# Double integral (iterated integral)

• Jun 7th 2009, 10:18 AM
arbolis
Double integral (iterated integral)
I must calculate the following integral in the 2 senses ( $dxdy$ and $dydx$).
$\int _0^2 \int _1^{e^x} dydx$.
My attempt : $\int _0^2 \int_1^{e^x} dydx=\int _0^2 y \big | _1^{e^x}dx=e^2-3$.

I'm not sure how to proceed when it comes to change the order of integration. Here's what I did : $y$ goes from $1$ to $e^2$.
$x$ goes from $0$ to $2$.
So I have the double integral $\int_{1}^{e^2} \int_{0}^{2} dxdy = \int_{1}^{e^2} 2dy=2(e^2-1) \neq e^2-3$. And I think they must be equal, so I made an error.
• Jun 7th 2009, 10:37 AM
Random Variable
$\int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy$
• Jun 7th 2009, 10:40 AM
arbolis
Quote:

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

Oh... nice. Let me think on this!
• Jun 7th 2009, 10:42 AM
Random Variable
You have to draw the region to figure out the limits.
• Jun 7th 2009, 10:50 AM
Spec
$1\leq y \leq e^x \implies x \textcolor{green}{\geq} \ln y$ (notice the part in green; that tells you it's the lower limit)

So we have $\ln y \leq x \leq 2$
• Jun 7th 2009, 05:01 PM
Scott H
Because $e^2$ is a constant, the limits

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

would define a rectangle, not a curvilinear region as in

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