009B Sample Midterm 3, Problem 4 - Revision history

MathAdmin at 01:35, 13 November 2017

2017-11-13T01:35:51Z

MathAdmin at 03:02, 14 April 2017

2017-04-14T03:02:18Z

MathAdmin at 19:20, 9 April 2017

2017-04-09T19:20:31Z

MathAdmin: Created page with "Evaluate the integral: :: $\int \sin (\ln x)~dx.$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |- |'..."

2016-04-19T00:53:47Z

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: |- |'..."

New page

<span class="exam">Evaluate the integral:

::<math>\int \sin (\ln x)~dx.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int vdu.</math>
|-
|'''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>
|-
|
::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>
|}

'''Solution:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We proceed using integration by parts.
|-
|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>
|-
|Therefore, we get
|-
|
::<math>\int \sin (\ln x)~dx=x\sin(\ln x)-\int \cos(\ln x)~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need to use integration by parts again.
|-
|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>
|-
|Therfore, we get
|-
|
::<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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Notice that the integral on the right of the last equation is the same integral that we had at the beginning.
|-
|So, if we add the integral on the right to the other side of the equation, we get
|-
|
::<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
|-
|
::<math>\int \sin(\ln x)~dx=\frac{x\sin(\ln x)}{2}-\frac{x\cos(\ln x)}{2}.</math>
|-
|Thus, the final answer is
|-
|
::<math>\int \sin(\ln x)~dx=\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   <math>\frac{x}{2}(\sin(\ln x)-\cos(\ln x))+C</math>
|}
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam"> The rate of reaction to a drug is given by:
+<span class="exam"> Compute the following integrals:
-::<math>r'(t)=2t^2e^{-t}</math>
+<span class="exam">(a) &nbsp; <math>\int x^2\sin (x^3) ~dx</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">(b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 60: / Line 60: @@
+|
 &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}}\\
+\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{\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{(-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}.}

@@ 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>
@@ Line 7: / Line 11: @@
 !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>
 |-
+|
-::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:'''
@@ Line 19: / Line 25: @@
 !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
 |-
+|
-::<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;
 |-
-|Now, we need to use integration by parts again.
+|We proceed using integration by parts.
 |-
-|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>
 |}
@@ Line 49: / Line 50: @@
 !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>
 |-
-|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}) \\
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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, Problem 4 - Revision history

MathAdmin at 01:35, 13 November 2017

MathAdmin at 03:02, 14 April 2017

MathAdmin at 19:20, 9 April 2017

MathAdmin: Created page with "Evaluate the integral: ::\int \sin (\ln x)~dx. {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |- |'..."

MathAdmin: Created page with "Evaluate the integral: :: $\int \sin (\ln x)~dx.$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Foundations: |- |'..."