Mathematics, 31.01.2020 21:42 homework2358
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.
Answers: 3
![\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}](/tpl/images/0491/2179/df1bf.png)
![\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}}](/tpl/images/0491/2179/69de3.png)
![\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}](/tpl/images/0491/2179/5f8de.png)
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Sketch the region in the plane bounded by the x-axis, the line x=2 and the curve y=(1/9)x^5. find th...
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