Converges or Diverges

[uneven dust]

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[gentle coral]

firstly you can split the interval at $x=2$ to deal with the abs value, then just integrate each piece

[wet breach]

Omegabet_

[gentle coral]

idk why you made limits going to 2

|2-x|/x^3 has no problems at x=2

the problem is x=0 for the denominator

(also wrong handling of the ||)

$=\int_0^2\frac{2-x}{x^3}\dd{x}+\int_2^4\frac{x-2}{x^3}\dd{x}$

[wet breach]

Omegabet_

[gentle coral]

it's clear the last integral converges (or compute it yourself and check), so it converges iff $\int_0^2$ does

[wet breach]

Omegabet_

[graceful raptor]

But there's a infinity there

[graceful raptor]

Is my working correct

[gentle coral]

pretty much

aside from the fact you wrote absurd stuff like 1/0^2

[graceful raptor]

I mean -1/0 technically infinity right

[gentle coral]

no

[graceful raptor]

Right

[gentle coral]

1/0 is complete garbage

$\int_0^2\frac{2}{x^3}-\frac{1}{x^2}\dd{x}:=\lim_{a\downarrow 0}\int_a^2\frac{2}{x^3}-\frac{1}{x^2}\dd{x}=\lim_{a\downarrow0}[\frac{-1}{x^2}+\frac{1}{x}]a^2=\lim{a\downarrow0}\frac{1}{4}+\frac{1}{a^2}-\frac{1}{a}$

[wet breach]

Omegabet_

[graceful raptor]

Owh okayh, should we plug in 'a' value?

[gentle coral]

well obviously you cant

the function isnt defined at a=0

you evaluate that limit

$=\frac{1}{4}+\lim_{a\downarrow 0}\frac{1-a}{a^2}$. Handwavily, numerator goes to $1$ from above, and denominator goes to $0$ from above, so net limit goes to $\infty$

[wet breach]

Omegabet_

[graceful raptor]

Owh, so then the limit is infinity?

[gentle coral]

yes

hence the integral diverges

[graceful raptor]

Great

I guess it's the same for the other one?

[gentle coral]

wdym other one

[graceful raptor]

The upper bound of 4 I mean

[gentle coral]

$\int_2^4$ clearly converges

[wet breach]

Omegabet_

[gentle coral]

nothing has the possibility to blow-up on [2,4]

[graceful raptor]

Owh, wait how do I know?

[gentle coral]

you can just compute it

but I just told you

it's a continuous bounded function defined on [2,4]

it converges

[graceful raptor]

Owh

[gentle coral]

cause you know from above what the primitive is (up to a - at least)

and you know from said primitive the integral will be (finite thing) - (finite thing) = finite thing

[graceful raptor]

I see

[gentle coral]

yeah

[graceful raptor]

So one limit is diverges, the other is converges. Hence, for the entire integral is divergence, correct?

[gentle coral]

one piece diverges and one piece converges

yes, their sum diverges

[graceful raptor]

Great

Alright, thank you very much