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AllQuestion and Answers: Page 1180

Question Number 97638 Answers: 0 Comments: 0

Question Number 97637 Answers: 0 Comments: 1

Given p,q∈R_+ ^∗ −{−1}/(1/p)+(1/q)=1 show that; ∀a,b ∈R ab≤(a^p /p)+(b^q /q)

$$\mathrm{Given}\:\mathrm{p},\mathrm{q}\in\mathbb{R}_{+} ^{\ast} −\left\{−\mathrm{1}\right\}/\frac{\mathrm{1}}{\mathrm{p}}+\frac{\mathrm{1}}{\mathrm{q}}=\mathrm{1}\:\mathrm{show}\:\mathrm{that}; \\ $$$$\forall\mathrm{a},\mathrm{b}\:\in\mathbb{R}\:\mathrm{ab}\leqslant\frac{\mathrm{a}^{\mathrm{p}} }{\mathrm{p}}+\frac{\mathrm{b}^{\mathrm{q}} }{\mathrm{q}} \\ $$

Question Number 97629 Answers: 1 Comments: 0

Question Number 97628 Answers: 1 Comments: 0

Question Number 97627 Answers: 2 Comments: 0

give ∫_0 ^∞ ((arctan(x))/((1+x^2 )^2 ))dx at form of serie

$$\mathrm{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{at}\:\mathrm{form}\:\mathrm{of}\:\mathrm{serie} \\ $$

Question Number 97626 Answers: 0 Comments: 2

solve y^(′′) −2y^′ +y =x^2 with y^′ (0) =y(0) =−1

$$\mathrm{solve}\:\mathrm{y}^{''} \:−\mathrm{2y}^{'} \:+\mathrm{y}\:\:=\mathrm{x}^{\mathrm{2}} \:\mathrm{with}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\mathrm{y}\left(\mathrm{0}\right)\:=−\mathrm{1} \\ $$

Question Number 97625 Answers: 1 Comments: 0

solve x^2 y^(′′) −(x+1)y^′ =x^2 sinx

$$\mathrm{solve}\:\mathrm{x}^{\mathrm{2}} \:\mathrm{y}^{''} −\left(\mathrm{x}+\mathrm{1}\right)\mathrm{y}^{'} \:\:=\mathrm{x}^{\mathrm{2}} \mathrm{sinx} \\ $$

Question Number 97624 Answers: 2 Comments: 0

calculate ∫_2 ^∞ (dx/((x+1)^3 (x^2 +1)^4 ))

$$\mathrm{calculate}\:\int_{\mathrm{2}} ^{\infty} \:\:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} } \\ $$

Question Number 97622 Answers: 0 Comments: 0

let f(x) =arctan(x^2 −3) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie 3) calculate ∫_0 ^1 f(x)dx

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 97620 Answers: 1 Comments: 0

calculate ∫_0 ^∞ ((sin(πx^2 ))/(x^4 −x^2 +1))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\pi\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx} \\ $$

Question Number 97619 Answers: 1 Comments: 0

calculate ∫_0 ^∞ ((cos(3x))/((x^2 +3)^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{3x}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 97617 Answers: 0 Comments: 0

give ∫_0 ^∞ ((arctan(2x))/((1+x^2 )^2 ))dx at form of serie

$$\mathrm{give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{2x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{at}\:\mathrm{form}\:\mathrm{of}\:\mathrm{serie} \\ $$

Question Number 97616 Answers: 1 Comments: 0

give ∫_0 ^∞ (e^(−x) /((1+x)^2 ))dx at form of serie

$$\mathrm{give}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{x}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\mathrm{dx}\:\mathrm{at}\:\mathrm{form}\:\mathrm{of}\:\mathrm{serie} \\ $$

Question Number 97615 Answers: 0 Comments: 0

Question Number 97606 Answers: 1 Comments: 0

given that the polynomial p(x)=(3x+2)(x−1)q(x)−2x−4 of degree 3 is exactly divisible by x−2 and that p(−1)=−12. find q(x).

$${given}\:{that}\:{the}\:{polynomial}\:{p}\left({x}\right)=\left(\mathrm{3}{x}+\mathrm{2}\right)\left({x}−\mathrm{1}\right){q}\left({x}\right)−\mathrm{2}{x}−\mathrm{4} \\ $$$${of}\:{degree}\:\mathrm{3}\:{is}\:{exactly}\:{divisible}\:{by}\:{x}−\mathrm{2}\:{and}\: \\ $$$${that}\:{p}\left(−\mathrm{1}\right)=−\mathrm{12}.\:{find}\:{q}\left({x}\right). \\ $$

Question Number 97600 Answers: 1 Comments: 0

Question Number 97597 Answers: 0 Comments: 3

Notification test ...

$${Notification}\:{test}\:... \\ $$

Question Number 97591 Answers: 1 Comments: 0

∫(x^4 /(x^3 −2x^2 −7x+4))dx

$$\int\frac{{x}^{\mathrm{4}} }{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}}{dx} \\ $$

Question Number 97590 Answers: 1 Comments: 0

find the integration ∫(√(sin(x)))dx

$$\mathrm{find}\:\mathrm{the}\:\mathrm{integration} \\ $$$$\int\sqrt{{sin}\left({x}\right)}{dx} \\ $$

Question Number 97577 Answers: 2 Comments: 1

lim_(x→1) ((4x lnx)/(x−1)) =??

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{\mathrm{4}{x}\:\mathrm{ln}{x}}{{x}−\mathrm{1}}\:=?? \\ $$

Question Number 97576 Answers: 2 Comments: 0

Show that RE[(1/(1−z))]=(1/2) where z = cos θ + i sinθ

$$\:\mathrm{Show}\:\mathrm{that}\:{RE}\left[\frac{\mathrm{1}}{\mathrm{1}−{z}}\right]=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{where}\:{z}\:=\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\theta \\ $$$$ \\ $$

Question Number 97569 Answers: 1 Comments: 0

∫_0 ^(π/4) (√(tan(x)))(√(1−tan(x))) dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{{tan}\left({x}\right)}\sqrt{\mathrm{1}−{tan}\left({x}\right)}\:{dx} \\ $$

Question Number 97564 Answers: 1 Comments: 0

Question Number 97563 Answers: 1 Comments: 0

∫_0 ^(+∞) e^(−x) cos(x^2 )dx. Discuss the convergence of this generalised intergral. Please help

$$\int_{\mathrm{0}} ^{+\infty} \mathrm{e}^{−\mathrm{x}} \mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}. \\ $$$$\mathrm{Discuss}\:\mathrm{the}\:\mathrm{convergence}\:\mathrm{of}\:\mathrm{this}\: \\ $$$$\mathrm{generalised}\:\mathrm{intergral}. \\ $$$$\mathrm{Please}\:\mathrm{help} \\ $$

Question Number 97557 Answers: 1 Comments: 0

Question Number 97555 Answers: 0 Comments: 8

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