Difference between revisions of "009B Sample Final 3, Problem 2"

Latest revision as of 17:41, 2 December 2017

Evaluate 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_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~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{x^2}{(1+x^3)^2}~dx}

(c) 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^e \frac{\cos(\ln(x))}{x}~dx}

Solution

Detailed Solution

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@@ Line 7: / Line 7: @@
 <span class="exam">(c) &nbsp;<math>\int_1^e \frac{\cos(\ln(x))}{x}~dx</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 3, Problem 2 Solution|'''<u>Solution</u>''']]
-|-
-|'''1.'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{1}{1+x^2}~dx=\arctan(x)+C</math>
-|-
-|'''2.''' How would you integrate &nbsp; <math style="vertical-align: -12px">\int \frac{\ln x}{x}~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: -5px">u=\ln(x).</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math style="vertical-align: -13px">du=\frac{1}{x}dx.</math>
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; Thus,
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\int \frac{\ln x}{x}~dx} & = & \displaystyle{\int u~du}\\
-&&\\
-& = & \displaystyle{\frac{u^2}{2}+C}\\
-&&\\
-& = & \displaystyle{\frac{(\ln x)^2}{2}+C.}
-\end{array}</math>
-|-
-|}
-'''Solution:'''
+[[009B Sample Final 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|First, we notice
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx=\int_0^{\frac{\sqrt{3}}{4}} \frac{1}{1+(4x)^2}~dx.</math>
-|-
-|Now, we use &nbsp;<math>u</math>-substitution.
-|-
-|Let &nbsp;<math style="vertical-align: -5px">u=4x.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: -5px">du=4dx</math>&nbsp; and &nbsp;<math>\frac{du}{4}=dx.</math>
-|-
-|Also, we need to change the bounds of integration.
-|-
-|Plugging in our values into the equation &nbsp;<math style="vertical-align: -5px">u=4x,</math>
-|-
-|we get &nbsp;<math style="vertical-align: -15px">u_1=4(0)=0</math>&nbsp; and &nbsp;<math style="vertical-align: -16px">u_2=4\bigg(\frac{\sqrt{3}}{4}\bigg)=\sqrt{3}.</math>
-|-
-|Therefore, the integral becomes
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -19px">\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~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^{\frac{\sqrt{3}}{4}} \frac{1}{1+16x^2}~dx} & = & \displaystyle{\frac{1}{4}\int_0^{\sqrt{3}} \frac{1}{1+u^2}~du}\\
-&&\\
-& = & \displaystyle{\frac{1}{4}\arctan(u)\bigg|_0^{\sqrt{3}}}\\
-&&\\
-& = & \displaystyle{\frac{1}{4}\arctan(\sqrt{3})-\frac{1}{4}\arctan(0)}\\
-&&\\
-& = & \displaystyle{\frac{1}{4}\bigg(\frac{\pi}{3}\bigg)-0}\\
-&&\\
-& = & \displaystyle{\frac{\pi}{12}.}
-\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=1+x^3.</math>
-|-
-|Then, &nbsp;<math style="vertical-align: 0px">du=3x^2dx</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">\frac{du}{3}=x^2dx.</math>
-|-
-|Therefore, the integral becomes
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\frac{1}{3}\int \frac{1}{u^2}~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 \frac{x^2}{(1+x^3)^2}~dx} & = & \displaystyle{\frac{1}{3}\int \frac{1}{u^2}~du}\\
-&&\\
-& = & \displaystyle{-\frac{1}{3u}+C}\\
-&&\\
-& = & \displaystyle{-\frac{1}{3(1+x^3)}+C.}
-\end{array}</math>
-|-
-|
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>\frac{\pi}{12}</math>
-|-
-|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>-\frac{1}{3(1+x^3)}+C</math>
-|-
-|&nbsp; &nbsp;'''(c)''' &nbsp; &nbsp;
-|}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 3, Problem 2"

Latest revision as of 17:41, 2 December 2017

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