#1 16. 03. 2020 08:12
- stuart clark
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product of definite integration
If ![$f:[0,1]\rightarrow (0,\infty)$](/mathtex/5c/5c142183ea3359f94450893ed04f67a2.gif "kopírovat do textarea")
is a continuous function such that 
Then maximum value of ![$\displaystyle \bigg(\int^{1}_{0}\sqrt[3]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[5]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[7]{f(x)}dx\bigg)$](/mathtex/bd/bdf6f819eb09f5bf10a45ab1b45fc290.gif "kopírovat do textarea")
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#2 16. 03. 2020 11:31
Re: product of definite integration
and 
, 
it follows that
, 
and 
we obtain
![$\displaystyle \bigg(\int^{1}_{0}\sqrt[3]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[5]{f(x)}dx\bigg)\bigg(\int^{1}_{0}\sqrt[7]{f(x)}dx\bigg) \leq \left(\int_0^1 f(x)\,\mathrm{d}x\right)^{\frac{1}{3}+\frac{1}{5}+\frac{1}{7}} = \left(\int_0^1 f(x)\,\mathrm{d}x\right)^{\frac{71}{105}} = 1^{\frac{71}{105}} = 1$](/mathtex/49/490086f7f973cb78eda37e5f3fa1837f.gif "kopírovat do textarea")
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#3 17. 03. 2020 10:01
- stuart clark
- Příspěvky: 1015
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Re: product of definite integration
Thanks ↑ laszky:
i solved like this way![$\bigg(\int^{1}_{0}\sqrt[3]{f(x)}dx\bigg)^3\leq \int^{1}_{0}f(x)dx$](/mathtex/46/469bba97d90aa01d18854abe05b5d516.gif "kopírovat do textarea")
![$\bigg(\int^{1}_{0}\sqrt[5]{f(x)}dx\bigg)^5\leq \int^{1}_{0}f(x)dx$](/mathtex/35/35b5407d30417f45fc6beb68d580f4c3.gif "kopírovat do textarea")
![$\bigg(\int^{1}_{0}\sqrt[7]{f(x)}dx\bigg)^7\leq \int^{1}_{0}f(x)dx$](/mathtex/77/77c34fb1a4b66488c5eb38a2a200463e.gif "kopírovat do textarea")
So ![$\int^{1}_{0}\sqrt[3]{f(x)}dx\cdot \int^{1}_{0}\sqrt[5]{f(x)}dx\cdot \int^{1}_{0}\sqrt[7]{f(x)}dx\leq \bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{3}}\bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{5}}\bigg(\int^{1}_{0}f(x)\bigg)^{\frac{1}{7}}=1$](/mathtex/46/46bcbff19f23a69f64167b8cd8a3f888.gif "kopírovat do textarea")
$
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