Does the integral converge or diverge? If it converges, find its value.
The integral from -3 to 3 1/√ (9-x^2) dx

Question

Does the integral converge or diverge? If it converges, find its value. 
The integral from -3 to 3     1/√ (9-x^2) dx

anonymous · Answer

$$\int\limits_{-3}^{3?} 1/ \sqrt{9-x^2}  dx$$

hartnn · Answer

16th in your list

anonymous · Answer

this is a new section, should of said that. lol were not using this list on this one.

hartnn · Answer

lol, but 16th is a standard integral! 
anyways 
you can then plug in x= 3 sin y

anonymous · Answer

i know we should start out by saying     lim t -> some number then solve normaly

anonymous · Answer

okay

anonymous · Answer

we couldn't use 3 in the denominator

anonymous · Answer

1/3cosy *3cosy  ?

hartnn · Answer

oh, for the convergent part, 
yes you need to show 
$ \large \lim \limits_{t \rightarrow -3} \int \limits_t^3 ...= finite \quad and \ \large \lim \limits_{t \rightarrow 3} \int \limits_{-3}^{t} ...= finite$

hartnn · Answer

yup, so, that =1 
and its integral = y
and y= sin^{-1} x ^_^

hartnn · Answer

sorry, here ,
y= sin^{-1} (x/3)
because 
x= 3 sin y

hartnn · Answer

doubts ?

anonymous · Answer

wouldnt the top and bottom cancel each other out?

hartnn · Answer

yeah, the top and bottom cancels out and gives you 1 !!

hartnn · Answer

integral of 1 dy is just y+c

anonymous · Answer

they are trying to be tricky aren't they! lol

anonymous · Answer

so it would converge since its a number?

hartnn · Answer

absolutely, finite number means convergent :)

hartnn · Answer

you solved convergent problems before, right ? and know the steps.... ? 
could find those 2 limits ?

anonymous · Answer

so how would you do the final subtraction if the answer is just 1..  or y

anonymous · Answer

cant plug in 3 there?

hartnn · Answer

ok, we have $\int 1dy =y+c \ y= \sin^{-1}(x/3)$
why not ? 
sin^{-1} 1 and sin^{-1} -1 are finite numbers (infact angles)!

anonymous · Answer

ohhh i see what your saying

hartnn · Answer

let me give you steps : 
$\large \lim \limits_{t \rightarrow -3} \int \limits_t^3 \dfrac{1}{\sqrt{9-x^2}}dx = \lim \limits_{t \rightarrow -3} \int \limits_{\sin^{-1}{(t/3)}}^{\sin^{-1}1}1dy  \ \large =\lim \limits_{t \rightarrow -3} y|_{{\sin^{-1}{(t/3)}}}^{\sin^{-1}1}=\lim \limits_{t \rightarrow -3}{\sin^{-1}{(t/3)}}-{\sin^{-1}1} \ \large ={\sin^{-1}{(-3/3)}}-{\sin^{-1}1}=....$

anonymous · Answer

thanks a lot, that really helped out!   its funny how someone on the internet can do a better job explaining then my teacher. lol

hartnn · Answer

lol, i get that alot! welcome ^_^