Double integral (iterated integral)

**arbolis** · June 7th 2009, 09:18 AM

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.

**Random Variable** · June 7th 2009, 09:37 AM

$\int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy$

**arbolis** · June 7th 2009, 09:40 AM

Originally Posted by Random Variable

$\int^{e^{2}}_{1} \int^{2}_{ln(y)} \ dx \ dy$

Oh... nice. Let me think on this!

**Random Variable** · June 7th 2009, 09:42 AM

You have to draw the region to figure out the limits.

**Spec** · June 7th 2009, 09:50 AM

$1\leq y \leq e^x \implies x \textcolor{green}{\geq} \ln y$ (notice the part in green; that tells you it's the lower limit)

So we have $\ln y \leq x \leq 2$

**Scott H** · June 7th 2009, 04:01 PM

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

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

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