Question

sketch the region in the plane bounded by the x-axis, the line x=2 and the curve y=(1/9)x^5. find the area of this area of this region.

Answer 1

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.

$\bf y=\cfrac{1}{9}x^5\implies 9y=x^5\implies \sqrt[5]{9y}=x\\\\ -------------------------------\\\\ \begin{cases} x=2\\ \sqrt[5]{9y}=x \end{cases}\implies 2=\sqrt[5]{9y}\implies 2^5=9y\implies \boxed{\cfrac{32}{9}=y}$

so, let's use that then.

$\bf \displaystyle \int\limits_{0}^{\frac{32}{9}}\ \left([2] - \left[ \sqrt[5]{9y} \right]\right)dy\implies \int\limits_{0}^{\frac{32}{9}}\ 2\cdot dy-9^{\frac{1}{5}}\int\limits_{0}^{\frac{32}{9}}\ y^{\frac{1}{5}}\cdot dy \\\\\\ \left.\cfrac{}{} 2y \right]_{0}^{\frac{32}{9}}-\left. \sqrt[5]{9}\cdot \cfrac{y^{\frac{6}{5}}}{\frac{6}{5}} \right]_{0}^{\frac{32}{9}}\implies \left.\cfrac{}{} 2y \right]_{0}^{\frac{32}{9}}-\left.\cfrac{5\sqrt[5]{9y^6}}{6} \right]_{0}^{\frac{32}{9}}$

$\bf \left[ \cfrac{64}{9} \right]-\left[ \cfrac{5\sqrt[5]{\frac{32^6}{9^5}}}{6} \right]\implies \cfrac{32}{27}\approx 1.\overline{185}$