anonymous
  • anonymous
determine if the integral 1 / (1+x^2)^1/2 converges or diverges (Hint. use comparison test) so far i compared it to integral 1/ (x^2)^1/2 but that diverges
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
cantor u there
anonymous
  • anonymous
It diverges. I used the p series test though. If you distribute the exponent of 1/2, you get x^1/2, and 1/2 < 1 so it is divergent. If you are required to use the ratio test, you must use a function that gives values smaller than the original.
anonymous
  • anonymous
I'm sorry not x^1/2 but it equals 1/x which if you use the integral test goes to the natural log, which is divergent. Also according to the p series test, the degree in the denominator must not equal one, so either way it diverges.

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anonymous
  • anonymous
If you need to use the comparison test, you'd compare it to 1/x because in the example you used, you must distribute the exponent.
anonymous
  • anonymous
but its the wrong comparison
anonymous
  • anonymous
how do you compare it
anonymous
  • anonymous
i am using the comparison test, so 0< integral g(x) < integral f(x)
anonymous
  • anonymous
if integral g(x) diverges, then integral f(x) diverges , find integral g(x)
anonymous
  • anonymous
Yes, the function you choose must be smaller. In this instance, the +1 doesn't matter, giving us 1/(x^2)^1/2, so if you distribute the one half, you get 1/x, which we can is the function g(x). We know g(x) diverges because of the several tests I mentioned above (both the integral, ratio, and p series tests) which means that the function f(x) which is 1/(1+x^2)^1/2, riveted as well. Therefore, because g(x) diverges, so does f(x). Does that answer your question?
anonymous
  • anonymous
no because 1/x^2 ^1/2 is greater than 1 / ( 1 + x^2 ) ^1/2
anonymous
  • anonymous
its in the wrong order
anonymous
  • anonymous
In this instance it doesn't matter, because if you take the limit as you approach infinity, the +1 doesn't matter. It's like 1/10000000 is hardly different from 1/10000001.
anonymous
  • anonymous
but youre not using the comparison test then
anonymous
  • anonymous
but i see your point
anonymous
  • anonymous
oh, hmmmm, dunno
anonymous
  • anonymous
You can check it by the Alternating series test. If it still diverges, it absolutely diverges.
anonymous
  • anonymous
its not an alternating series though
anonymous
  • anonymous
If there was an x! Or another variable, then it would change things, but a constant is irrelevant in this instance.
anonymous
  • anonymous
yo cantor
anonymous
  • anonymous
I know it's not alternating. You use the alternating test to check for absolute divergence or convergence. For example, 1/x is conditionally divergent because if it alternates, it converges. If there is no alternating piece, it diverges. Hence, you can use the alternating test to check if it always diverges or only sometimes.
anonymous
  • anonymous
hmmm, how is that relevant
anonymous
  • anonymous
It allows you to check convergence or divergence. That was your question originally.
anonymous
  • anonymous
cantor
anonymous
  • anonymous
???????
anonymous
  • anonymous
yeah, this question im getting tutored for im sorry
anonymous
  • anonymous
i promise i will answer your question
anonymous
  • anonymous
I'm saying, if you have a choice, I wouldn't use the comparison test. If you don't, then just do the same thing with the denominator that is larger and use one of the methods I discussed earlier to see.
anonymous
  • anonymous
can yuo give me a formal method
anonymous
  • anonymous
You mean show you the steps? To which one? The comparison test? The p series? The integral test?
anonymous
  • anonymous
any , the comparison test fails, as i explained earlier
anonymous
  • anonymous
the direct integral comparison test
anonymous
  • anonymous
cantor can u help me now, I have 2 parts left and I'm done. first question at the top
anonymous
  • anonymous
one sec

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