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| | <span class="exam"> Evaluate the improper integrals: | | <span class="exam"> Evaluate the improper integrals: |
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| − | ::<span class="exam">a) <math>\int_0^{\infty} xe^{-x}~dx</math>
| + | <span class="exam">(a) <math>\int_0^{\infty} xe^{-x}~dx</math> |
| − | ::<span class="exam">b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(b) <math>\int_1^4 \frac{dx}{\sqrt{4-x}}</math> |
| − | !Foundations:
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| − | |-
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| − | |'''1.''' How could you write <math style="vertical-align: -14px">\int_0^{\infty} f(x)~dx</math> so that you can integrate?
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| − | |
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| − | ::You can write <math>\int_0^{\infty} f(x)~dx=\lim_{a\rightarrow\infty} \int_0^a f(x)~dx.</math>
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| − | |-
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| − | |'''2.''' How could you write <math>\int_{-1}^1 \frac{1}{x}~dx</math> ?
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| − | ::The problem is that  <math>\frac{1}{x}</math>  is not continuous at <math style="vertical-align: 0px">x=0</math>.
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| − | ::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>.
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| − | |-
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| − | |'''3.''' How would you integrate <math style="vertical-align: -13px">\int xe^x\,dx</math> ?
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| − | ::You can use integration by parts.
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| − | |
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| − | ::Let <math style="vertical-align: 0px">u=x</math> and <math style="vertical-align: 0px">dv=e^xdx</math>.
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| − | |}
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| − | '''Solution:''' | + | <hr> |
| | + | [[009B Sample Final 1, Problem 6 Solution|'''<u>Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009B Sample Final 1, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 1:
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| − | |-
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| − | |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>.
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| − | |- | |
| − | |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>.
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| − | |-
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| − | |Thus, the integral becomes
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| − | |-
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| − | |
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| − | ::<math>\int_0^{\infty} xe^{-x}~dx=\lim_{a\rightarrow \infty} \left.-xe^{-x}\right|_0^a-\int_0^a-e^{-x}\,dx.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |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>.
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| − | |-
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| − | |Since the integral is a definite integral, we need to change the bounds of integration.
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| − | |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>.
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| − | |-
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| − | |Thus, the integral becomes
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-\int_0^{-a}e^{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} -xe^{-x}\bigg|_0^a-e^{u}\bigg|_0^{-a}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}.\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, we evaluate to get
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} -ae^{-a}-(e^{-a}-1)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a}{e^a}-\frac{1}{e^a}+1}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-a-1}{e^a}+1}.\\
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| − | \end{array}</math>
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| − | |-
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| − | |Using L'Hôpital's Rule, we get
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int_0^{\infty} xe^{-x}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{-1}{e^a}+1}\\
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| − | &&\\
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| − | & = & \displaystyle{0+1}\\
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| − | &&\\
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| − | & = & \displaystyle{1}.\\
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| − | \end{array}</math>
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| − | |-
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| − | |
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| − | |}
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| − |
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| − | == 3 ==
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |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>.
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| − | |-
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| − | |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>.
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| − | |-
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| − | |Since the integral is a definite integral, we need to change the bounds of integration.
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| − | |-
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| − | |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>.
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| − | |-
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| − | |Thus, the integral becomes
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| − | |-
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| − | |
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| − | ::<math>\int_1^4 \frac{dx}{\sqrt{4-x}}\,=\,\lim_{a\rightarrow 4} \int_3^{4-a}\frac{-1}{\sqrt{u}}~du.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |We integrate to get
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^4 \frac{dx}{\sqrt{4-x}}} & = & \displaystyle{\lim_{a\rightarrow 4} -2u^{\frac{1}{2}}\bigg|_{3}^{4-a}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow 4}-2\sqrt{4-a}+2\sqrt{3}}\\
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| − | &&\\
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| − | & = & \displaystyle{2\sqrt{3}}.\\
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| − | \end{array}</math>
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| − | |}
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| − | == 4 ==
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | |'''(a)''' <math style="vertical-align: -3px">1</math>
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| − | |-
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| − | |'''(b)''' <math style="vertical-align: -4px">2\sqrt{3}</math>
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| − | |}
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| | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
