hi how do you know for 0 to 2 (1st int) and x^2 to 2x (2nd int) (x^3) dydx if you should have x now going from 0 to y^.5 or 0 to y/2 and likewise y going from 0 to 2x or 0 to x^2?

Printable View

- Mar 8th 2008, 06:41 PMcandireversing order of integration
hi how do you know for 0 to 2 (1st int) and x^2 to 2x (2nd int) (x^3) dydx if you should have x now going from 0 to y^.5 or 0 to y/2 and likewise y going from 0 to 2x or 0 to x^2?

- Mar 8th 2008, 06:52 PMmr fantastic
Have you drawn a sketch of the region of integration?

It is*very*clear from such a sketch that the reversed integral limits are x = y/2 to $\displaystyle x = +\sqrt{y}$ and y = 0 to y = 4:

$\displaystyle I = \int_{y=0}^{y=4} \int_{x = y/2}^{x = \sqrt{y}} x^3 \,dx \, dy$. - Mar 8th 2008, 06:58 PMcandi
i did draw a sketch. and the graphs are both at the origin. but i'll look at what you wrote again to see if i'll understand it better. thanks.

- Mar 9th 2008, 07:44 AMKrizalid
It's pretty clear having a sketch, you'll see mr fantastic's result is correct.