1. ## Double integral problem, help please

Double integral - e^(x/y) dA D = 1<y<2, y<x<y^3 (all the inequalities have _ under them)

2. Seems fairly standard to me. What have you done so far? If you start integrating dx first then it will all work out okay.

3. $\displaystyle \int_{1<y<2}\int_{y<x<y^3}-e^{\frac{x}{y}}dxdy$

it is the same as

$\displaystyle \int_{1}^{2}\int_{y}^{y^3}-e^{\frac{x}{y}} dx dy$

first we integrate with respect to x and y constant

$\displaystyle \int_{1}^{2}-ye^{\left(\dfrac{x}{y}\right)}\mid_{y}^{y^3}$

sub now

$\displaystyle \int_{1}^{2}-y(e^{\left(\frac{y}{y}\right)}-e^{\left(\frac{y^3}{y}\right)})dy$

$\displaystyle \int_{1}^{2}-y(e^{1}-e^{y^2})dy$

$\displaystyle \int_{1}^{2}-y(e) dy+ \int_{1}^{2}ye^{y^2} dy$

the first one is easy the second let u=y^2 and you continue

4. Originally Posted by Amer
$\displaystyle \int_{1<y<2}\int_{y<x<y^3}-e^{\frac{x}{y}}dxdy$

it is the same as

$\displaystyle \int_{1}^{2}\int_{y}^{y^3}-e^{\frac{x}{y}} dx dy$

first we integrate with respect to x and y constant

$\displaystyle \int_{1}^{2}-ye^{\left(\dfrac{x}{y}\right)}\mid_{y}^{y^3}$

sub now

$\displaystyle \int_{1}^{2}-y(e^{\left(\frac{y}{y}\right)}-e^{\left(\frac{y^3}{y}\right)})dy$

$\displaystyle \int_{1}^{2}-y(e^{1}-e^{y^2})dy$

$\displaystyle \int_{1}^{2}-y(e) dy+ \int_{1}^{2}ye^{y^2} dy$

the first one is easy the second let u=y^2 and you continue

Edit: deleted

I got it, thanks!

5. look I integrate not derive it is possible that I make mistakes and typos so feel free

$\displaystyle \int e^{\frac{x}{y}} dx$

$\displaystyle \dfrac{e^{\frac{x}{y}}}{\left(\frac{d}{dx}(\frac{x }{y})\right)}$

as you can see the denominator of this is 1/y

$\displaystyle \dfrac{e^{\frac{x}{y}}}{\left(\frac{1}{y}\right)}$

$\displaystyle \frac{a}{\left(\frac{c}{d}\right)}=\frac{ad}{c}$

I think it is clear