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| − | <span class="exam"> The rate of reaction to a drug is given by: | + | <span class="exam"> Compute the following integrals: |
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| − | ::<math>r'(t)=2t^2e^{-t}</math>
| + | <span class="exam">(a) <math>\int x^2\sin (x^3) ~dx</math> |
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| − | <span class="exam">where <math style="vertical-align: 0px">t</math> is the number of hours since the drug was administered. | + | <span class="exam">(b) <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math> |
| | + | <hr> |
| | + | [[009B Sample Midterm 3, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | <span class="exam">Find the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: 0px">t=6.</math>
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| | + | [[009B Sample Midterm 3, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |If we calculate <math style="vertical-align: -14px">\int_a^b r'(t)~dt,</math> what are we calculating?
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| − | We are calculating <math style="vertical-align: -5px">r(b)-r(a).</math> This is the total reaction to the
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| − | drug from <math style="vertical-align: 0px">t=a</math> to <math style="vertical-align: 0px">t=b.</math>
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |To calculate the total reaction to the drug from <math style="vertical-align: -1px">t=1</math> to <math style="vertical-align: -4px">t=6,</math>
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| − | |we need to calculate
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| − | <math>\int_1^6 r'(t)~dt=\int_1^6 2t^2e^{-t}~dt.</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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| − | |We proceed using integration by parts.
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| − | |Let <math style="vertical-align: 0px">u=2t^2</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math>
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| − | |Then, <math style="vertical-align: -1px">du=4t~dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
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| − | |Then, we have
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| − | | <math style="vertical-align: -14px">\int_1^62t^2e^{-t}~dt=\left. -2t^2e^{-t}\right|_1^6+\int_1^6 4te^{-t}~dt.</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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| − | |Now, we need to use integration by parts again.
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| − | |Let <math style="vertical-align: 0px">u=4t</math> and <math style="vertical-align: 0px">dv=e^{-t}dt.</math>
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| − | |Then, <math style="vertical-align: -1px">du=4dt</math> and <math style="vertical-align: 0px">v=-e^{-t}.</math>
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| − | |Thus, we get
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^62t^2e^{-t}~dt} & = & \displaystyle{\left. (-2t^2e^{-t}-4te^{-t})\right|_1^6+\int_1^6 4e^{-t}}\\
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| − | &&\\
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| − | & = & \displaystyle{\left. (-2t^2e^{-t}-4te^{-t}-4e^{-t})\right|_1^6}\\
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| − | &&\\
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| − | & = & \displaystyle{(-2(6)^2e^{-6}-4(6)e^{-6}-4e^{-6})-(-2(1)^2e^{-1}-4(1)e^{-1}-4e^{-1})} \\
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| − | &&\\
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| − | & = & \displaystyle{\frac{-100+10e^5}{e^6}.}
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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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| − | | <math>\frac{-100+10e^5}{e^6}</math>
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| − | |}
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| | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
