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The double integral over the region: $$ R = \left\{ \left( x,\: y \right) : a \leqslant x \leqslant b,\: g\left( x \right) \leqslant y \leqslant h\left( x \right) \right\} $$ is expressed as $$ \iint_R f\left( x,\: y \right) \mathrm{d}A = \int_a^b \left[ \int_{g\left( x \right)}^{h\left( x \right)} f\left( x, \: y \right) \mathrm{d}y \right] \mathrm{d}x. $$ By Fubini's theorem, any permutation of the order of integration is equivalent if the function $f$ is integrable.

Assuming $f$ is integrable, if $g$ and $h$ are invertible on the interval $a \leqslant x \leqslant b$, is it correct to say that in general, the reversed double integral has the form: $$ \iint_R f\left( x, \: y \right) \mathrm{d}A = \int_{g\left( a \right)}^{h\left( b \right)} \left[ \int_{h^{-1}\left( y \right)}^{g^{-1}\left( y \right)} f\left( x, \: y \right) \mathrm{d}x \right] \mathrm{d}y. $$ Would there be any other restrictions on the functions $g$ and $h$?

Thank you for your help.

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    $\begingroup$ Look up the following adjectives in relationship to changing order of integration and Fubini's theorem- simple, type I, type II & connected. $\endgroup$
    – nickalh
    Commented Apr 23 at 10:35

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The area of the double integral with the reversed bounds differs from that of the first integral, which can be easily seen from the following figure; compare the two ranges $(a,b)$ and $\left (h^{-1}(g(a)),g^{-1}(h(b))\right).$

enter image description here

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  • $\begingroup$ Thanks. This was basically how I arrived at the equation in my post. I guess a plot suffices. $\endgroup$ Commented Apr 29 at 14:39
  • $\begingroup$ I will award you the bounty as it's about to expire. $\endgroup$ Commented Apr 29 at 14:41
  • $\begingroup$ @LightninBolt74 Thank you! $\endgroup$
    – Amir
    Commented Apr 29 at 15:04

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