# Double Integral help needed

• Nov 2nd 2009, 06:16 PM
Diggidy
Double Integral help needed
Ok if i am given the double integral (sorry about the format i dont knw how to enter integral signs and what not... i typed it as it looks)

Integral from 1-3 Integral from 0-lnx of 2x dydx

How do i write an equivalent integral with the order of integration reversed.

I know that in order to reverse the order you simply switch the bounded integrals then switch dydx to dxdy and then integrate accordingly. but in my notes it says that you have to integrate the non-constant part first and that is where i get confused. If i switched the bounded integrals then the one iwth the lnx bound would have to integrated second.. How do i get around this?

Also just to make sure if i am doing these right could somone please check these solutions.

inegral from 0-4 integral from 0-7 of (x+y)dxdy

For this i got 98

and

Integral from 1-5 Integral from 0-lnx of e^y dydx

for this one i got 8

Any help is greatly appreciated.
Thank you,
Diggidy
• Nov 2nd 2009, 06:41 PM
Prove It
Quote:

Originally Posted by Diggidy
Ok if i am given the double integral (sorry about the format i dont knw how to enter integral signs and what not... i typed it as it looks)

Integral from 1-3 Integral from 0-lnx of 2x dydx

How do i write an equivalent integral with the order of integration reversed.

I know that in order to reverse the order you simply switch the bounded integrals then switch dydx to dxdy and then integrate accordingly. but in my notes it says that you have to integrate the non-constant part first and that is where i get confused. If i switched the bounded integrals then the one iwth the lnx bound would have to integrated second.. How do i get around this?

Also just to make sure if i am doing these right could somone please check these solutions.

inegral from 0-4 integral from 0-7 of (x+y)dxdy

For this i got 98

and

Integral from 1-5 Integral from 0-lnx of e^y dydx

for this one i got 8

Any help is greatly appreciated.
Thank you,
Diggidy

Is this what you're trying to write?

$\displaystyle \int_{x = 1}^{x = 3}{\int_{y = 0}^{y = \ln{x}}{2x\,dy}\,dx}$?

If so, look at the terminals - reversing the order of integration simply involves rearranging the resulting inequalities.

It should be pretty obvious that

$\displaystyle 1 \leq x \leq 3$ and $\displaystyle 0 \leq y \leq \ln{x}$

Taking the logarithm of everything in the first inequality gives

$\displaystyle \ln{1} \leq \ln{x} \leq \ln{3}$

$\displaystyle 0 \leq \ln{x} \leq \ln{3}$.

Exponentiating everything in the second inequality gives

$\displaystyle e^0 \leq e^y \leq e^{\ln{x}}$

$\displaystyle 1 \leq e^y \leq x$

Therefore

$\displaystyle e^y \leq x \leq 3$ and $\displaystyle 0 \leq y \leq \ln{3}$.

So the double integral becomes

$\displaystyle \int_{x = 1}^{x = 3}{\int_{y = 0}^{y = \ln{x}}{2x\,dy}\,dx} = \int_{y = 0}^{y = \ln{3}}{\int_{x = e^y}^{x = 3}{2x\,dx}\,dy}$.
• Nov 2nd 2009, 06:56 PM
Prove It
Quote:

Originally Posted by Diggidy
Also just to make sure if i am doing these right could somone please check these solutions.

inegral from 0-4 integral from 0-7 of (x+y)dxdy

For this i got 98

and

Integral from 1-5 Integral from 0-lnx of e^y dydx

for this one i got 8

I get 42 for the first.

I get the same as you for the second.