# Double integral (iterated integral)

• Jun 7th 2009, 09:18 AM
arbolis
Double integral (iterated integral)
I must calculate the following integral in the 2 senses ($\displaystyle dxdy$ and $\displaystyle dydx$).
$\displaystyle \int _0^2 \int _1^{e^x} dydx$.
My attempt : $\displaystyle \int _0^2 \int_1^{e^x} dydx=\int _0^2 y \big | _1^{e^x}dx=e^2-3$.

I'm not sure how to proceed when it comes to change the order of integration. Here's what I did : $\displaystyle y$ goes from $\displaystyle 1$ to $\displaystyle e^2$.
$\displaystyle x$ goes from $\displaystyle 0$ to $\displaystyle 2$.
So I have the double integral $\displaystyle \int_{1}^{e^2} \int_{0}^{2} dxdy = \int_{1}^{e^2} 2dy=2(e^2-1) \neq e^2-3$. And I think they must be equal, so I made an error.
• Jun 7th 2009, 09:37 AM
Random Variable
$\displaystyle \int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy$
• Jun 7th 2009, 09:40 AM
arbolis
Quote:

Originally Posted by Random Variable
$\displaystyle \int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy$

Oh... nice. Let me think on this!
• Jun 7th 2009, 09:42 AM
Random Variable
You have to draw the region to figure out the limits.
• Jun 7th 2009, 09:50 AM
Spec
$\displaystyle 1\leq y \leq e^x \implies x \textcolor{green}{\geq} \ln y$ (notice the part in green; that tells you it's the lower limit)

So we have $\displaystyle \ln y \leq x \leq 2$
• Jun 7th 2009, 04:01 PM
Scott H
Because $\displaystyle e^2$ is a constant, the limits

$\displaystyle \int_1^{e^2}\int_0^2$

would define a rectangle, not a curvilinear region as in

$\displaystyle \int_0^2\int_1^{e^x}.$