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