# Comaprison tests for convergence

#### 99.95

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!

#### richard1234

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

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

Therefore

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

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99.95

#### 99.95

^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?