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$$ \begin{array}{l}{\text { (3) Let } f \text { be an integrable function. Express the integral }} \\ {\qquad \int_{0}^{1} \int_{y}^{2 y} \int_{0}^{x+y} f(x, y, z)\, d z\, d x \,d y} \\ {\text { as a sum of iterated integrals in the order } d x d y d z \text { . }}\end{array} $$

For this to be in order $dx\, dy\, dz$, we need $f$ to be a continuous function and partial derivative need to exist but no condition is given as such so how to tackle such kind of function?

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  • $\begingroup$ No? If $f\in L^1(\mathbb{R}^3)$ (or $L^1([0,1]\times [0,2]\times [0,3])$ as seems to be your case), then the interchangability of integration order is part of the Fubini theorem. $\endgroup$ – WoolierThanThou Oct 10 '19 at 6:52
  • $\begingroup$ @WoolierThanThou so what should be answer can you explain please? $\endgroup$ – maths student Oct 10 '19 at 7:22
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The answer is $\int_0^{3}\int_0^{1} \int_{\max \{z-y,y\}}^{2y} f(x,y,z)\,dx\,dy\,dz$. Justification is by Fubini's Theoorem. [ Note that the inequalities $0<z<x+y, y<x<2y, 0<y<1$ are equivalent to the inequalities $\max \{z-y,y\} <x <2y, 0<y<1$ and $0<z<3$].

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  • $\begingroup$ Isn't it the case that x range from 0 to 2 $\endgroup$ – maths student Oct 10 '19 at 7:31
  • $\begingroup$ $x$ can on;y go up to $2y$. Of course, $2y <2$ but the limit of integration w.r..t $x$ has to go up to $2y$ not $2$. @mathsstudent $\endgroup$ – Kavi Rama Murthy Oct 10 '19 at 7:35
  • $\begingroup$ Then for z why the limit of integration from 0 to 3? $\endgroup$ – maths student Oct 10 '19 at 7:37
  • $\begingroup$ @mathsstudent Once you have taken care of all the restrictions on $x$ and $y$ you have to integrate w.r.t. $z$ from its minimum value to its maximum value. [No other variable can appear in the final integral]. The maximum value of $z$ is $3$ because the maximum value of $y$ is $2$ and the maximum value of $x$ is $1$. $\endgroup$ – Kavi Rama Murthy Oct 10 '19 at 7:40

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