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I know how to do it of course I am just wondering why it is done. Are some problems unsolvable without doing that or is it just to make it faster to solve?

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    $\begingroup$ Yes, some problems are very hard to solve one way, but switching the order makes them easier. $\endgroup$
    – wgrenard
    Nov 21 '17 at 16:58
  • $\begingroup$ Double counting is just the main trick of Mathematics. To know the conditions ensuring $\sum\int=\int\sum$ or $\int_A \int_B = \int_B \int_A$ is important since such trick can be applied to almost every Calculus (and not only) problem. $\endgroup$ Nov 21 '17 at 17:05
  • $\begingroup$ @wgrenard I understand that problems can be very hard. But no one directly answered the OP's question if there exists any problem which is unsolvable via a certain order of integration. I think maybe math.stackexchange.com/q/2497664/308392 is relevant here. $\endgroup$
    – Aditya P
    Aug 7 '18 at 11:59
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Consider the integral$$\int_0^1\int_x^1e^{y^2}\,\mathrm dy\,\mathrm dx.$$If you try to compute directly it's… hard, to say the least. But it is equal to$$\int_0^1\int_0^ye^{y^2}\,\mathrm dx\,\mathrm dy.$$And this one is quite easy.

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Here is a simple example. Consider integrating over the triangle between the points $(0,0), (2,0)$ and $(1,1)$ in the $xy$-plane.

If you integrate $dydx$, you will need 2 integrals since the upper boundary changes from $y=x$ to $y=2-x$ at $(1,1)$, so you end up with $$ \int_{x=0}^{x=1} \int_{y=0}^{y=x} dydx + \int_{x=1}^{x=2} \int_{y=0}^{y=2-x} dydx $$

However, if you integrate $dxdy$, only one integral is fine and you end up with $$\int_{y=0}^{y=1} \int_{x=y}^{x=2-y} dydx$$

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Here is another less obvious reason as to why one might want to change the order of integration in a double integral. It relates to its use in evaluating some single variable definite integrals which are otherwise difficult to find by providing one with additional room in which to manoeuvre.

As an example, consider the problem of evaluating the rather difficult integral $$I = \int^1_0 \frac{x^3 - 1}{\ln x} \, dx.$$ If we note that $$\int^3_0 x^t \, dt = \left. \frac{x^t}{\ln x} \right |^{t = 3}_{t = 0} = \frac{x^3 - 1}{\ln x},$$ the original integral can be rewritten as a double integral as follows: $$I = \int^1_0 \int^3_0 x^t \, dt dx.$$

For the double integral, without changing the order of the integration, performing the $t$-integration first would immediately lead us back to where we started. The double integral however provides us with some much needed room in which to manoeuvre. Changing the order of integration allows us to gain this extra room by allowing one to perform the $x$-integration first rather than the $t$-integration which, as we saw, only brings us back to where we started.

On changing the order of integration we have $$I = \int^3_0 \int^1_0 x^t \, dx dt =\int^3_0 \left [\frac{x^{t + 1}}{t + 1} \right ]^1_0 dt = \int^3_0 \frac{dt}{t + 1},$$ giving $$\int^1_0 \frac{x^3 - 1}{\ln x} \, dx = \ln (4).$$

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