009B Sample Midterm 1, Problem 3 - Revision history

Kayla Murray at 17:27, 20 November 2017

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Kayla Murray at 00:57, 27 February 2017

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Kayla Murray at 23:48, 7 February 2017

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Kayla Murray at 16:13, 7 February 2017

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Kayla Murray at 03:54, 7 February 2017

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Kayla Murray at 16:11, 6 February 2017

2017-02-06T16:11:56Z

Kayla Murray at 16:05, 6 February 2017

2017-02-06T16:05:44Z

← Older revision		Revision as of 17:27, 20 November 2017
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−	~~[[009B_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]~~
−
−
−	~~{\| 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 v~du.</math>~~
−	\|-
−	~~\|'''2.''' How would you integrate  <math style="vertical-align: -12px">\int x\ln x~dx?</math>~~
−	\|-
−	\|
−	~~        You could use integration by parts.~~
−	\|-
−	\|
−	~~        Let  <math style="vertical-align: -1px">u=\ln x</math>  and  <math style="vertical-align: 0px">dv=x~dx.</math>~~
−	\|-
−	~~\|        Then,  <math style="vertical-align: -13px">du=\frac{1}{x}dx</math>  and  <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int x\ln x~dx} & = & \displaystyle{\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}~~
−	~~\end{array}</math>~~
−	\|}
−
−
−	~~'''Solution:'''~~
−
−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We proceed using integration by parts.~~
−	\|-
−	~~\|Let  <math style="vertical-align: 0px">u=x^2</math>  and  <math style="vertical-align: 0px">dv=e^xdx.</math>~~
−	\|-
−	~~\|Then,  <math style="vertical-align: 0px">du=2xdx</math>  and  <math style="vertical-align: 0px">v=e^x.</math>~~
−	\|-
−	~~\|Therefore, we have~~
−	\|-
−	~~\|         <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^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: 0px">u=2x</math>  and  <math style="vertical-align: 0px">dv=e^xdx.</math>~~
−	\|-
−	~~\|Then,  <math style="vertical-align: 0px">du=2dx</math>  and  <math style="vertical-align: 0px">v=e^x.</math>~~
−	\|-
−	~~\|Building on the previous step, we have~~
−	\|-
−	~~\|         <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We proceed using integration by parts.~~
−	\|-
−	~~\|Let  <math style="vertical-align: -1px">u=\ln x</math>  and  <math style="vertical-align: 0px">dv=x^3dx.</math>~~
−	\|-
−	~~\|Then,  <math style="vertical-align: -13px">du=\frac{1}{x}dx</math>  and  <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>~~
−	\|-
−	~~\|Therefore, we have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right\|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right\|_{1}^{e}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we evaluate to get~~
−	\|-
−	~~\|         <math>\begin{array}{rcl}~~
−	~~\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{3e^4+1}{16}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     <math>x^2e^x-2xe^x+2e^x+C</math>~~
−	\|-
−	~~\|    '''(b)'''     <math>\frac{3e^4+1}{16}</math>~~
−	\|}
	[[009B_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]		[[009B_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Evaluate the indefinite and definite integrals.
+<span class="exam"> A population grows at a rate
-<span class="exam">(a) &nbsp; <math>\int x^2 e^x~dx</math>
+::<math>P'(t)=500e^{-t}</math>
-<span class="exam">(b) &nbsp; <math>\int_{1}^{e} x^3\ln x~dx</math>
+<span class="exam">where &nbsp;<math style="vertical-align: -5px">P(t)</math>&nbsp; is the population after &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; months.

@@ Line 19: / Line 19: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -1px">u=\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=x~dx.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and &nbsp;<math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
 |-
+|
@@ Line 40: / Line 40: @@
 |We proceed using integration by parts.
 |-
-|Let &nbsp;<math style="vertical-align: 0px">u=x^2</math> and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=x^2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: 0px">du=2xdx</math> and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=2xdx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
 |Therefore, we have
@@ Line 54: / Line 54: @@
 |Now, we need to use integration by parts again.
 |-
-|Let &nbsp;<math style="vertical-align: 0px">u=2x</math> and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=2x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: 0px">du=2dx</math> and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=2dx</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
 |Building on the previous step, we have

@@ Line 13: / Line 13: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 |-
-|'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
+|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -12px">\int x\ln x~dx?</math>
 |-
+|
@@ Line 19: / Line 19: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and &nbsp;<math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
 |-
+|
@@ Line 40: / Line 40: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=x^2</math> and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
-|Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=2xdx</math> and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
 |Therefore, we have
@@ Line 54: / Line 54: @@
 |Now, we need to use integration by parts again.
 |-
-|Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
+|Let &nbsp;<math style="vertical-align: 0px">u=2x</math> and &nbsp;<math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
-|Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|Then, &nbsp;<math style="vertical-align: 0px">du=2dx</math> and &nbsp;<math style="vertical-align: 0px">v=e^x.</math>
 |-
 |Building on the previous step, we have
@@ Line 73: / Line 73: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math>
+|Let &nbsp;<math style="vertical-align: -1px">u=\ln x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=x^3dx.</math>
 |-
-|Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>
+|Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>
 |-
 |Therefore, we have

