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Double Integral - Can I do it in any order?

Let's say I have a dobule integral of the form:

I=f(x)g(x)abf(x,y)dxdy I = \displaystyle \int_{f(x)}^{g(x)} \int_a^b f(x,y) \, dx \, dy

Clearly, if I integrate as is, then I'll end up with I=h(x) I = h(x).

So, can I just change the order of integration without changing the limits to end up with a numerical answer, like so:

I=abf(x)g(x)f(x,y)dydx I = \displaystyle \int_a^b \int_{f(x)}^{g(x)} f(x,y) \, dy \, dx

?

After all, the function being integrated hasn't changed, and neither has the region being integrated over, so surely this is a valid move, even though these integrals give vastly different answers, no?
You can directly interchange the limits if each of the limits is constant (i.e., if the range you integrate over is a rectangle of finite area). However, the situation is more complex than you described in general - though it seems plausible that it would be valid, unfortunately it isn't true (as you say, the integrals give different answers in general - it's not too hard to provide a counterexample). Fubini's theorem deals with exchanging the order of integration, but it is quite a high-level result to understand and prove.

There are other ways to exchange the limits of integration in some cases - though they're difficult to explain. Have you covered any such material? If you're interested, there will no doubt be some good resources online somewhere.

Edit: At a quick glance, this looks like a reasonable source http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-double-integrals/session-49-exchanging-the-order-of-integration/
(edited 11 years ago)
Reply 2
Original post by dantheman1261
You can directly interchange the limits if each of the limits is constant (i.e., if the range you integrate over is a rectangle of finite area). However, the situation is more complex than you described in general - though it seems plausible that it would be valid, unfortunately it isn't true (as you say, the integrals give different answers in general - it's not too hard to provide a counterexample). Fubini's theorem deals with exchanging the order of integration, but it is quite a high-level result to understand and prove.

There are other ways to exchange the limits of integration in some cases - though they're difficult to explain. Have you covered any such material? If you're interested, there will no doubt be some good resources online somewhere.

Edit: At a quick glance, this looks like a reasonable source http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-double-integrals/session-49-exchanging-the-order-of-integration/


I completely understand and I am familiar with double integration and changing the order of integration.

However, it is usually the case that you get an integral of the form:

I=abf(x)g(x)f(x,y)dydx I = \displaystyle \int_a^b \int_{f(x)}^{g(x)} f(x,y) \, dy \, dx

i.e. the numerical limits are in the outer integral whilst the variable limits are in the inner integral. This produces a numerical result. Sometimes it is desirable to change the order of integration because of the difficulty involved with integrating the integrand with respect to the inner variable. However, when you make that change of order, you still get an integral which has numerical limits on the outer integral and variable limits on the inner integral - so you get a numerical result either way.

I'm actually tutoring somebody, and one past paper question that has come up in their exam involves an integral where the variable limits are on the outside and the numerical limits are on the inside. I.e.:

I=f(x)g(x)abf(x,y)dxdy I = \displaystyle \int_{f(x)}^{g(x)} \int_a^b f(x,y) \, dx \, dy

Am I to assume, then, that this is not a mistake and that the answer they want will, indeed, just be a function of xx? Either that or there is another way to deal with this in order to get a numerical result (if so, what?) ... OR the lecturer has simply made an arse of it and it's an erroneous question.

Speaking of erroneous questions, would anybody be willing to vouch for the fact that this is a nonsensical question:

I=132y0(yx2+xy2)dxdy I = \displaystyle \int_{1}^{3} \int_{2y}^{0} \left(yx^2 + xy^2\right) \, dx \, dy

The region here doesn't make any sense, right?
Original post by ThisIsTheLife
I completely understand and I am familiar with double integration and changing the order of integration.

However, it is usually the case that you get an integral of the form:

I=abf(x)g(x)f(x,y)dydx I = \displaystyle \int_a^b \int_{f(x)}^{g(x)} f(x,y) \, dy \, dx

i.e. the numerical limits are in the outer integral whilst the variable limits are in the inner integral. This produces a numerical result. Sometimes it is desirable to change the order of integration because of the difficulty involved with integrating the integrand with respect to the inner variable. However, when you make that change of order, you still get an integral which has numerical limits on the outer integral and variable limits on the inner integral - so you get a numerical result either way.

I'm actually tutoring somebody, and one past paper question that has come up in their exam involves an integral where the variable limits are on the outside and the numerical limits are on the inside. I.e.:

I=f(x)g(x)abf(x,y)dxdy I = \displaystyle \int_{f(x)}^{g(x)} \int_a^b f(x,y) \, dx \, dy

Am I to assume, then, that this is not a mistake and that the answer they want will, indeed, just be a function of xx? Either that or there is another way to deal with this in order to get a numerical result (if so, what?) ... OR the lecturer has simply made an arse of it and it's an erroneous question.


I see - apologies. The notation is the confusion here - you don't get a function of x after the integration. It's misleading as it makes you think

f(x)g(x)abf(x,y)dxdy=f(x)g(x)(abf(x,y)dx)dy \displaystyle \int_{f(x)}^{g(x)} \int_a^b f(x,y) \, dx \, dy = \displaystyle \int_{f(x)}^{g(x)} (\int_a^b f(x,y) \, dx) \, dy

which is not the case. It is more obvious to write

Rf(x,y)dA \displaystyle \int \int_R f(x,y) \, dA

where R is the region defined in the question. You can't integrate out the x first as you want to - purely because it doesn't make sense in terms of the region that you're given.

However, in this case, you can just simply exchange the limits (as you posted - apologies again for my unnecessary first post) and come out with the correct (numerical) answer. This is easily checked by sketching the region.

Original post by ThisIsTheLife

Speaking of erroneous questions, would anybody be willing to vouch for the fact that this is a nonsensical question:

I=132y0(yx2+xy2)dxdy I = \displaystyle \int_{1}^{3} \int_{2y}^{0} \left(yx^2 + xy^2\right) \, dx \, dy

The region here doesn't make any sense, right?


I suppose, again, this is kind of misleading. But it's still a valid question - just the same as

abf(x)dx=baf(x)dx \displaystyle \int_a^b f(x) \, dx = -\displaystyle \int_b^a f(x) \, dx

You can swap the limits and negate the integrand if necessary - or, when looking at the region of integration (apologies if this is not the method you are used to), you would draw the lines "down" rather than "up".

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