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α_1 ^3 [((Π_(i=2) ^n (x−α_i ))/(Π_(i=2) ^n (α_1 −α_i )))]+Σ_(j=2) ^n (α_j ^3 [((Π_(i=1) ^(j−1) (x−α_i )Π_(i=j+1) ^n (x−α_j ))/(Π_(i=1) ^(j−1) (α_j −α_i )Π_(i=j+1) ^n (α_j −α_i )))]+α_n ^3 [((Π_(i=1) ^(n−1) (x−α_i ))/(Π_(i=1) ^(n−1) (α_n −α_i )))]−x^3 =0 solve for x . [ where n≥5 ]

$$ \\ $$$$\alpha_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\alpha_{\mathrm{1}} −\alpha_{{i}} \right)}\right]+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\alpha_{{j}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{j}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)}\right]+\alpha_{{n}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\alpha_{{n}} −\alpha_{{i}} \right)}\right]−{x}^{\mathrm{3}} =\mathrm{0}\right. \\ $$$${solve}\:{for}\:{x}\:.\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{where}\:{n}\geqslant\mathrm{5}\:\right] \\ $$

Question Number 195224 Answers: 2 Comments: 0

∫_1 ^e 0 dx = ??

$$\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{0}\:{dx}\:=\:?? \\ $$

Question Number 195208 Answers: 4 Comments: 0

Question Number 195202 Answers: 0 Comments: 0

Question Number 195203 Answers: 1 Comments: 0

Calculer ∫^( +∞) _( 0) (dt/((e^t −e^(−t) )^2 +a^2 ))

$$\mathrm{Calculer}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{dt}}{\left(\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} \right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} } \\ $$

Question Number 195195 Answers: 1 Comments: 1

find the limit: _(x→a ) ^(lim) (x^(1/3) /x^(1/2) ) − (a^(1/3) /a^(1/2) )

$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{limit}: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{a}\:} {\overset{\mathrm{lim}} {\:}}\:\:\:\frac{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\frac{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{3}}} }{\boldsymbol{\mathrm{a}}^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$

Question Number 195194 Answers: 2 Comments: 0

Question Number 195192 Answers: 1 Comments: 0

lim_(x→∞) (sinx+(π/2))^x =?

$$ \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{sinx}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{x}} =? \\ $$

Question Number 195206 Answers: 1 Comments: 0

Question Number 195185 Answers: 0 Comments: 0

1/ Montrer que ∫^( +∞) _( 0) (((1−x^2 )^(2p−1) )/(1−x^(4p) ))dx=((2^(2p−3) /p))π[1+2Σ_(k=1) ^(p−1) cos^(2p−1) (((kπ)/(2p)))] 2/ En de^ duire ∫^( 1) _( 0) (((1−x^2 )^(2p−1) )/(1−x^(4p) ))dx

$$\mathrm{1}/\:\:\mathrm{Montrer}\:\mathrm{que}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \:\frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx}=\left(\frac{\mathrm{2}^{\mathrm{2}{p}−\mathrm{3}} }{{p}}\right)\pi\left[\mathrm{1}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{p}−\mathrm{1}} {\sum}}{cos}^{\mathrm{2}{p}−\mathrm{1}} \left(\frac{{k}\pi}{\mathrm{2}{p}}\right)\right] \\ $$$$\mathrm{2}/\:\:\:\:\mathrm{En}\:\mathrm{d}\acute {\mathrm{e}duire}\:\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \frac{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}{p}−\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{4}{p}} }{dx} \\ $$

Question Number 195180 Answers: 1 Comments: 0

1. f(x)= { ((sinx , (π/2)<x≤2π)),((cosx , 0≤x≤(π/2))) :} then find the f^′ ((π/2)) =? 2. f(x)= { ((sinx , (π/2)<x≤2π)),((cosx , 0≤x≤(π/2))) :} then find the f′(2π) =?

$$ \\ $$$$\mathrm{1}.\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}^{'} \left(\frac{\pi}{\mathrm{2}}\right)\:=? \\ $$$$\mathrm{2}.\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\begin{cases}{\mathrm{sinx}\:\:\:\:\:\:,\:\:\:\:\frac{\pi}{\mathrm{2}}<\mathrm{x}\leqslant\mathrm{2}\pi}\\{\mathrm{cosx}\:\:\:\:\:\:,\:\:\:\:\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{f}'\left(\mathrm{2}\pi\right)\:=? \\ $$$$ \\ $$

Question Number 195178 Answers: 2 Comments: 0