Question

Comparison theorem - Calc 2

Answer 1

rewrite then evaluate

Answer 2

how?

Answer 3

Your integral is divergent
Notice
\[
\frac{1}{x^3-3 x}=\frac{x}{3 \left(x^2-3\right)}-\frac{1}{3 x}
\]
The integral for \[ \frac{1}{3 x}\] is infinite for x between 0 and 1

Answer 4

how can i do it using the comparison theorem though?

Answer 5

Notice that for x> and x very close to zero, we have
\[
x^3-3 x<x^3+3 x<3 x+x<4 x
\]
Hence
\[
\frac {1}{x^3-3x} > \frac 1{4x}
\]
Since the integral of \[ \frac 1{4x}\] is divergent near zero, the original one is also divergent

Answer 6

This should have been
\[
x^3-3 x<x^3+3 x<3 x+x=4 x
\]

Answer 7

Did you understand it?

Answer 8

no im stuck on some parts

Answer 9

I dont get get where the 4x came from?

Answer 10

\[
x\le 1 \implies x^3 \le x\\
x^3 + 3x \le x +3x=4x
\]

Answer 11

why is x^3+3x<3x+x ?

Answer 12

Read my previous post

Answer 13

Is P used at all in this explanation?

Answer 14

What is P?

Answer 15

I recall if P . 1 it is convergent and if it is less than equal to 1 it is divergent

Answer 16

Yes p here is equal to 1 in \[ \frac 1 4 \frac 1{ x^1} \]

Answer 17

Oh ok, thanks