Double integral - reversing order

[exidez]

Homework Statement

<br /> \displaystyle\int^1_0 \int^{e^x}_{1}dydx<br />

Homework Equations

none

The Attempt at a Solution

the above integral i can do with no problem, but changing the order of integration give me a totally different answer and need to know if i am doing it correct

First off
<br /> \displaystyle\int^1_0 \int^{e^x}_{1}dydx = e^1 - 2<br />

To reverse the order of integration i get:
<br /> \displaystyle\int^{e^1}_1 \int^{ln(y)}_{0}dxdy<br />
which gives me 1 which is wrong

Before i post how i went about my solution I want to know if i am doing my limit right?

 

[Dick]

Did you draw a graph of the integration region? x doesn't go from 0 to ln(y).

 

[exidez]

it goes from 0 to ln(e^1) which is 0 to 1

considering y=e^x then x = ln(y)

but I am guessing my understanding is wrong

 

[exidez]

Ok, these are two questions alike. Can you please tell me if i am completely misunderstanding the region of integration.
I need to show by reversing the order that i still get the same answer.

[URL]http://www.users.on.net/~rdenker/assign1.jpg[/URL]

EDIT: ok, another quick question. For Question 4 in the image here, is my region of integration on the wrong side of the line?

 

[Dick]

Your graph is on the correct side of the line. But look at it and imagine integrating dx. Isn't ln(y) the LOWER bound for x?

 

[exidez]

that seemed so hard to wrap my head around at the time but is so simple now
thanks!