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Question Number 30582 Answers: 0 Comments: 0
$${x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}\:} \:{are}\:{roots}\:{of}\:{the}\:{polynomial}\:{x}^{\mathrm{3}} \:−{x}+\mathrm{1}\:{find} \\ $$$${the}\:{polynomial}\:{wich}\:{have}\:{for}\:{roots}\:{x}_{\mathrm{1}} ^{\mathrm{3}} \:,{x}_{\mathrm{2}} ^{\mathrm{3}} \:{and}\:{x}_{\mathrm{3}} ^{\mathrm{3}} \:\:. \\ $$
Question Number 30581 Answers: 0 Comments: 0
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\:\frac{\mathrm{1}}{\left({x}+{iy}\right)^{{n}} }\:. \\ $$
Question Number 30580 Answers: 0 Comments: 1
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}\:{with}\:{m}\geqslant{n}+\mathrm{2} \\ $$$${then}\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}{dx}. \\ $$
Question Number 30579 Answers: 0 Comments: 0
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} −\mathrm{1}}{{x}^{\mathrm{2}{n}} −\mathrm{1}}\:. \\ $$
Question Number 30578 Answers: 0 Comments: 0
$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:{on}\:{C}\left[{x}\right].{with}\:{n}\:{fromN}. \\ $$
Question Number 30576 Answers: 0 Comments: 1
$${let}\:{consider}\:{the}\:{equation}\:{x}^{\mathrm{3}} \:+{px}\:+{q}\: \\ $$$${find}\:{S}=\:\sum_{{i}\neq{j}} \:\frac{{x}_{{i}} }{{x}_{{j}} }\:. \\ $$
Question Number 30575 Answers: 0 Comments: 0
$${find}\:\int\int_{{D}} \:\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){dxdy}\:\:{with} \\ $$$${D}=\left\{\left({x},{y}\right)/\:{x}\leqslant\mathrm{1}\:{and}\:{x}^{\mathrm{2}} \leqslant{y}\leqslant\mathrm{2}\:\right\}. \\ $$
Question Number 30574 Answers: 0 Comments: 0
$${find}\:\int\int_{\left[\mathrm{1},{e}\right]^{\mathrm{2}} } \:\:\:{ln}\left({xy}\right){dxdy}. \\ $$
Question Number 30573 Answers: 0 Comments: 0
$${find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right]} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dxdy}. \\ $$
Question Number 30572 Answers: 0 Comments: 1
$${find}\:\:{I}=\int\int_{\left[\mathrm{3},\mathrm{4}\right]×\left[\mathrm{1},\mathrm{2}\right]} \:\:\frac{{dxdy}}{\left({x}+{y}\right)^{\mathrm{2}} }\:. \\ $$
Question Number 30571 Answers: 0 Comments: 0
$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{{n}} }\:{on}\:{C}\left[{x}\right]. \\ $$
Question Number 30570 Answers: 0 Comments: 0
$${find}\:\int\int_{{U}} \:\frac{{dxdy}}{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} }\:{with}\:{U}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\mathrm{1}\leqslant{x}^{\mathrm{2}} \:+\mathrm{2}{y}^{\mathrm{2}} \leqslant\mathrm{4}\right\} \\ $$
Question Number 30569 Answers: 0 Comments: 0
$${find}\:{I}=\:\int\int_{{D}} \:\sqrt{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\:\:{dxdy}\:\:{with}\:{D}\:{is}\:{the}\:{interior} \\ $$$${of}\:{ellipce}\:\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\mathrm{1}. \\ $$
Question Number 30568 Answers: 0 Comments: 0
$${find}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}}\:. \\ $$
Question Number 30567 Answers: 0 Comments: 0
$${integrate}\:{xy}^{,} \:+\left({x}−\mathrm{1}\right){y}\:+{y}^{\mathrm{2}} =\mathrm{0} \\ $$
Question Number 30566 Answers: 0 Comments: 0
$${study}\:{the}\:{convergence}\:{of}\:{A}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\alpha−\mathrm{1}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{and} \\ $$$${find}\:{its}\:{value}. \\ $$
Question Number 30565 Answers: 0 Comments: 0
$${let}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{n}!}\left({px}−{qx}^{\mathrm{2}} \right)^{{n}} \:\:\:{find}\:{maxf}\:\:. \\ $$
Question Number 30564 Answers: 0 Comments: 0
$${f}\:{and}\:{g}\:{are}\:\mathrm{2}\:{function}\:\:{C}^{{n}} \:{on}\:\left[{a},{b}\right]\:{prove}\:{that} \\ $$$$\int_{{a}} ^{{b}} \:{f}^{\left({n}\right)} \left({x}\right){g}\left({x}\right){dx}=\left[\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} \:{f}^{\left({k}\right)} {g}^{\left({n}−{k}\right)} \right]_{{a}} ^{{b}} \:+\left(−\mathrm{1}\right)^{{n}} \int_{{a}} ^{{b}} {f}\left({x}\right){g}^{\left({n}\right)} \left({x}\right){dx} \\ $$
Question Number 30563 Answers: 0 Comments: 0
$${let}\:{f}\left({x}\right)=\mid{x}−\mathrm{2}\:\left[\frac{{x}+\mathrm{1}}{\mathrm{2}}\right]\mid \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{f}\:{is}\:{periodic} \\ $$$$\left.\mathrm{2}\right)\:{simplify}\:{f}\left({x}\right)\:{if}\:{p}\leqslant{x}+\mathrm{1}\:{and}\:{p}\in{Z}\:. \\ $$
Question Number 30560 Answers: 0 Comments: 0
$${study}\:{the}\:{roots}\:{of}\:{f}_{{n}} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{{x}^{{k}} }{{k}!}\:. \\ $$
Question Number 30559 Answers: 0 Comments: 0
$${find}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{{n}} {dt}\:\:. \\ $$
Question Number 30558 Answers: 0 Comments: 0
$${find}\:{I}=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:{dx}\:\:.\right. \\ $$
Question Number 30557 Answers: 0 Comments: 0
$${if}\:\varphi\:{convexe}\:{and}\:{f}\:{continue}\:{on}\:\left[{a},{b}\right]\:{prove}\:{that} \\ $$$$\varphi\left(\:\frac{\mathrm{1}}{{b}−{a}}\:\int_{{a}} ^{{b}} \:{f}\left({t}\right){dt}\right)\leqslant\:\frac{\mathrm{1}}{{b}−{a}}\:\int_{{a}} ^{{b}} \:\varphi{of}\left({t}\right){dt}. \\ $$
Question Number 30556 Answers: 0 Comments: 0
$${let}\:\:{S}_{{n}} \left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{{sin}\left({kx}\right)}{{k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)}\:\:{find}\:{lim}_{{n}\rightarrow\infty} {S}_{{n}} \left({x}\right). \\ $$
Question Number 30555 Answers: 0 Comments: 0
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}\:. \\ $$
Question Number 30554 Answers: 0 Comments: 0
$${find}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{xcos}\theta\:+\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xcos}\theta\:+\mathrm{1}}{dx}\:. \\ $$
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