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 Calculus Calculus Math Forum

May 18th, 2018, 06:43 PM   #1
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Improper integrals

Hello everyone,
Can anyone help me with the antiderivative of this integtral?

I graphically proved it is convergent to zero but algebraicly I could't find the antiderivative of the integral.

I tried u-sub and then arctan. but couldn't get out of it. Attached Images imp.jpg (7.5 KB, 1 views) May 18th, 2018, 07:24 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra Are you trying to evaluate the integral of just prove convergence? In the latter case you can use $$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x+2}}$$ on the interval. May 19th, 2018, 06:01 AM   #3
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Quote:
 Originally Posted by v8archie Are you trying to evaluate the integral of just prove convergence? In the latter case you can use $$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x+2}}$$ on the interval.
maybe this comparison instead ...

$$\frac1{\sqrt{x^5+2}} < \frac1{\sqrt{x^4+2}}$$

since $\displaystyle \int_1^\infty \frac{1}{\sqrt{x+2}} \, dx$ diverges May 19th, 2018, 10:29 AM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 before you all rush off to try and show this converges to zero... Wolfram|Alpha: Computational Intelligence May 19th, 2018, 11:24 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra The integral clearly doesn't, the integrand does. May 19th, 2018, 01:19 PM #6 Senior Member   Joined: Apr 2017 From: New York Posts: 155 Thanks: 6 smaller function diverges mean larger function (above that line) also diverges or larger function converges mean smaller function (line below that) also converges. How can I prove that original function is diverging by proving a larger function is diverging? Shouldn't I pick smaller function then that and prove it diverges instead of picking larger one which diverges?! Last edited by skipjack; May 19th, 2018 at 10:02 PM. May 19th, 2018, 04:14 PM   #7
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Quote:
 Originally Posted by Leonardox smaller function diverges mean larger function (above that line) also diverges or larger function converges mean smaller function (line below that) also converges. How can I prove that original function is diverging by proving a larger function is diverging? Shouldn't I pick smaller function then that and prove it diverges instead of picking larger one which diverges?!
$\displaystyle \int_1^\infty \frac{1}{\sqrt{x^5+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4+2}} \, dx < \int_1^\infty \frac{1}{\sqrt{x^4}} \, dx = \int_1^\infty \frac{1}{x^2} \, dx = 1$

Last edited by skipjack; May 19th, 2018 at 10:13 PM. Tags improper, integrals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post contrattivo Real Analysis 3 December 21st, 2012 03:56 AM aaron-math Calculus 2 October 7th, 2011 01:05 AM tulipe Calculus 2 March 11th, 2011 11:59 PM ChloeG Calculus 2 March 1st, 2011 02:10 PM Dr Grey Calculus 2 July 27th, 2010 12:30 PM

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