1. ## 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????

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

Thus:
$\displaystyle \int_0^1 \int_{2x}^2 e^{y^2} \, dy\, dx =$$\displaystyle \int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\, dy$

3. so the new integral is what??? it says syntax error

4. 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.

5. 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.

$\displaystyle \int_0^2 \int_{0}^{\frac{y}{2}} e^{y^2} \, dx\, dy = \int_0^2 \frac{y}{2} e^{y^2}\, dy = \frac14 e^{y^2} \bigg{|}_0^2 = \frac{e^4 - 1}{4}$