# double integral

• Jul 7th 2008, 10:08 AM
chris25
double integral
evaluate double integral...

integral from 0 to 1 integral from 2x to 2 of e^y^2 dy dx.

i don't know how to show that other than words...

the part that is messing me up is evaluating the e^y^2. i think you have to maybe change the bounds and change it to dx dy but i have no idea...can any one help me reset up the integral????
• Jul 7th 2008, 10:43 AM
Isomorphism
Quote:

Originally Posted by chris25
evaluate double integral...

integral from 0 to 1 integral from 2x to 2 of e^y^2 dy dx.

i don't know how to show that other than words...

the part that is messing me up is evaluating the e^y^2. i think you have to maybe change the bounds and change it to dx dy but i have no idea...can any one help me reset up the integral????

$2x \le y \le 2 , 0 \le x \le 1 \Leftrightarrow 0 \le 2x \le y \le 2 \Leftrightarrow 0 \le x \le \frac{y}2 \le 1$

Thus:
$\int_0^1 \int_{2x}^2 e^{y^2} \, dy\, dx =$ $\int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\, dy$
• Jul 7th 2008, 10:46 AM
chris25
so the new integral is what??? it says syntax error
• Jul 7th 2008, 05:40 PM
Krizalid
Read again Isomorphism's post, he just played with inequalities.

First two determine the region where the double integral is taken, then after some algebraic manipulations he got the new region where the double integral is taken in reverse order.
• Jul 7th 2008, 05:46 PM
mr fantastic
Quote:

Originally Posted by chris25
so the new integral is what??? it says syntax error

You should draw a sketch graph that shows the region of integration.

Getting the new integral limits for reversing the order of integration is then much easier to see and it will then be very clear to you where Isomorphism's integral limits have come from.
• Jul 8th 2008, 11:04 PM
jme44
$\int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\, dy = \int_0^2 \frac{y}{2} e^{y^2}\, dy = \frac14 e^{y^2} \bigg{|}_0^2 = \frac{e^4 - 1}{4}$