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AllQuestion and Answers: Page 1205

Question Number 87902 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ∫_0 ^∞ (e^(−(x^2 +y^2 )) /((x+y)^2 ))dxdy

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} }{\left({x}+{y}\right)^{\mathrm{2}} }{dxdy} \\ $$

Question Number 87901 Answers: 0 Comments: 0

calculate ∫∫_([0,1]^2 ) ((arctan(x+y))/(x+y))dxdy

$${calculate}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\frac{{arctan}\left({x}+{y}\right)}{{x}+{y}}{dxdy} \\ $$

Question Number 87893 Answers: 1 Comments: 0

∫ (√((sin x)/(sin x−cos x))) dx

$$\int\:\sqrt{\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}}\:\:\mathrm{dx}\: \\ $$

Question Number 87886 Answers: 1 Comments: 2

Question Number 87884 Answers: 1 Comments: 0

Question Number 87881 Answers: 1 Comments: 1

∫_(−∞) ^( +∞) (1/x) dx =

$$\:\int_{−\infty} ^{\:+\infty} \frac{\mathrm{1}}{{x}}\:{dx}\:=\: \\ $$

Question Number 87878 Answers: 1 Comments: 0

posons (1+2(√3))^n =a_n +b_n (√3) montre que pgcd(a_n ;b_n )=1

$${posons}\: \\ $$$$\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} =\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} +\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \sqrt{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{montre}}\:\boldsymbol{\mathrm{que}}\:\boldsymbol{\mathrm{pgcd}}\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{{b}}_{\boldsymbol{{n}}} \right)=\mathrm{1} \\ $$

Question Number 87877 Answers: 0 Comments: 0

posons (1+2(√3))^n =a_n +b_n (√3) montre que pgcd(a_n ;b_n )=1

Question Number 87876 Answers: 0 Comments: 1

prove that Γ(z)=∫_0 ^∞ e^(−x) x^(z−1) dx,Re(z)>0

$${prove}\:{that} \\ $$$$\Gamma\left({z}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} \:{x}^{{z}−\mathrm{1}} \:{dx},{Re}\left({z}\right)>\mathrm{0} \\ $$

Question Number 87870 Answers: 0 Comments: 0

x amd y are imtegers. how many possible solitions do the eqiation has x^2 −10y^2 = ±1

$$\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{amd}}\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{imtegers}}. \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{many}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{solitions}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{eqiation}}\:\boldsymbol{\mathrm{has}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{10}\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:=\:\pm\mathrm{1} \\ $$

Question Number 87862 Answers: 0 Comments: 3

Evaluate ∫_(−1) ^1 (1/(x−1)) dx

$$\mathrm{Evaluate}\:\:\:\:\overset{\mathrm{1}} {\int}_{−\mathrm{1}} \frac{\mathrm{1}}{{x}−\mathrm{1}}\:{dx}\: \\ $$

Question Number 87861 Answers: 0 Comments: 4

∫_0 ^(π/4) tanh 2x dx

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\:\mathrm{tanh}\:\mathrm{2}{x}\:{dx} \\ $$

Question Number 87860 Answers: 0 Comments: 3

Question Number 87843 Answers: 0 Comments: 2

Question Number 87839 Answers: 1 Comments: 0

I = ∫_0 ^(π/4) ((sin 4x)/(cos^2 x (√(tan^4 x+1)))) dx

$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\:\frac{\mathrm{sin}\:\mathrm{4x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:\sqrt{\mathrm{tan}\:^{\mathrm{4}} \mathrm{x}+\mathrm{1}}}\:\mathrm{dx} \\ $$

Question Number 87833 Answers: 1 Comments: 5

1)find ∫2^(ln(x)) dx 2)prove (i)^(1/i) = resl number

$$\left.\mathrm{1}\right){find}\:\int\mathrm{2}^{{ln}\left({x}\right)} \:{dx} \\ $$$$\left.\mathrm{2}\right){prove}\:\sqrt[{{i}}]{{i}}\:=\:{resl}\:{number} \\ $$

