# Thread: Determine the value of the improper integral

1. ## Determine the value of the improper integral

$\displaystyle I = \displaystyle 3 \int_0^{2} \frac{1}{\sqrt{2-x}} dx$

$\displaystyle I = \displaystyle -6\sqrt{2-x} \right]_0^{2}$

$\displaystyle I = \displaystyle -6\sqrt{0} + 6\sqrt{2}$

$\displaystyle I = \displaystyle 6\sqrt{2}$

is this correct

OR

$\displaystyle \displaystyle \lim_{\epsilon \to 2}\left[ -6\sqrt{2-\epsilon} \right]_0^{\epsilon}$

2. I don't understand how you got a factor of 6... This is solved using a $\displaystyle \displaystyle u$ substitution...

3. Originally Posted by bijosn
$\displaystyle I = \displaystyle \int_0^{2} \frac{1}{\sqrt{2-x}} dx$

$\displaystyle I = \displaystyle -6\sqrt{2-x} \right]_0^{2}$

$\displaystyle I = \displaystyle -6\sqrt{0} + 6\sqrt{2}$

$\displaystyle I = \displaystyle 6\sqrt{2}$

is this correct
No. You can make sure by differentiating the anti-derivative.

4. Originally Posted by Prove It
I don't understand how you got a factor of 6... This is solved using a $\displaystyle \displaystyle u$ substitution...
sorry, I forgot to add the 3 before the integral sogn in the first post...now fixed

$\displaystyle I = \displaystyle 3 \int_0^{2} \frac{1}{\sqrt{2-x}} dx$

5. Then yes, it is correct.