Mathematics: HELP HELP HELP HELP HELP HELP

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cancerbaby209

21.10.2021

$\left(- \frac{x^{3}}{2} + x\right)\cdot \left(\frac{x^{2}}{4} + 1\right)= \\ x- \frac{x^{5}}{8} - \frac{x^{3}}{4} \\ \displaystyle\int\limits^{{\infty}}_{{0}} \left(x-\dfrac{x^5}{8}-\dfrac{x^3}{4}\right){\mathrm{d}x} \\ = -\int\limits^{{\infty}}_{{0}}\dfrac{x^6}{48}{\mathrm{d}x}-\int\limits^{{\infty}}_{{0}}\dfrac{x^4}{16}{\mathrm{d}x}+\int\limits^{{\infty}}_{{0}}\dfrac{x^2}{2}{\mathrm{d}x}$

The integral is divergent...

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Ответ:

jm832098

09.02.2020

Mathematics

He made 149 hits during the regular season.

Step-by-step explanation:

there u go my friend :D

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HELP HELP HELP HELP HELP HELP

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