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Question Number 124555 by mnjuly1970 last updated on 04/Dec/20

Answered by mnjuly1970 last updated on 05/Dec/20

$$solution:$$$$f:[a,b]\to {\mathbb{R}}^{+}\cup \left\{0\right\}iscontinuous:$$$${lim}_{n\to \mathrm{\infty }}{\left({\int }_a^b{f}^n\left(x\right)dx\right)}^{\frac{1}{n}}=\mathrm{M}$$$$where:\mathrm{M}:=sup\{f\left(x\right)\mid x\in [a,b]\}$$$$\mathrm{M}={sup}_{[0,\frac{\pi }{2}]}\left\{(sin\left(x\right)+\sqrt{3}cos\left(x\right))\right\}=2$$$$\therefore {lim}_{n\to \mathrm{\infty }}\sqrt[2n]{{\int }_0^{\frac{\pi }{2}}{\left\{(sin\left(x\right)+\sqrt{3}cos\left(x\right))\right\}}^n}dx$$$$={lim}_{n\to \mathrm{\infty }}{\left[{\left({\int }_0^{\frac{\pi }{2}}{\{sin\left(x\right)+\sqrt{3}cos\left(x\right)\}}^n\right)}^{\frac{1}{n}}\right]}^{\frac{1}{2}}=\sqrt{2}✓$$

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