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Question-124555




Question Number 124555 by mnjuly1970 last updated on 04/Dec/20
Answered by mnjuly1970 last updated on 05/Dec/20
solution:    f :[a,b]→R^+ ∪{0} is continuous :      lim_(n→∞) (∫_a ^( b) f^( n) (x)dx)^(1/n) =M       where  : M:=sup{f(x)∣x ∈[a,b]}                  M=  sup_([0 ,(π/2)]) {(sin(x)+(√3) cos(x))}=2         ∴lim_(n→∞) ((∫_0 ^( (π/2)) {(sin(x)+(√3) cos(x))}^n  ))^(1/(2n))  dx             =lim_(n→∞) [(∫_0 ^( (π/2)) {sin(x)+(√3) cos(x)}^n )^(1/n) ]^(1/2) =(√2) ✓
$${solution}: \\ $$$$\:\:{f}\::\left[{a},{b}\right]\rightarrow\mathbb{R}^{+} \cup\left\{\mathrm{0}\right\}\:{is}\:{continuous}\:: \\ $$$$\:\:\:\:{lim}_{{n}\rightarrow\infty} \left(\int_{{a}} ^{\:{b}} {f}^{\:{n}} \left({x}\right){dx}\right)^{\frac{\mathrm{1}}{{n}}} =\mathrm{M} \\ $$$$\:\:\:\:\:{where}\:\::\:\mathrm{M}:={sup}\left\{{f}\left({x}\right)\mid{x}\:\in\left[{a},{b}\right]\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{M}=\:\:{sup}_{\left[\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right]} \left\{\left({sin}\left({x}\right)+\sqrt{\mathrm{3}}\:{cos}\left({x}\right)\right)\right\}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\therefore{lim}_{{n}\rightarrow\infty} \sqrt[{\mathrm{2}{n}}]{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left\{\left({sin}\left({x}\right)+\sqrt{\mathrm{3}}\:{cos}\left({x}\right)\right)\right\}^{{n}} \:}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:={lim}_{{n}\rightarrow\infty} \left[\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left\{{sin}\left({x}\right)+\sqrt{\mathrm{3}}\:{cos}\left({x}\right)\right\}^{{n}} \right)^{\frac{\mathrm{1}}{{n}}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} =\sqrt{\mathrm{2}}\:\checkmark \\ $$

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