Reversing the order of integration to solve the double integral

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:

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

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.

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

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