Question

Express the integral int(0 to 1)int(e^x to e^2x) xlnydydx with reverse order of integration (don't solve)

Answer 1

\[\int\limits_{1}^{e}\int\limits_{(1/2)lny}^{lny}xlnydxdy+ \int\limits_{e}^{e^2}\int\limits_{(1/2)lny}^{1}xlnydxdy \]

Answer 2

can anyone confirm that this is right?

Answer 3

Im new to double integrals (in fact, we discussed them on monday) but should the new borders be 1<= y <= e^2 , 0.5ln(y) <= x <= ln y ? why do you split them up at y = e ?

Answer 4

if your simply switching out the bounds then you start with

y = [e^x, e^(2x)] ; x = [0,1] as given.

-------------------------

y = e^x; x = ln(y)

y = e^(2x); x = [ln(y)]/2

-----------------------------

y = e^(0) = 1

y = e^(2(1)) = e^2

--------------------------

\[\large\int_{1}^{e^2}\int_{\frac{ln(y)}{2}}^{ln(y)}f(x)dxdy\]

Answer 5

i think I see why you split it now ...

Answer 6

If we continue on without redefining the split; you gain extra that wasnt included to begin with

Answer 7

so do you think I was right to split it?

Answer 8

i think you did good; it is wasnt for the split there would be no control over the inegration within the region .

Answer 9

okay cool, I was a little unsure. thanks!

Answer 10

Can sby explain me why the split is necessary? How do you know if you have to "split" ?

Answer 11

Amistre, your graph is slightly incorrect since both \(e^x\) and \(e^{2x}\) intersect the y axis at
y = 1. However you do need to split it since when integrating with respect to x first, you have two different curves you are dealing with on the right most edge of the domain.. \(x = ln(y)\) for y from 1 to e, and \(x = 1\) for y from e to \(e^2\).

\[y = e^x|_{x=0} = 1\]\[y = e^{2x}|_{x=1} = e^2\]\[y = e^x \implies x = ln(y)\]\[y = e^{2x} \implies x = \frac{ln(y)}{2}\]
\[\int_0^1\int_{e^x}^{e^2x}x(ln\ y)\ dydx = \int_{1}^{e} \int_{\frac{1}{2}(ln\ y)}^{(ln\ y)}x(ln\ y)\ dxdy + \int_{e}^{e^2} \int_{\frac{1}{2}(ln\ y)}^{1}x(ln\ y)\ dxdy\]

Answer 12

\[1 < ln(y)\text{, } \forall y > e\]

Answer 13

..... just a bit off on the graph :) thnx

Answer 14

Yep, otherwise it was fine. I was attempting to explain to the other person why the split was needed.

Answer 15

@polpak: So, if we would integrate like amistre proposed, we would include that certain area where x>=1 and y is between our two curves, like this(purple area)?

Answer 16

Correct.

Answer 17

Without breaking it into two integrals you'll get a much larger domain when you switch the order of integration.

Answer 18

Thanks a lot, i got it now!