Question

Test the integral for convergence or divergence, if it converges evaluate it
int (1/sqrt(x-1)) from 1 to 10

Answer 1

\[ \int_1^{10} \frac 1{\sqrt{x - 1}}dx = \left [ \frac{\sqrt{x - 1} }{\frac 12 }\right ]_1^{10}\]

Answer 2

how did you come to that? I thought that the integral diverges since its similar to 1/(p^x) and when p <=1 it diverges 1/(x-1)^1/2)

Answer 3

this one diverges
http://www.wolframalpha.com/input/?i=integrate+1%2F%28x-1%29+from+1+to+10
and the one you asked converges.
just take a limit and see if the improper integral is finite or infinite.
\[ \lim_{a \rightarrow 1} \int_a^{10} \frac1{\sqrt{x - 1}}dx\]

Answer 4

sorry still not quite getting it. whats the difference from the equation with and without the square root. You still get 1/sqrt(0) or either 1/0 which both are undefined.

Answer 5

without square root you will get
log(x-1)
now log(1-1) = infinity ... which is divergent

Answer 6

you are not evaluating 1/sqrt(0)
you are evaluating 1/sqrt(0) x dx
as you can see this value is not infinity ... it has rather some finite value.

some infinities are bigger that some infinities ... eg, x lnx = 0 as x->0
it's because log x = infinity < 1/x = infinity.

same case is going here on 1/sqrt(x) times dx

plus while integrating you do not require to evaluate it exactly at that point. rather you are evaluating it at limit point of discontuinity

Answer 7

you are evaluating 1/sqrt(0) \times dx

Answer 8

also note that I'm not quite sure about this ...

Answer 9

lol your last message. Okay I am going to try to work this out some more, I'm understanding it a bit better. I'm going to see what I come up with.

Answer 10

I'll update if I find some more info

Answer 11

What I did was I got
int dx/sqrt(x-1) from 1 to 10 =
lim t->1+, int dx/sqrt(x-1) from t to 10 =
lim t->1+, 2sqrt(x-1) from t to 10 =
lim t->1+, (6-2sqrt(t-1) = 6

Answer 12

Hmm ... that's right procedure.

Answer 13

Here is what we call the p-integral test.
Let \( a < \infty \) and \( c < \infty\)
\[
\int_a^c \frac {dx}{ (x-a)^p}
\]
converges if \( p <1\) and diverges for \(p\ge 1\).
Let b>a then

\[
\int_b^\infty \frac {dx}{ (x-a)^p}
\]
converges if \( p>1\) and diverges for \(p\le 1\).

Answer 14

@experimentX is right.

Answer 15

ty prof