Matematické Fórum

stuart clark · 16. 03. 2020 08:12

If $f:[0,1]\rightarrow (0,\infty)$ is a continuous function such that $\displaystyle \int^{1}_{0}f(x)dx = 1.$

Then maximum value of $\displaystyle \bigg(\int^{1}_{0}\sqrt[3]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[5]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[7]{f(x)}dx\bigg)$

laszky · 16. 03. 2020 11:31

↑ stuart clark:

Hi

Autor skryl část textu. Zobrazit/skrýt!

stuart clark · 17. 03. 2020 10:01

Thanks ↑ laszky:

i solved like this way

$\bigg(\int^{1}_{0}\sqrt[3]{f(x)}dx\bigg)^3\leq \int^{1}_{0}f(x)dx$

$\bigg(\int^{1}_{0}\sqrt[5]{f(x)}dx\bigg)^5\leq \int^{1}_{0}f(x)dx$

$\bigg(\int^{1}_{0}\sqrt[7]{f(x)}dx\bigg)^7\leq \int^{1}_{0}f(x)dx$

So $\int^{1}_{0}\sqrt[3]{f(x)}dx\cdot \int^{1}_{0}\sqrt[5]{f(x)}dx\cdot \int^{1}_{0}\sqrt[7]{f(x)}dx\leq \bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{3}}\bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{5}}\bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{7}}=1$ $

Matematické Fórum

#1 16. 03. 2020 08:12

product of definite integration

#2 16. 03. 2020 11:31

Re: product of definite integration

#3 17. 03. 2020 10:01

Re: product of definite integration

Zápatí