Is switching the order of integration helpful in evaluating double integrals?

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Switching the order of integration in double integrals can simplify calculations, particularly when the region of integration is rectangular, as demonstrated with the integral of the function x/(1 + xy). The discussion highlights that for the region R defined as {(x, y)| 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}, both orders of integration yield the same result, but the process may be easier in one order depending on the function. The use of substitution, such as letting u = 1 + xy, is also discussed as a method to facilitate integration.

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harpazo
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I just started learning double integrals. It is interestingly difficult. I know that switching dxdy to dydx can simplify the integration. I am not too clear why switching dxdy to dydx or vice-versa can make things easier.

Let S = integral symbol

SS [x/(1 + xy)] dxdy

Which is easier: SS [x/(1 + xy)] dxdy or
SS [x/(1 + xy)] dydx?

The region R is given to be:

{(x, y)| 0 less than or equal to x less than or equal to 1, 0 less than or equal to y less than or equal to 1 }.

What's the difference?
 
Last edited:
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Hi Harpazo,

What is your region of integration?
 
Euge said:
Hi Harpazo,

What is your region of integration?

I forgot to include the region. I will now edit the question.

- - - Updated - - -

Euge said:
Hi Harpazo,

What is your region of integration?

The question has been edited.
 
Isn't the region of integration just a square? Since there's no variation in either variable there won't be any difference in reversing the order of integration...
 
To integrate with respect to y first, let u= 1+ xy. Then du= xdy . When y= 0 u= 1 and when y= 1, u= 1+ x. The first integral becomes \int_1^x \frac{du}{u}du= \left[ln(u)\right]_1^x= ln(x+1). We now have \int_0^1 ln(x+1)dx.

The other order is a little harder because of that "x" in the numerator.
 
Last edited by a moderator:
HallsofIvy said:
To integrate with respect to y first, let u= 1+ xy. Then du= xdy . When y= 0 u= 1 and when y= 1, u= 1+ x. The first integral becomes \int_1^x \frac{du}{u}du= \left[ln(u)\right]_1^x= ln(x+1). We now have \int_0^1 ln(x+1)dx.<br /> <br /> The other order is a little harder because of that &quot;x&quot; in the numerator.
<br /> <br /> Part of your LaTex reply did not display.
 
Harpazo said:
Part of your LaTex reply did not display.

I've edited that post to correct the problem.
 
greg1313 said:
I've edited that post to correct the problem.

What's the difference between integrating over dxdy as opposed to dydx for this problem?
 
Harpazo said:
What's the difference between integrating over dxdy as opposed to dydx for this problem?

What Prove It was trying to say above is that it makes no difference, because the region over which you are integrating is rectangular. If you have a triangle, or some region where $x$ and $y$ are interacting, then it can make a great deal of difference, and interchanging the order of integration is one of the standard tricks you should have in your toolbox.
 
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Ackbach said:
What Prove It was trying to say above is that it makes no difference, because the region over which you are integrating is rectangular. If you have a triangle, or some region where $x$ and $y$ are interacting, then it can make a great deal of difference, and interchanging the order of integration is one of the standard tricks you should have in your toolbox.

Thank you. I know it is a great tool to have in terms of multiple integrals. It takes a complicated or almost impossible double integral, in this case, and makes it easy to integrate.
 

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