Question Number 87831 Answers: 1 Comments: 1

Question Number 87817 Answers: 1 Comments: 3

f(((x−3)/(x+1)))+f(((x+3)/(x−1)))=x find f(x)

$${f}\left(\frac{{x}−\mathrm{3}}{{x}+\mathrm{1}}\right)+{f}\left(\frac{{x}+\mathrm{3}}{{x}−\mathrm{1}}\right)={x} \\ $$$${find}\:{f}\left({x}\right) \\ $$

Question Number 87815 Answers: 1 Comments: 3

I = ∫_0 ^(π/2) cos 2x(cos^4 x+sin^4 x) dx

$$\mathrm{I}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cos}\:\mathrm{2x}\left(\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\right)\:\mathrm{dx} \\ $$

Question Number 87803 Answers: 2 Comments: 4

Question Number 87799 Answers: 0 Comments: 2

f(x)= { ((ax^2 +bx −1≤x≤0)),((cx^2 +d 0<x≤(1/2))),((bx+d (1/2)<x≤1)) :} f(x) is continuous on[−1,1] prove d=0 c=2b

$${f}\left({x}\right)=\begin{cases}{{ax}^{\mathrm{2}} +{bx}\:\:\:\:\:\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0}}\\{{cx}^{\mathrm{2}} +{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}}\\{{bx}+{d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}}\end{cases} \\ $$$${f}\left({x}\right)\:{is}\:{continuous}\:{on}\left[−\mathrm{1},\mathrm{1}\right] \\ $$$${prove}\:{d}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}=\mathrm{2}{b} \\ $$

Question Number 87793 Answers: 2 Comments: 0

show that ∫e^(sin(x)) dx= −Σ_(n=0) ^∞ (1/(n!))[ cos(x)∗(sin(x))^(n+1) ∗[(sin(x))^2 ]^((((−n)/2)−(1/2))) ∗ 2F_1 [(1/2),((1−n)/2);(3/2);(cos(x))^2 ] ]+c notice\2F_1 is special function called hypergeometric function

$${show}\:{that} \\ $$$$\int{e}^{{sin}\left({x}\right)} \:{dx}= \\ $$$$−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left[\:{cos}\left({x}\right)\ast\left({sin}\left({x}\right)\right)^{{n}+\mathrm{1}} \ast\left[\left({sin}\left({x}\right)\right)^{\mathrm{2}} \right]^{\left(\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \ast\:\mathrm{2}{F}_{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}−{n}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\left({cos}\left({x}\right)\right)^{\mathrm{2}} \right]\:\right]+{c} \\ $$$$ \\ $$$${notice}\backslash\mathrm{2}{F}_{\mathrm{1}} \:{is}\:{special}\:{function}\:{called}\:{hypergeometric}\:{function} \\ $$

Question Number 87790 Answers: 2 Comments: 0

How many handshakes are exchanged betwen 27 boys?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{handshakes}\:\mathrm{are}\:\mathrm{exchanged} \\ $$$$\mathrm{betwen}\:\mathrm{27}\:\mathrm{boys}? \\ $$

Question Number 87789 Answers: 1 Comments: 1

(D^3 −D^2 )y = x^2 +1 ,y(0)=1 y ′(0)=−1 ,y ′′(0) = 0

$$\left(\mathrm{D}^{\mathrm{3}} −\mathrm{D}^{\mathrm{2}} \right)\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}\:,\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\mathrm{y}\:'\left(\mathrm{0}\right)=−\mathrm{1}\:,\mathrm{y}\:''\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$

Question Number 87785 Answers: 0 Comments: 0

Question Number 87784 Answers: 1 Comments: 1

Given f(x,y) = y f(y,x) +x find the value of f(1,2) .

$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{y}\:\mathrm{f}\left(\mathrm{y},\mathrm{x}\right)\:+\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{1},\mathrm{2}\right)\:. \\ $$

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