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