@@ Line 1: / Line 1: @@
 <span class="exam">Evaluate the indefinite and definite integrals.
-::<span class="exam">a) &nbsp; <math>\int x^2 e^x~dx</math>
+<span class="exam">(a) &nbsp; <math>\int x^2 e^x~dx</math>
-::<span class="exam">b) &nbsp; <math>\int_{1}^{e} x^3\ln x~dx</math>

@@ Line 10: / Line 10: @@
 |'''1.''' Integration by parts tells us that
 |-
-|&nbsp; &nbsp; <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
 |-
+|
-&nbsp; &nbsp; You could use integration by parts.
+&nbsp; &nbsp; &nbsp; &nbsp; You could use integration by parts.
 |-
+|
-&nbsp; &nbsp; Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math>
 |-
+|
-&nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int x\ln x~dx=\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 33: / Line 39: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
 |Therefore, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math>
 |}
@@ Line 45: / Line 53: @@
 |Now, we need to use integration by parts again.
 |-
-|Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math>
 |-
 |Building on the previous step, we have
 |-
-| &nbsp; &nbsp; <math>\begin{array}{rcl}
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int x^2 e^x~dx} & = & \displaystyle{x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)}\\
 &&\\
@@ Line 62: / Line 72: @@
 |We proceed using integration by parts.
 |-
-|Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>
+|Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math>
 |-
 |Therefore, we have
 |-
+|
-&nbsp; &nbsp; <math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx}\\
 &&\\
@@ Line 79: / Line 91: @@
 |Now, we evaluate to get
 |-
-| &nbsp; &nbsp; <math>\begin{array}{rcl}
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\
 &&\\
@@ Line 92: / Line 104: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>x^2e^x-2xe^x+2e^x+C</math>
 |-
-|'''(b)''' &nbsp; <math>\frac{3e^4+1}{16}</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{3e^4+1}{16}</math>
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
+|'''1.''' Integration by parts tells us that
 |-
 |'''2.''' How would you integrate <math style="vertical-align: -12px">\int x\ln x~dx?</math>
 |-
+|
-::You could use integration by parts.
+&nbsp; &nbsp; You could use integration by parts.
 |-
+|
-::Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
+&nbsp; &nbsp; Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x~dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -12px">v=\frac{x^2}{2}.</math>
 |-
+|
-::Thus, <math style="vertical-align: -15px">\int x\ln x~dx=\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.</math>
+&nbsp; &nbsp; Thus, <math style="vertical-align: -15px">\int x\ln x~dx=\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.</math>
 |}
@@ Line 29: / Line 31: @@
 !Step 1: &nbsp;
 |-
-|We proceed using integration by parts. Let <math style="vertical-align: 0px">u=x^2</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2xdx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|We proceed using integration by parts.
 |-
 |Therefore, we have
@@ Line 39: / Line 43: @@
 !Step 2: &nbsp;
 |-
-|Now, we need to use integration by parts again. Let <math style="vertical-align: 0px">u=2x</math> and <math style="vertical-align: 0px">dv=e^xdx.</math> Then, <math style="vertical-align: 0px">du=2dx</math> and <math style="vertical-align: 0px">v=e^x.</math>
+|Now, we need to use integration by parts again.
 |-
 |Building on the previous step, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -15px">\int x^2 e^x~dx=x^2e^x-\bigg(2xe^x-\int 2e^x~dx\bigg)=x^2e^x-2xe^x+2e^x+C.</math>
+| &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 50: / Line 60: @@
 !Step 1: &nbsp;
 |-
-|We proceed using integration by parts. Let <math style="vertical-align: -1px">u=\ln x</math> and <math style="vertical-align: 0px">dv=x^3dx.</math> Then, <math style="vertical-align: -13px">du=\frac{1}{x}dx</math> and <math style="vertical-align: -14px">v=\frac{x^4}{4}.</math>
+|We proceed using integration by parts.
 |-
 |Therefore, we have
 |-
-| &nbsp;&nbsp; <math style="vertical-align: -20px">\int_{1}^{e} x^3\ln x~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx=\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}.</math>
-|-
+&nbsp; &nbsp; <math>\begin{array}{rcl}
+\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)\right|_{1}^{e}-\int_1^e \frac{x^3}{4}~dx}\\
 |}
@@ Line 64: / Line 79: @@
 |Now, we evaluate to get
 |-
-| &nbsp; &nbsp; <math style="vertical-align: -17px">\int_{1}^{e} x^3\ln x~dx=\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)=\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}=\frac{3e^4+1}{16}.</math>
+| &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{\int_{1}^{e} x^3\ln x~dx} & = & \displaystyle{\bigg((\ln e) \frac{e^4}{4}-\frac{e^4}{16}\bigg)-\bigg((\ln 1) \frac{1^4}{4}-\frac{1^4}{16}\bigg)}\\
+&&\\
-|-
+& = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\
+&&\\
 |}

@@ Line 21: / Line 21: @@
 ::Thus, <math style="vertical-align: -15px">\int x\ln x~dx=\frac{x^2\ln x}{2}-\int \frac{x}{2}~dx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.</math>
 |}
 '''Solution:'''
@@ Line 69: / Line 70: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"