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