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:

$\displaystyle \int_0^9 \frac {1} {\sqrt{9-x}} \, dx $

discontinuous at x=9, so rewrite the right endpoint in terms of a limit:

$\displaystyle \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:

$\displaystyle \lim_{b \to 9-} \int_8^{9-b} \frac {1} {\sqrt{u}} \, du $

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

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

$\displaystyle = 2 \sqrt{8} $

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