Question

from 1 to 9
1/(x-9)^1/3
Determine whether the integral is convergent or divergent.

Answer 1

Is it divergent? Because when x is 9 it isn't a real number. Right?..

Answer 2

have u integrated the function?

Answer 3

\[\large \int\limits_{}^{}\frac{ 1 }{ \sqrt[3]{x-9} }=\frac{ 3 }{ 2 }*(x-9)^{2/3}\]
now put the limits

Answer 4

I think so, I got 3(x-9)^(3/2)/2

Answer 5

http://www.wolframalpha.com/input/?i=integrate+1%2F%28x-9%29%5E%281%2F3%29+from+x%3D1+to+x%3D9

Answer 6

it converges at 6??

Answer 7

I need a value it converges to

Answer 8

I used u = x - 9.. it is improper at u = 0, so I plugged values back in for 'u' to find the new bounds. For 9 u is 0... for 1 u is -8. Is this right? Please help

Answer 9

if the integral have a finite value between the limits then it will converge

Answer 10

So is this right? integrate from -8 to 0. I am so confused.

Answer 11

medal for whoever helps me solve this D:

Answer 12

yes limits will be -8 to 0

Answer 13

okay. I got 9.917234

Answer 14

is this what you got too?

Answer 15

I just want to learn how to solve this so I don't spend so much time do it wrong

Answer 16

so it produces a finite value then it should converge

Answer 17

yes, but I need the exact value. Is this right?

Answer 18

my question is, what IS the value it converges at

Answer 19

\[\int_1^9\frac{dx}{(x-9)^{1/3}}\]
Substitute \(u=x-9\), so \(du=dx\) and you have
\[\int_{-8}^0\frac{du}{u^{1/3}}\]
The integral is improper, so treat as a limit:
\[\lim_{c\to0^+}\int_{-8}^c\frac{du}{u^{1/3}}\]
Power rule:
\[\lim_{c\to0^+}\left[\frac{3u^{2/3}}{2}\right]_{-8}^c\]
Fundamental theorem:
\[\frac{3}{2}\lim_{c\to0^+}\left[c^{2/3}-(-8)^{2/3}\right]=\frac{3}{2}\left[0-(-8)^{2/3}\right]=\frac{3}{2}(-4)=-6\]

Answer 20

Should be POSITIVE 6, right? Double negatives. Just making sure.