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

Latest revision as of 17:16, 23 November 2017

Find a curve 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 y=f(x)} with the following properties:

(i) 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 f''(x)=6x}

(ii) Its graph passes through the point 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 (0,1)} and has a horizontal tangent there.

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam"> Compute the following integrals:
+<span class="exam"> Find a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; with the following properties:
-<span class="exam">(a) &nbsp; <math>\int x^2\sin (x^3) ~dx</math>
+<span class="exam">(i) &nbsp; <math style="vertical-align: -5px">f''(x)=6x</math>
-<span class="exam">(b) &nbsp; <math>\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx</math>
+<span class="exam">(ii) &nbsp; Its graph passes through the point &nbsp;<math style="vertical-align: -5px">(0,1)</math>&nbsp; and has a horizontal tangent there.
+<hr>
+[[009B Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
-!Foundations: &nbsp;
-|-
-|How would you integrate &nbsp;<math style="vertical-align: -5px">2x(x^2+1)^3~dx?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; You could use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -3px">u=x^2+1.</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2x~dx.</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Thus,
-|-
-|
-&nbsp; &nbsp;&nbsp;&nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int 2x(x^2+1)^3~dx} & = & \displaystyle{\int u^3~du}\\
-&&\\
-& = & \displaystyle{\frac{u^4}{4}+C}\\
-&& \\
-& = & \displaystyle{\frac{(x^2+1)^4}{4}+C.}\\
-\end{array}</math>
-|}
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -1px">u=x^3.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -1px">du=3x^2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{3}=x^2~dx.</math>
-|-
-|Therefore, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\int x^2\sin (x^3) ~dx=\int \frac{\sin(u)}{3}~du.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We integrate to get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int x^2\sin (x^3) ~dx} & = & \displaystyle{-\frac{1}{3}\cos(u)+C}\\
-&&\\
-& = & \displaystyle{-\frac{1}{3}\cos(x^3)+C.}\\
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We proceed using u substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -5px">u=\cos(x).</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -5px">du=-\sin(x)~dx.</math>
-|-
-|Since this is a definite integral, we need to change the bounds of integration.
-|-
-|We have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -15px">u_1=\cos\bigg(-\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">u_2=\cos\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Therefore, we get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(x)\sin (x)~dx} & = & \displaystyle{\int_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}} -u^2~du}\\
-&&\\
-& = & \displaystyle{\left.\frac{-u^3}{3}\right|_{\frac{\sqrt{2}}{2}}^{\frac{\sqrt{2}}{2}}}\\
-&&\\
-& = & \displaystyle{0.} \\
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>-\frac{1}{3}\cos(x^3)+C</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>0</math>
-|}
 [[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

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

Latest revision as of 17:16, 23 November 2017

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