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| | ::<math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math> | | ::<math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 3, Problem 7 Solution|'''<u>Solution</u>''']] |
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| − | |'''Direct Comparison Test for Improper Integrals'''
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| − | | 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>
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| − | | 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>
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| − | | '''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.
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| − | | '''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.
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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 3, Problem 7 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |We use the Direct Comparison Test for Improper Integrals.
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| − | |For all <math style="vertical-align: 0px">x</math> in <math style="vertical-align: -5px">[1,\infty),</math>
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| − | | <math>0\le \frac{\sin^2(x)}{x^3} \le \frac{1}{x^3}.</math>
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| − | |Also,
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| − | | <math style="vertical-align: -15px">\frac{\sin^2(x)}{x^3}</math> and <math style="vertical-align: -15px">\frac{1}{x^3}</math>
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| − | |-
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| − | |are continuous on <math style="vertical-align: -5px">[1,\infty).</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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| − | |Now, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\int_1^\infty \frac{1}{x^3}~dx} & = & \displaystyle{\lim_{a\rightarrow \infty} \int_1^a \frac{1}{x^3}~dx}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{1}{-2x^2}\bigg|_1^a}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{a\rightarrow \infty} \frac{1}{-2a^2}+\frac{1}{2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Since <math style="vertical-align: -15px">\int_1^\infty \frac{1}{x^3}~dx</math> converges,
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| − | | <math>\int_1^\infty \frac{\sin^2(x)}{x^3}~dx</math>
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| − | |-
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| − | |converges by the Direct Comparison Test for Improper Integrals.
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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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| − | | converges (by the Direct Comparison Test for Improper Integrals)
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| − | |}
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| | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
