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2017-04-10T16:55:33Z

Created page with "<span class="exam">Does the following integral converge or diverge? Prove your answer! ::<math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math> {| class="mw-collapsible mw-coll..."

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<span class="exam">Does the following integral converge or diverge? Prove your answer!

::<math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''Direct Comparison Test for Improper Integrals'''
|-
|        Let  <math style="vertical-align: -5px">f</math>  and  <math style="vertical-align: -5px">g</math>  be continuous on  <math style="vertical-align: -5px">[a,\infty)</math>
|-
|        where  <math style="vertical-align: -5px">0\le f(x)\le g(x)</math>  for all  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -5px">[a,\infty).</math>
|-
|       '''1.'''  If  <math style="vertical-align: -14px">\int_a^\infty g(x)~dx</math>  converges, then  <math style="vertical-align: -14px">\int_a^\infty f(x)~dx</math>  converges.
|-
|       '''2.'''  If  <math style="vertical-align: -14px">\int_a^\infty f(x)~dx</math>  diverges, then  <math style="vertical-align: -14px">\int_a^\infty g(x)~dx</math>  diverges.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We use the Direct Comparison Test for Improper Integrals.
|-
|For all  <math style="vertical-align: 0px">x</math>  in  <math style="vertical-align: -5px">[1,\infty),</math>
|-
|        <math>0\le \frac{\sin^2(x)}{x^3} \le \frac{1}{x^3}.</math>
|-
|Also,
|-
|       <math style="vertical-align: -15px">\frac{\sin^2(x)}{x^3}</math>  and  <math style="vertical-align: -15px">\frac{1}{x^3}</math>
|-
|are continuous on  <math style="vertical-align: -5px">[1,\infty).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_1^\infty \frac{1}{x^3}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \int_1^a \frac{1}{x^3}~dx}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{1}{-2x^2}\bigg|_1^a}\\
&&\\
& = & \displaystyle{\lim_{a\rightarrow \infty} \frac{1}{-2a^2}+\frac{1}{2}}\\
&&\\
& = & \displaystyle{\frac{1}{2}.}
\end{array}</math>
|-
|Since  <math style="vertical-align: -15px">\int_1^\infty \frac{1}{x^3}~dx</math>  converges,
|-
|       <math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math>
|-
|converges by the Direct Comparison Test for Improper Integrals.
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       converges (by the Direct Comparison Test for Improper Integrals)
|}
[[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

009B Sample Final 3, Problem 7 - Revision history

MathAdmin: Created page with "Does the following integral converge or diverge? Prove your answer! ::\int_1^\infty \frac{\sin^2(x)}{x^3}~dx {| class="mw-collapsible mw-coll..."

MathAdmin: Created page with "Does the following integral converge or diverge? Prove your answer! :: $\int_1^\infty \frac{\sin^2(x)}{x^3}~dx$ {| class="mw-collapsible mw-coll..."