1. ## Improper integral converge/diverge

Can anyone check my work here and if I went wrong, tell where and how and why etc etc. Thanks!

Question:
Decide whether the improper integral converges or diverges and if it converges, give it's value:
$\int_0^9 \frac {1} {\sqrt{9-x}} \, dx$

discontinuous at x=9, so rewrite the right endpoint in terms of a limit:
$\lim_{b \to 9-} \int_0^b \frac {1} {\sqrt{9-x}} \, dx$

substitute u=9-x, and since this is a definite integral, we need to change the limits of integration:
$\lim_{b \to 9-} \int_8^{9-b} \frac {1} {\sqrt{u}} \, du$

$\lim_{b \to 9-} [ -2\sqrt{u}]_8^{9-b} \,$

$\lim_{b \to 9-} (-2\sqrt{9-b} + 2\sqrt{8}) \,$

$= 2 \sqrt{8}$
Thus the limit equals a finite number, and the integral converges to $2 \sqrt{8}$.

2. Originally Posted by Dr Zoidburg
Can anyone check my work here and if I went wrong, tell where and how and why etc etc. Thanks!

Question:
Decide whether the improper integral converges or diverges and if it converges, give it's value:
$\int_0^9 \frac {1} {\sqrt{9-x}} \, dx$

discontinuous at x=9, so rewrite the right endpoint in terms of a limit:
$\lim_{b \to 9-} \int_0^b \frac {1} {\sqrt{9-x}} \, dx$

substitute u=9-x, and since this is a definite integral, we need to change the limits of integration:
$\lim_{b \to 9-} \int_8^{9-b} \frac {1} {\sqrt{u}} \, du$

$\lim_{b \to 9-} [ -2\sqrt{u}]_8^{9-b} \,$

$\lim_{b \to 9-} (-2\sqrt{9-b} + 2\sqrt{8}) \,$

$= 2 \sqrt{8}$
Thus the limit equals a finite number, and the integral converges to $2 \sqrt{8}$.
You are correct that you make the substitution $u = 9 - x$, but $\frac{du}{dx} = -1$. So $du = -dx$.

So $\int{\frac{1}{\sqrt{9 - x}}\,dx} = -\int{\frac{1}{\sqrt{u}}\,du}$

$= -\int{u^{-\frac{1}{2}}\,du}$

$= -2u^{\frac{1}{2}} + C$

$= -2\sqrt{9 - x} + C$.

Now inputting the terminals $0 \leq x \leq 9$

$\int_0^9{\frac{1}{\sqrt{9 - x}}\,dx} = \lim_{b \to 9^{-}}[-2\sqrt{9 - b}] - [-2\sqrt{9 - 0}]$

$= 0 + 2\sqrt{9}$

$= 6$.

And yes, obviously the integral is convergent.

3. ah. missed a step. I didn't feel like it was quite right but couldn't work out where I'd gone wrong.
thanks for that.

4. Originally Posted by Dr Zoidburg
Can anyone check my work here and if I went wrong, tell where and how and why etc etc. Thanks!

Question:
Decide whether the improper integral converges or diverges and if it converges, give it's value:
$\int_0^9 \frac {1} {\sqrt{9-x}} \, dx$

discontinuous at x=9, so rewrite the right endpoint in terms of a limit:
$\lim_{b \to 9-} \int_0^b \frac {1} {\sqrt{9-x}} \, dx$

substitute u=9-x, and since this is a definite integral, we need to change the limits of integration:
$\lim_{b \to 9-} \int_8^{9-b} \frac {1} {\sqrt{u}} \, du$
You did not change the limits of integation correctly. the integral should be
$-\int_9^{9-b}\frac{1}{\sqrt{u}}du= \int_{9-b}^9 \frac{1}{\sqrt{u}}du$

$\lim_{b \to 9-} [ -2\sqrt{u}]_8^{9-b} \,$

$\lim_{b \to 9-} (-2\sqrt{9-b} + 2\sqrt{8}) \,$

$= 2 \sqrt{8}$
Thus the limit equals a finite number, and the integral converges to $2 \sqrt{8}$.

5. So my original answer was correct despite my mistakes? how odd!
Thanks all for your help. I'm doing this paper by correspondence so this is the only place (other than the textbooks) I can turn for advice/help when I get stuck or just when I want confirmation I'm on the right track.

6. Originally Posted by Dr Zoidburg
So my original answer was correct despite my mistakes? how odd!
Thanks all for your help. I'm doing this paper by correspondence so this is the only place (other than the textbooks) I can turn for advice/help when I get stuck or just when I want confirmation I'm on the right track.
No it wasn't. The integral converges to $6$, not $2\sqrt{8} = 4\sqrt{2}$.