anonymous
  • anonymous
Use an appropriate change of variable to evaluate the convergent improper integral 1/[(x^0.5)(1+x)] what should i use?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
the answer is (pi) but i dunno how should i get to the answer
anonymous
  • anonymous
Is 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
  • anonymous
sorry..my bad... 1st time use this website...but i found this fun ^ ^ the question is based on improper integral. \[\int\limits\limits_{0}^{\infty} 1\div [ \sqrt{x} (1+x) ]\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Okay, for the right hand side, you're going to use a u-substitution (try u = sqrt(x)) and after the sub, you should see a pretty clean inverse trig function that pops out.
anonymous
  • anonymous
Then, finding evaluating at the limits will be an easy task. :)
anonymous
  • anonymous
*eliminate "finding" :P typo.
anonymous
  • anonymous
thanks 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
  • anonymous
I'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
  • anonymous
hm...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
  • anonymous
Exactly, 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/2-0) after applying the limits and the final value is π.
anonymous
  • anonymous
damn 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.. ^ ^

Looking for something else?

Not the answer you are looking for? Search for more explanations.