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


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