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Question Number 219515 Answers: 2 Comments: 0

Find: 𝛀 = ∫_0 ^( 1) x^x^2 dx = ?

$$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:?\: \\ $$

Question Number 219503 Answers: 1 Comments: 0

Question Number 219506 Answers: 2 Comments: 0

Q. Integrate ((x^2 +x+1)/(2x^2 −x−3)).

$${Q}.\:{Integrate}\:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}. \\ $$$$ \\ $$

Question Number 219496 Answers: 0 Comments: 0

p(t)=−(1/(iπ))∫_(−∞i+𝛄) ^( ∞i+𝛄) ((e^(st) (ln(s)+𝛄_0 ))/s) ds q(t)=(1/(iπ))∫_(−∞i+𝛄) ^( ∞i+𝛄) {−(π/(2s))L_0 (s)+(π/(2s))iY_0 (−is)}e^(st) ds g(s)=∫_0 ^( ∞) J_ν (t)J_ν (st)dt h(s)=∫_0 ^( ∞) cos(t)J_ν (st)dt e^t is exponential function ln(t) is natural logarithm cos(t) is cosine function J_ν (t) is Bessel function of the First kind Y_ν (t) is Bessel function of the Second kind L_ν (t) is modified StruveH function 𝛄_0 is Euler-mascheroni Const.(0.5772156649015.....) π is pi (3.141592653589793238.........) i is (√(−1)) ∞ is infinity

$${p}\left({t}\right)=−\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:\frac{{e}^{{st}} \left(\mathrm{ln}\left({s}\right)+\boldsymbol{\gamma}_{\mathrm{0}} \right)}{{s}}\:\mathrm{d}{s} \\ $$$${q}\left({t}\right)=\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\left\{−\frac{\pi}{\mathrm{2}{s}}\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left({s}\right)+\frac{\pi}{\mathrm{2}{s}}\boldsymbol{{i}}{Y}_{\mathrm{0}} \left(−\boldsymbol{{i}}{s}\right)\right\}{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{g}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu} \left({t}\right){J}_{\nu} \left({st}\right)\mathrm{d}{t} \\ $$$${h}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{cos}\left({t}\right){J}_{\nu} \left({st}\right)\mathrm{d}{t} \\ $$$${e}^{{t}} \:\mathrm{is}\:\mathrm{exponential}\:\mathrm{function} \\ $$$$\mathrm{ln}\left({t}\right)\:\mathrm{is}\:\mathrm{natural}\:\mathrm{logarithm} \\ $$$$\mathrm{cos}\left({t}\right)\:\mathrm{is}\:\mathrm{cosine}\:\mathrm{function}\: \\ $$$${J}_{\nu} \left({t}\right)\:\mathrm{is}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{First}\:\mathrm{kind} \\ $$$${Y}_{\nu} \left({t}\right)\:\mathrm{is}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Second}\:\mathrm{kind} \\ $$$$\boldsymbol{\mathrm{L}}_{\nu} \left({t}\right)\:\mathrm{is}\:\mathrm{modified}\:\mathrm{StruveH}\:\mathrm{function} \\ $$$$\boldsymbol{\gamma}_{\mathrm{0}} \:\mathrm{is}\:\mathrm{Euler}-\mathrm{mascheroni}\:\mathrm{Const}.\left(\mathrm{0}.\mathrm{5772156649015}.....\right) \\ $$$$\pi\:\mathrm{is}\:\mathrm{pi}\:\left(\mathrm{3}.\mathrm{141592653589793238}.........\right) \\ $$$$\boldsymbol{{i}}\:\mathrm{is}\:\sqrt{−\mathrm{1}}\: \\ $$$$\infty\:\mathrm{is}\:\mathrm{infinity}\: \\ $$

Question Number 219509 Answers: 2 Comments: 0

I = ∫^( 16) _( 1) (((x +(√x))^(1/4) )/x^(3/4) ) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{I}\:=\:\underset{\:\mathrm{1}} {\int}^{\:\mathrm{16}} \:\frac{\left({x}\:+\sqrt{{x}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }{{x}^{\frac{\mathrm{3}}{\mathrm{4}}} }\:{dx} \\ $$$$ \\ $$

