Question

Determine whether or not the improper integral converges:

Answer 1

\[\int\limits_{4}^{\infty}1/x \sqrt{x} \] dx

Answer 2

Moya, can you integrate that from 4 to some variable, say "t"?

Answer 3

Is this it?
\[\int\limits_{4}^{\infty}\frac{1}{x \sqrt{x}}\]

Answer 4

kropot2 made a good point. Please clarify the integrand.

Answer 5

yes thats it kropot. @ gpfreitas isnt it the infity thtat is replaced with t instead of the 4?

Answer 6

@Moya. Yes, you should first replace the infinity by t. That is why I said "from 4 to t".

Answer 7

Ok, what is the indefinite integral of
\[\frac{1}{x \sqrt{x}}\]

?

Answer 8

Hint: write this integrand as \(x^k\) for some fraction \(k\).

Answer 9

should i rewrite the funtion so it becomes a product that is lnx*x^-1/2??

Answer 10

i dont understand

Answer 11

Simpler than that. \(1/x\) is \(x^{-1}\). \(1/\sqrt{x}\) is \(x^{-1/2}\). The integrand is the product of those two expressions, which is...

Answer 12

Anyway, you will get something that is of the form \(x^{k}\) where \(k\) will be some fraction. I am guessing you know how to integrate something of that form. Integrate it from 4 to \(t\). You will get a formula that depends only on \(t\). Take the limit of that formula when \(t \to \infty\). That is the answer. Do you see what you have to do?

Answer 13

x^1/2

Answer 14

Hmmmm... are you sure? Check your calculation again. Make sure you got the signs right.

Answer 15

oh x^-3/2

Answer 16

Yep. Next step?

Answer 17

(Quick aside; try to get into the habit of writing x^(-3/2), with parentheses, or use the math notation \(x^{-3/2}\). What you wrote could be interpreted as
\[\frac{x^{-3}}{2}\]
which is not what you meant.)

Answer 18

thanks i think i have the idea now, ill make x tend to t and substitute in the t and 4 then find the limit of that answer right?

Answer 19

I'm not sure I understand you. To solve it, you don't really make \(x\) tend to \(t\)... You integrate \(x^{-3/2}\) from 4 to some variable t. From that you will get an expression that depends on \(t\), but not on \(x\). Take the limit of that when \(t \to \infty\). That is all.

Sorry, but I need to go now! Good luck!