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I am trying to solve the double integral:

$$\int_{0}^1\int_{1-x}^{\sqrt{(1-x)}}e^{\left({\frac{y^2}{2}}-{\frac{y^3}{3}}\right)}\ dydx$$

by reversing the order of integration, however, I am unsure how to go about doing it. Is it right to say that initially:

$\sqrt(1-x)$$≤y≤(1-x)$ and $0≤x≤1$. After reversing the order, we get $1-y^2≤x≤1-y$ and $0≤y≤1$, hence the reversed order of integration will be:

$$\int_{0}^1\int_{1-y}^{1-y^2}e^{{y^2/2}-{y^3/3}}\ dxdy?$$

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Your new integral is fine; careful though, this:

$\sqrt{1-x} \le y \le (1-x)$

and this:

$1-y^2 \le x \le 1-y$

should be the other way around:

$${1-x} \le y \le \sqrt{1-x} \quad \mbox{and} \quad 1-y \le x \le 1-y^2$$

You have the order right in the integrals though.

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Initially, we have $$1-x \le y \le \sqrt{1-x}$$

Hence we have $$1-y \le x \le 1-y^2.$$

Your final expression is correct.

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