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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

cantor u there

- 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

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.

Looking for something else?

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

## More answers

- 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

but its the wrong comparison

- anonymous

how do you compare it

- anonymous

i am using the comparison test, so 0< integral g(x) < integral f(x)

- anonymous

if integral g(x) diverges, then integral f(x) diverges , find integral g(x)

- 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

no because 1/x^2 ^1/2 is greater than 1 / ( 1 + x^2 ) ^1/2

- anonymous

its in the wrong order

- 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

but youre not using the comparison test then

- anonymous

but i see your point

- anonymous

oh, hmmmm, dunno

- anonymous

You can check it by the Alternating series test. If it still diverges, it absolutely diverges.

- anonymous

its not an alternating series though

- anonymous

If there was an x! Or another variable, then it would change things, but a constant is irrelevant in this instance.

- anonymous

yo cantor

- 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

hmmm, how is that relevant

- anonymous

It allows you to check convergence or divergence. That was your question originally.

- anonymous

cantor

- anonymous

???????

- anonymous

yeah, this question im getting tutored for
im sorry

- anonymous

i promise i will answer your question

- 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

can yuo give me a formal method

- anonymous

You mean show you the steps?
To which one? The comparison test? The p series? The integral test?

- anonymous

any , the comparison test fails, as i explained earlier

- anonymous

the direct integral comparison test

- anonymous

cantor can u help me now, I have 2 parts left and I'm done. first question at the top

- anonymous

one sec

Looking for something else?

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