http://img528.imageshack.us/img528/7545/scan0003yo.jpg

am I on the right track here? :)

Printable View

- Dec 13th 2012, 12:49 AMtheunforgivenFind the area of the region bounded by y and the x-axis
http://img528.imageshack.us/img528/7545/scan0003yo.jpg

am I on the right track here? :) - Dec 13th 2012, 12:56 AMMarkFLRe: Find the area of the region bounded by y and the x-axis
Yes. I would write:

$\displaystyle A=\int_1^9 u^{\frac{1}{2}}\,du=\frac{2}{3}\left[u^{\frac{3}{2}} \right]_1^9=\frac{2}{3}(27-1)=\frac{52}{3}$

edit: Don't forget the differential in your definite integral and avoid using a rounded decimal approximation (at least if you have the same professor I did).(Nod) - Dec 13th 2012, 12:56 AMa tutorRe: Find the area of the region bounded by y and the x-axis
That's fine. You can check this sort of thing here integral (-2,6) (x+3)^(1/2) dx - Wolfram|Alpha

- Dec 13th 2012, 01:00 AMtheunforgivenRe: Find the area of the region bounded by y and the x-axis
I see. So we assume u = (x+3) and go from there. Looks more professional to me. Thanks! On thing I'm not sure about though is, how did you get the 9 and 1?

EDIT: Oh nevermind. I get it now. (Happy)

Oh that tool is amazing! So I take it the graph shown there is the sketch of the region.