   This is the total reaction to the

        drug from  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=a}   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=b.}


Solution:

Step 1:  
To calculate the total reaction to the drug from  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=1}   to  
we need to calculate

        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 \int_1^6 r'(t)~dt=\int_1^6 2t^2e^{-t}~dt.}

Step 2:  
We proceed using integration by parts.
Let  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 u=2t^2}   and  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 dv=e^{-t}dt.}
Then,  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 du=4t~dt}   and  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 v=-e^{-t}.}
Then, we have
       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 \int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.}
Step 3:  
Now, we need to use integration by parts again.
Let  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 u=4t}   and  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 dv=e^{-t}dt.}
Then,  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 du=4dt}   and  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 v=-e^{-t}.}
Thus, we get

        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 \begin{array}{rcl} \displaystyle{\int_1^62t^2e^{-t}~dt} & = & \displaystyle{\left. -2t^2e^{-t}-4te^{-t}\right|_1^6+\int_1^6 4e^{-t}}\\ &&\\ & = & \displaystyle{\left. -2t^2e^{-t}-4te^{-t}-4e^{-t}\right|_1^6}\\ &&\\ & = & \displaystyle{-2(6)^2e^{-6}-4(6)e^{-6}-4e^{-6}}-(-2(1)^2e^{-1}-4(1)e^{-1}-4e^{-1}) \\ &&\\ & = & \displaystyle{\frac{-100+10e^5}{e^6}.} \end{array}}


Final Answer:  
       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 \frac{-100+10e^5}{e^6}}

Return to Sample Exam

Retrieved from "https://wiki.math.ucr.edu/index.php?title=009B_Sample_Midterm_3,_Problem_4&oldid=1506"