Difference between revisions of "009B Sample Midterm 3, Problem 4"

Revision as of 17:26, 4 November 2017

Compute the following integrals:

(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^2\sin (x^3) ~dx}

(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx}

(insert picture of handwritten solution)

Detailed Solution for this Problem

Return to Sample Exam

@@ 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) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
-<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the number of hours since the drug was administered.
+<span class="exam">(b) &nbsp; <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 &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<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: &nbsp;
-|-
-|If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b r'(t)~dt,</math>&nbsp; what are we calculating?
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; We are calculating &nbsp;<math style="vertical-align: -5px">r(b)-r(a).</math>&nbsp; This is the total reaction to the
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; drug from &nbsp;<math style="vertical-align: 0px">t=a</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=b.</math>
-|}
-'''Solution:'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|To calculate the total reaction to the drug from &nbsp;<math style="vertical-align: -1px">t=1</math>&nbsp; to &nbsp;<math style="vertical-align: -4px">t=6,</math>
-|-
-|we need to calculate
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|We proceed using integration by parts.
-|-
-|Let &nbsp;<math style="vertical-align: 0px">u=2t^2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -1px">du=4t~dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
-|-
-|Then, we have
-|-
-|&nbsp;&nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|Now, we need to use integration by parts again.
-|-
-|Let &nbsp;<math style="vertical-align: 0px">u=4t</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-t}dt.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -1px">du=4dt</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">v=-e^{-t}.</math>
-|-
-|Thus, we get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
-|-
-|&nbsp; &nbsp; &nbsp;&nbsp; <math>\frac{-100+10e^5}{e^6}</math>
-|}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 3, Problem 4"

Revision as of 17:26, 4 November 2017

Navigation menu

Search