Check the picture below... so the function looks like so.
now, bear in mind, the function x = 2, is just a vertical line, so we couldn't use the [ceiling] - [floor] type of function arrangement, thus let's use [right] - [left].
as you can see from the graph, which one is on the left side, and thus the left-function and which is on the right, or the right-function.
so, we have to have both in "y" terms, and the bound points are coming from the y-axis. From the graph, we can tell the lower-bound is 0, what's the upper-bound? let's check by seeing where those functions meet.
![\bf y=\cfrac{1}{9}x^5\implies 9y=x^5\implies \sqrt[5]{9y}=x\\\\ -------------------------------\\\\ \begin{cases} x=2\\ \sqrt[5]{9y}=x \end{cases}\implies 2=\sqrt[5]{9y}\implies 2^5=9y\implies \boxed{\cfrac{32}{9}=y}](https://tex.z-dn.net/?f=%5Cbf%20y%3D%5Ccfrac%7B1%7D%7B9%7Dx%5E5%5Cimplies%209y%3Dx%5E5%5Cimplies%20%5Csqrt%5B5%5D%7B9y%7D%3Dx%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Ax%3D2%5C%5C%0A%5Csqrt%5B5%5D%7B9y%7D%3Dx%0A%5Cend%7Bcases%7D%5Cimplies%202%3D%5Csqrt%5B5%5D%7B9y%7D%5Cimplies%202%5E5%3D9y%5Cimplies%20%5Cboxed%7B%5Ccfrac%7B32%7D%7B9%7D%3Dy%7D)
so, let's use that then.
![\bf \displaystyle \int\limits_{0}^{\frac{32}{9}}\ \left([2] - \left[ \sqrt[5]{9y} \right]\right)dy\implies \int\limits_{0}^{\frac{32}{9}}\ 2\cdot dy-9^{\frac{1}{5}}\int\limits_{0}^{\frac{32}{9}}\ y^{\frac{1}{5}}\cdot dy \\\\\\ \left.\cfrac{}{} 2y \right]_{0}^{\frac{32}{9}}-\left. \sqrt[5]{9}\cdot \cfrac{y^{\frac{6}{5}}}{\frac{6}{5}} \right]_{0}^{\frac{32}{9}}\implies \left.\cfrac{}{} 2y \right]_{0}^{\frac{32}{9}}-\left.\cfrac{5\sqrt[5]{9y^6}}{6} \right]_{0}^{\frac{32}{9}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D%5C%20%5Cleft%28%5B2%5D%20-%20%5Cleft%5B%20%5Csqrt%5B5%5D%7B9y%7D%20%5Cright%5D%5Cright%29dy%5Cimplies%20%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D%5C%202%5Ccdot%20dy-9%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5Cint%5Climits_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D%5C%20y%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%5Ccdot%20dy%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%5Ccfrac%7B%7D%7B%7D%202y%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D-%5Cleft.%20%5Csqrt%5B5%5D%7B9%7D%5Ccdot%20%5Ccfrac%7By%5E%7B%5Cfrac%7B6%7D%7B5%7D%7D%7D%7B%5Cfrac%7B6%7D%7B5%7D%7D%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D%5Cimplies%20%0A%5Cleft.%5Ccfrac%7B%7D%7B%7D%202y%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D-%5Cleft.%5Ccfrac%7B5%5Csqrt%5B5%5D%7B9y%5E6%7D%7D%7B6%7D%20%5Cright%5D_%7B0%7D%5E%7B%5Cfrac%7B32%7D%7B9%7D%7D)
![\bf \left[ \cfrac{64}{9} \right]-\left[ \cfrac{5\sqrt[5]{\frac{32^6}{9^5}}}{6} \right]\implies \cfrac{32}{27}\approx 1.\overline{185}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%20%5Ccfrac%7B64%7D%7B9%7D%20%5Cright%5D-%5Cleft%5B%20%5Ccfrac%7B5%5Csqrt%5B5%5D%7B%5Cfrac%7B32%5E6%7D%7B9%5E5%7D%7D%7D%7B6%7D%20%5Cright%5D%5Cimplies%20%5Ccfrac%7B32%7D%7B27%7D%5Capprox%201.%5Coverline%7B185%7D)
now, bear in mind, the function x = 2, is just a vertical line, so we couldn't use the [ceiling] - [floor] type of function arrangement, thus let's use [right] - [left].
as you can see from the graph, which one is on the left side, and thus the left-function and which is on the right, or the right-function.
so, we have to have both in "y" terms, and the bound points are coming from the y-axis. From the graph, we can tell the lower-bound is 0, what's the upper-bound? let's check by seeing where those functions meet.
so, let's use that then.