Find g(y) and b so that $\int_0^2 (\int_{x^3}^{4x}\sin y^2\space dy)\space dx=\int_0^b\sin y^2g(y)\space dy$ This came from a Calculus II exam:  

Find $b$ and $g(y)$ so that:
$$\int_0^2 \left(\int_{x^3}^{4x}\sin y^2\space dy\right)\space dx=\int_0^b\sin y^2g(y)\space dy$$

I have no idea how to tackle this problem.
If $D=\{(x,y): 0\leq x\leq 2, x^3\leq y\leq 4x\}$ we have
$$ \iint_{D} \sin(y^2)\,dx\,dy = \int_{0}^{8}\sin(y^2)\cdot\left(\sqrt[3]{y}-\frac{y}{4}\right) \,dy $$
by Fubini's theorem, hence a solution is given by $b=8$ and $g(y)=\sqrt[3]{y}-\frac{y}{4}$.

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You just have to describe the region between two curves in two different ways, as $f_1(x)\leq y\leq f_2(x)$ or as $g_1(y)\leq x \leq g_2(y)$. The region between $y=4x$ and $y=x^3$ on the interval $x\in[0,2]$ is a normal domain with respect to both the $x$ and the $y$ variable.
Formally, this is an underconstrained problem, You was wanting to seek for a value and for a function at the same time for giving an specific value, and there are infinitely many solutions. 

A not so trivial guessing will be $b=7.19=\text{fresnels}^{-1}(0.46)$ and $g(y)=1$.