Question Number 219492 Answers: 3 Comments: 0

a + b = 1 a^2 + b^2 = 2 Find: a^(11) + b^(11) = ?

$$\mathrm{a}\:+\:\mathrm{b}\:=\:\mathrm{1} \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}^{\mathrm{11}} \:+\:\mathrm{b}^{\mathrm{11}} \:=\:? \\ $$

Question Number 219491 Answers: 1 Comments: 0

if x′′−2x′+10x=e^(2t) , at t=0,x=0 and x′=1 find x(t) using laplace transform

$$\boldsymbol{{if}}\:\boldsymbol{{x}}''−\mathrm{2}\boldsymbol{{x}}'+\mathrm{10}\boldsymbol{{x}}=\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{t}}} ,\:\boldsymbol{{at}}\:\boldsymbol{{t}}=\mathrm{0},\boldsymbol{{x}}=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{x}}'=\mathrm{1} \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)\:\boldsymbol{{using}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}} \\ $$

Question Number 219488 Answers: 1 Comments: 0

solve the initial value problem y′−2e^(−t^2 ) +2ty=0 y(0)=1

$$\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{initial}}\:\boldsymbol{{value}}\:\boldsymbol{{problem}}\: \\ $$$$\boldsymbol{{y}}'−\mathrm{2}\boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } +\mathrm{2}\boldsymbol{{ty}}=\mathrm{0}\:\:\boldsymbol{{y}}\left(\mathrm{0}\right)=\mathrm{1} \\ $$

Question Number 219482 Answers: 0 Comments: 1

∫ ((e^(−x^2 ) cos (x))/(x^2 +1)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{{e}^{−{x}^{\mathrm{2}} } \mathrm{cos}\:\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$$$ \\ $$

Question Number 219720 Answers: 4 Comments: 0

Question Number 219481 Answers: 0 Comments: 0

pls Help.... 1. −(1/(iπ)) ∫_( −∞+i𝛇_0 ) ^( ∞+i𝛇_0 ) ((e^(st) (ln(t)+𝛄))/t) dt 2. ∫_0 ^( ∞) t∙J_ν (t)J_ν (rt) dt 3. ∫_0 ^( ∞) t∙cos(t)J_ν (rt) dt

$$\mathrm{pls}\:\mathrm{Help}.... \\ $$$$\mathrm{1}.\:−\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\:\int_{\:−\infty+\boldsymbol{{i}\zeta}_{\mathrm{0}} } ^{\:\infty+\boldsymbol{{i}\zeta}_{\mathrm{0}} } \:\frac{{e}^{{st}} \left(\mathrm{ln}\left({t}\right)+\boldsymbol{\gamma}\right)}{{t}}\:\mathrm{d}{t}\: \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\infty} \:\:{t}\centerdot{J}_{\nu} \left({t}\right){J}_{\nu} \left({rt}\right)\:\mathrm{d}{t} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}} ^{\:\infty} \:{t}\centerdot\mathrm{cos}\left({t}\right){J}_{\nu} \left({rt}\right)\:\mathrm{d}{t} \\ $$

Question Number 219476 Answers: 1 Comments: 1

let a_1 = 1 ; (n+1)a_(n+1) +na_n = 2n−3 find nth term of a_(n )

$$\:\mathrm{let}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:;\:\left(\mathrm{n}+\mathrm{1}\right)\mathrm{a}_{\mathrm{n}+\mathrm{1}} +\mathrm{na}_{\mathrm{n}} \:=\:\mathrm{2n}−\mathrm{3}\: \\ $$$$\:\:\mathrm{find}\:\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}_{\mathrm{n}\:} \\ $$

