009B Sample Midterm 1, Problem 3 - Revision history

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MathAdmin: Created page with "Evaluate the indefinite and definite integrals. ::a) $\int x^2 e^x~dx$ ::b) $\int_{1}^{e} x^3\ln x~d..."$

2016-02-01T22:37:06Z

Created page with "Evaluate the indefinite and definite integrals. ::a) <math>\int x^2 e^x~dx</math> ::b) <math>\int_{1}^{e} x^3\ln x~d..."

New page

Evaluate the indefinite and definite integrals.

::a) <math>\int x^2 e^x~dx</math>
::b) <math>\int_{1}^{e} x^3\ln x~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|Review integration by parts.
|}

'''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 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>.
|}

'''(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 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>.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we evaluate to get
|-
|    <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>.
|-
|
|-
|
|}

{| 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|'''Return to Sample Exam''']]

@@ 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.
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<span class="exam">(b) &nbsp; Use your answer to part (a) to find the size of the population after one month.
-!Foundations: &nbsp;
+<hr>
-|-
+[[009B Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</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) <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) <math>\int_{1}^{e} x^3\ln x~dx</math>
@@ Line 8: / Line 9: @@
 !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>
+|'''2.''' How would you integrate &nbsp;<math style="vertical-align: -12px">\int x\ln x~dx?</math>
 |-
+|
-::You could use integration by parts.
+&nbsp; &nbsp; &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> Thus,
+&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>
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp; <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.}\\
+& = & \displaystyle{\frac{x^2\ln x}{2}-\frac{x^2}{4}+C.}
 \end{array}</math>
 |}
 '''Solution:'''
@@ Line 32: / Line 38: @@
 !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
 |-
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -12px">\int x^2 e^x~dx=x^2e^x-\int 2xe^x~dx.</math>
 |}
@@ Line 43: / Line 52: @@
 !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; &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)}\\
 &&\\
-& = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}\\
+& = & \displaystyle{x^2e^x-2xe^x+2e^x+C.}
 \end{array}</math>
 |}
@@ Line 59: / Line 71: @@
 !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
 |-
+|
-::<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 }\\
+\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}.}\\
+& = & \displaystyle{\left.\ln x \bigg(\frac{x^4}{4}\bigg)-\frac{x^4}{16}\right|_{1}^{e}.}
 \end{array}</math>
 |}
@@ Line 76: / Line 92: @@
 |Now, we evaluate to get
 |-
+| &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)}\\
 &&\\
 & = & \displaystyle{\frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}}\\
 &&\\
-& = & \displaystyle{\frac{3e^4+1}{16}.}\\
+& = & \displaystyle{\frac{3e^4+1}{16}.}
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&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>
 |-
-|&nbsp;&nbsp; '''(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 16: / Line 16: @@
 |-
+|
-::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>
+::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,
 |-
+|
-::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>
+::<math>\begin{array}{rcl}
 |}

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|Review integration by parts.
+|Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du.</math>
 |}