double integral

**chris25** · July 7th 2008, 10:08 AM

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????

**Isomorphism** · July 7th 2008, 10:43 AM

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

**chris25** · July 7th 2008, 10:46 AM

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

**Krizalid** · July 7th 2008, 05:40 PM

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.

**mr fantastic** · July 7th 2008, 05:46 PM

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.

**jme44** · July 8th 2008, 11:04 PM

Is the answer for the integral 2e^4 or did I do something wrong?

**Isomorphism** · July 8th 2008, 11:27 PM

Originally Posted by jme44

Is the answer for the integral 2e^4 or did I do something wrong?

Well I get

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

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