Question Number 219465 Answers: 0 Comments: 0

prove ∫_0 ^( ∞) (√(r^2 −t^2 ))e^(−pt) dt=((−rπL_1 (rp)+πrI_1 (up)+2irK_1 (rp))/(2p)) L_ν (x) is Modified Struve function I_ν (x) is Modified Bessel function of the First kind K_ν (x) is Modified Bessel function of the Second kind

$${prove} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\sqrt{{r}^{\mathrm{2}} −{t}^{\mathrm{2}} }{e}^{−{pt}} \mathrm{d}{t}=\frac{−{r}\pi\boldsymbol{\mathrm{L}}_{\mathrm{1}} \left({rp}\right)+\pi{rI}_{\mathrm{1}} \left({up}\right)+\mathrm{2}\boldsymbol{{i}}{rK}_{\mathrm{1}} \left({rp}\right)}{\mathrm{2}{p}} \\ $$$$\boldsymbol{\mathrm{L}}_{\nu} \left({x}\right)\:\mathrm{is}\:\mathrm{Modified}\:\mathrm{Struve}\:\mathrm{function} \\ $$$${I}_{\nu} \left({x}\right)\:\mathrm{is}\:\mathrm{Modified}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{First}\:\mathrm{kind} \\ $$$${K}_{\nu} \left({x}\right)\:\mathrm{is}\:\mathrm{Modified}\:\mathrm{Bessel}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Second}\:\mathrm{kind} \\ $$

Question Number 219461 Answers: 1 Comments: 0

I_n = ∫_0 ^( 1) (x/((x + 1)^n ))dx find I_0 and I_1 express I_n interms of n for all n ≥ 2

$${I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}}{\left({x}\:+\:\mathrm{1}\right)^{{n}} }{dx} \\ $$$${find}\:{I}_{\mathrm{0}} \:{and}\:{I}_{\mathrm{1}} \\ $$$${express}\:{I}_{{n}} \:{interms}\:{of}\:{n}\:{for}\:{all}\:{n}\:\geqslant\:\mathrm{2} \\ $$

Question Number 219459 Answers: 0 Comments: 0

Two cups m and n contains the same mass of water, m is at 25°c while n is at the temperature of 102°c. If both cups are placed in the same freezer of internal temperature - 45°c. Which of the content of m and n freezes first ? Hence, show that tₘ⁻ tₙ = - wc In(²¹/₁₀). where tₘ and tₙ are the time taken for m and n to freezes and w is the mass of water and c is specific heat capacity of water.

Question Number 219458 Answers: 1 Comments: 0

∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) x_1 ^( 2) x_2 ^( 3) x_3 ^( 4) x_4 ^( 5) dx_1 dx_2 dx_3 dx_(4 )

$$ \\ $$$$\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \:{x}_{\mathrm{1}} ^{\:\mathrm{2}} {x}_{\mathrm{2}} ^{\:\mathrm{3}} {x}_{\mathrm{3}} ^{\:\mathrm{4}} {x}_{\mathrm{4}} ^{\:\mathrm{5}} \:{dx}_{\mathrm{1}} {dx}_{\mathrm{2}} {dx}_{\mathrm{3}} {dx}_{\mathrm{4}\:} \\ $$$$\: \\ $$

Question Number 219456 Answers: 0 Comments: 2

∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ∫_1 ^( 3) ((x_1 +x_2 +x_3 +x_4 −x_5 )/(x_1 +x_2 +x_3 +x_4 +x_5 )) dx_1 dx_2 dx_3 dx_4 dx_5

$$ \\ $$$$\:\int_{\mathrm{1}} ^{\:\mathrm{3}} \int_{\mathrm{1}} ^{\:\mathrm{3}} \int_{\mathrm{1}} ^{\:\mathrm{3}} \int_{\mathrm{1}} ^{\:\mathrm{3}} \int_{\mathrm{1}} ^{\:\mathrm{3}} \:\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} −{x}_{\mathrm{5}} }{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} +{x}_{\mathrm{5}} }\:{dx}_{\mathrm{1}} {dx}_{\mathrm{2}} {dx}_{\mathrm{3}} {dx}_{\mathrm{4}} {dx}_{\mathrm{5}} \:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219454 Answers: 1 Comments: 3

a, b, c are the roots of the equation x^3 −3x+1=0. find (a)^(1/3) +(b)^(1/3) +(c)^(1/3) =? & (1/( (a)^(1/3) ))+(1/( (b)^(1/3) ))+(1/( (c)^(1/3) ))=?

