Difference between revisions of "009B Sample Midterm 2, Problem 2"

Revision as of 18:31, 4 November 2017

Evaluate

(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_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt}

(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_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx}

(insert picture of handwritten solution)

Detailed Solution for this Problem

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@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
+<hr>
+(insert picture of handwritten solution)
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009B Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
-!Foundations: &nbsp;
-|-
-|How would you integrate &nbsp;<math style="vertical-align: -12px">\int (2x+1)\sqrt{x^2+x}~dx?</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; You can use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -2px">u=x^2+x.</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -4px">du=(2x+1)~dx.</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int (2x+1)\sqrt{x^2+x}~dx} & = & \displaystyle{\int \sqrt{u}~du}\\
-&&\\
-& = & \displaystyle{\frac{2}{3}u^{3/2}+C}\\
-&&\\
-& = & \displaystyle{\frac{2}{3}(x^2+x)^{3/2}+C.}
-\end{array}</math>
-|}
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We multiply the product inside the integral to get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt}\\
-&&\\
-& = & \displaystyle{\int_1^2 (8t^3+2-15t^{-3})~dt.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We integrate to get
-|-
-| &nbsp;&nbsp; &nbsp; &nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2.</math>
-|-
-|We now evaluate to get
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt} & = & \displaystyle{2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)}\\
-&&\\
-& = & \displaystyle{36+\frac{15}{8}-4-\frac{15}{2}}\\
-&&\\
-& = & \displaystyle{\frac{211}{8}.}
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -2px">u=x^4+2x^2+4.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -5px">du=(4x^3+4x)dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx.</math>
-|-
-|Also, we need to change the bounds of integration.
-|-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -4px">u=x^4+2x^2+4,</math>&nbsp; we get
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=2^4+2(2)^2+4=28.</math>
-|-
-|Therefore, the integral becomes
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|We now have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx} & = & \displaystyle{\frac{1}{4}\int_4^{28}\sqrt{u}~du}\\
-&&\\
-& = & \displaystyle{\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}}\\
-&&\\
-& = & \displaystyle{\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})}\\
-&&\\
-& = & \displaystyle{\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)}\\
-&&\\
-& = & \displaystyle{\frac{1}{6}((2\sqrt{7})^3-2^3).}
-\end{array}</math>
-|-
-|Therefore,
-|-
-| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{211}{8}</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
-|}
 [[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Midterm 2, Problem 2"

Revision as of 18:31, 4 November 2017

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