Having trouble confirming that these two values are equal.
$$1 - \int_{1}^{2} \ln(y)^{\frac{1}{3}} \, dy + 1 = \int_{0}^{1} e^{x^3} \, dx$$
For context, $$\ln(y)^{\frac{1}{3}}$$ is the inverse of $$e^{x^3}$$
I saw that there was no simple solution for f(x) and decided to try to convert to f(y) to find other ways to measure the value under the curve. If you look at both equations in a graphing calculator you will better understand my rational.
Am I wrong in thinking these would be equal?
The LHS and RHS of your equality are not equal unfortunately.
If $f(x)=e^{x^{3}}$ then $f^{-1}(x)=\ln(x)^{\frac{1}{3}}$ and you can derive, $$\int_{0}^{1}e^{x^{3}}dx=e-\int_{1}^{e}\ln\left(y\right)^{\frac{1}{3}}dy.$$ Is this what you were looking for?