1. Comaprison tests for convergence

Hi, I'm having some difficulty in recognising what new improper integral to create when doing a comparison test.
IS there a step-by-step logical approach that could be used?
My understanding is, we do comparison tests because the improper integral being dealt with is too difficult to integrate. I understand the rules in that if f(x)>g(x)
then f(x) is divergent if g(x) is divergent and g(x) is convergent if f(x) is convergent.

But, I don't understand a few problems:

So, for example,

http://i.imgur.com/OF9Ip.jpg

If you could look at 6b, and the solutions to 6b.

I don't understand what the solutions mean when they say "Use x< x^1/4"

The same with 6c. Why have they said 'use 1/3y^4' ?

I managed to work out 6a myself. are there flaws in the solutions? What am I missing here..

Additionally, in regards to the matrices question at the top, what method is being referred to there? Is it asking for me to use row operations?

Thanks!

2. Re: Comaprison tests for convergence

They use $x \le x^{\frac{1}{4}}$ so that $1 - x^{\frac{1}{4}} \le 1 - x$

$\frac{1}{1 - x^{\frac{1}{4}}} \ge \frac{1}{1-x}$ (along the domain)

Therefore

$\int_{0}^{1} \frac{1}{1 - x^{\frac{1}{4}}}\, dx \ge \int_{0}^{1} \frac{1}{1-x}\, dx$

3. Re: Comaprison tests for convergence

^Thanks!

makes sense

although, I'm still not sure about why they use 3y^4 ?
Would it not make more sense to use y^4?