009B Sample Final 1, Problem 6 - Revision history

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MathAdmin: Created page with " Evaluate the improper integrals: ::a) $\int_0^{\infty} xe^{-x}~dx$ ::b) $\int_1^4 \frac{dx}{\sqrt{4-..."$

2016-02-27T05:41:32Z

Created page with " Evaluate the improper integrals: ::a) <math>\int_0^{\infty} xe^{-x}~dx</math> ::b) <math>\int_1^4 \frac{dx}{\sqrt{4-..."

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Evaluate the improper integrals:

::a) <math>\int_0^{\infty} xe^{-x}~dx</math>
::b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' How could you write <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
|-
|
::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
|-
|'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx</math> ?
|-
|
::The problem is that  <math>\frac{1}{x}</math>  is not continuous at <math style="vertical-align: 0px">x=0</math>.
|-
|
::So, you can write <math style="vertical-align: -15px">\int_{-1}^1 \frac{1}{x}~dx=\lim_{a\rightarrow 0^-} \int_{-1}^a \frac{1}{x}~dx+\lim_{a\rightarrow 0^+} \int_a^1 \frac{1}{x}~dx</math>.
|-
|'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx</math> ?
|-
|
::You can use integration by parts.
|-
|
::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>.
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write <math style="vertical-align: -14px">\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \int_0^a xe^{-x}~dx</math>.
|-
|Now, we proceed using integration by parts. Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^{-x}dx</math>. Then, <math style="vertical-align: 0px">du=dx</math> and <math style="vertical-align: 0px">v=-e^{-x}</math>.
|-
|Thus, the integral becomes
|-
|
::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: 0px">u=-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
|-
|Since the integral is a definite integral, we need to change the bounds of integration.
|-
|Plugging in our values into the equation <math style="vertical-align: 0px">u=-x</math>, we get <math style="vertical-align: -5px">u_1=0</math> and <math style="vertical-align: -3px">u_2=-a</math>.
|-
|Thus, the integral becomes
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-\int_0^{-a}e^{u}~du}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-e^{u}\bigg|_0^{-a}}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we evaluate to get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a}{e^a}-\frac{1}{e^a}+1}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1}.\\
\end{array}</math>
|-
|Using L'Hôpital's Rule, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
&&\\
& = & \displaystyle{0+1}\\
&&\\
& = & \displaystyle{1}.\\
\end{array}</math>
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we write <math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}</math>.
|-
|Now, we proceed by <math style="vertical-align: 0px">u</math>-substitution. We let <math style="vertical-align: -1px">u=4-x</math>. Then, <math style="vertical-align: 0px">du=-dx</math>.
|-
|Since the integral is a definite integral, we need to change the bounds of integration.
|-
|Plugging in our values into the equation <math style="vertical-align: -1px">u=4-x</math>, we get <math style="vertical-align: -5px">u_1=4-1=3</math>  and <math style="vertical-align: -3px">u_2=4-a</math>.
|-
|Thus, the integral becomes
|-
|
::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We integrate to get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\
&&\\
& = & \displaystyle{2\sqrt{3}}.\\
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''  <math style="vertical-align: -3px">1</math>
|-
|'''(b)'''  <math style="vertical-align: -4px">2\sqrt{3}</math>
|}
[[009B_Sample_Final_1|'''Return to Sample Exam''']]

@@ Line 110: / Line 110: @@
 !Step 1: &nbsp;
 |-
-|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
+|First, we write &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}=\lim_{a\rightarrow 4^-} \int_1^a\frac{dx}{\sqrt{4-x}}.</math>
 |-
 |Now, we proceed by &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
@@ Line 125: / Line 125: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4^-} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
 |}
@@ Line 135: / Line 135: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
+\displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4^-} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
 &&\\
-& = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\
+& = & \displaystyle{\lim_{a\rightarrow 4^-}-2\sqrt{4-a}+2\sqrt{3}}\\
 &&\\
 & = & \displaystyle{2\sqrt{3}}.\\

@@ Line 138: / Line 138: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math style="vertical-align: -3px">1</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp;<math style="vertical-align: -3px">1</math>
 |-
-|'''(b)''' &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp;<math style="vertical-align: -4px">2\sqrt{3}</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]