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anonymous
 5 years ago
Use an appropriate change of variable to evaluate the convergent improper integral
1/[(x^0.5)(1+x)]
what should i use?
anonymous
 5 years ago
Use an appropriate change of variable to evaluate the convergent improper integral 1/[(x^0.5)(1+x)] what should i use?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is (pi) but i dunno how should i get to the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is it dy/dx = 1/[(x^0.5)(1+x)] ? Move the dx to the right side, integrate the left and you get "y", integrate the right side now and try using a trig substitution...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry..my bad... 1st time use this website...but i found this fun ^ ^ the question is based on improper integral. \[\int\limits\limits_{0}^{\infty} 1\div [ \sqrt{x} (1+x) ]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, for the right hand side, you're going to use a usubstitution (try u = sqrt(x)) and after the sub, you should see a pretty clean inverse trig function that pops out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then, finding evaluating at the limits will be an easy task. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*eliminate "finding" :P typo.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for helping but i cant get through with the trig part.. i cant see which i use to substitute i stuck at \[1\div (u + u ^{3})\] :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've been scratching my head at this one for a while, but I just got it. :) You've got du=1/(2sqrt(x)) dx, right? You've gotta multiply the left by 2sqrt(x) to get dx = 2√x du! Plug THAT into your integral and you'll get 2 * ∫ 1/(1+u^2) du. That should be much easier to solve!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hm...i getting 0 at the end .. hm... odd.. the answer i getting from my lecture is Pi what is the answer u getting? i get in the end \[\tan^{1} u \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Exactly, you're just forgetting to multiply by 2. So, your final expression in the end is 2 * arctan√x from 0 to ∞. The value of arctan(√x) as x approaches infinity is π/2, and as x approaches 0, is 0! So you get 2* (π/20) after applying the limits and the final value is π.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0damn i keep doing so stupid mistake...i might need my brain rest for a while..XD..it's been 5 hours for me ... thanks..u r awesome.. ^ ^
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