$${a},\:{b},\:{c}\:{are}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0}. \\ $$$${find}\:\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{c}}=?\:\&\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{b}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{c}}}=? \\ $$

Question Number 219453 Answers: 1 Comments: 0

∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ∫_1 ^( 2) ((x_1 +x_(2 ) +x_3 −x_4 )/(x_1 +x_2 +x_3 +x_4 )) dx_1 dx_2 dx_3 dx_(4 )

$$ \\ $$$$\:\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}\:} +{x}_{\mathrm{3}} −{x}_{\mathrm{4}} }{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} }\:{dx}_{\mathrm{1}} {dx}_{\mathrm{2}} {dx}_{\mathrm{3}} {dx}_{\mathrm{4}\:} \:\:\:\: \\ $$$$ \\ $$

Question Number 219451 Answers: 0 Comments: 1

Question Number 219450 Answers: 1 Comments: 3

Question Number 219449 Answers: 0 Comments: 0

L{sinx}=∫_0 ^∞ e^(−sx) sinx dx=∫_0 ^∞ e^(−sx) ((e^(ix) −e^(−ix) )/(2i))dx =(1/(2i))[∫_0 ^∞ e^(−(s−i)x) dx −∫_0 ^∞ e^(−(s+i)x) dx] =(1/(2i))[((−1)/(s−i))e^(−(s−i)x) +(1/(s+i))e^(−(s+i)x) ]_0 ^∞ =(1/(2i))[(1/(s−i))−(1/(s+i))]=(1/(2i))×((s+i−s+i)/((s−i)(s+i)))=(1/(2i))×((2i)/(s^2 +1))=(1/(s^2 +1))

$${L}\left\{{sinx}\right\}=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} {sinx}\:{dx}=\int_{\mathrm{0}} ^{\infty} {e}^{−{sx}} \frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}−{i}\right){x}} {dx}\:\:−\int_{\mathrm{0}} ^{\infty} {e}^{−\left({s}+{i}\right){x}} {dx}\right]\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{−\mathrm{1}}{{s}−{i}}{e}^{−\left({s}−{i}\right){x}} +\frac{\mathrm{1}}{{s}+{i}}{e}^{−\left({s}+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[\frac{\mathrm{1}}{{s}−{i}}−\frac{\mathrm{1}}{{s}+{i}}\right]=\frac{\mathrm{1}}{\mathrm{2}{i}}×\frac{{s}+{i}−{s}+{i}}{\left({s}−{i}\right)\left({s}+{i}\right)}=\frac{\mathrm{1}}{\mathrm{2}{i}}×\frac{\mathrm{2}{i}}{{s}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$

Question Number 219448 Answers: 0 Comments: 0

Question Number 219447 Answers: 1 Comments: 0

Lim_(x→∞) Σ_(i=1) ^∞ (((−x)/i))^i

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{Lim}_{{x}\rightarrow\infty} \underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{−{x}}{{i}}\right)^{{i}} \\ $$$$ \\ $$

Question Number 219446 Answers: 0 Comments: 0

∫tan(((1/n)/(sec(n)+(((1−sec(n))/(sec(n))))))dn

$$ \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\int{tan}\left(\frac{\frac{\mathrm{1}}{{n}}}{{sec}\left({n}\right)+\left(\frac{\mathrm{1}−{sec}\left({n}\right)}{{sec}\left({n}\right)}\right.}\right){dn} \\ $$$$ \\ $$

Question Number 219445 Answers: 0 Comments: 1

prove ∫_0 ^( ∞) (√(r^2 −t^2 ))e^(−wt) dt=((−rπL_1 (rw)+rπI_1 (rw)+2riK_1 (rw))/(2w))

$${prove} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\sqrt{{r}^{\mathrm{2}} −{t}^{\mathrm{2}} }{e}^{−{wt}} \mathrm{d}{t}=\frac{−{r}\pi\boldsymbol{\mathrm{L}}_{\mathrm{1}} \left({rw}\right)+{r}\pi{I}_{\mathrm{1}} \left({rw}\right)+\mathrm{2}{r}\boldsymbol{{i}}{K}_{\mathrm{1}} \left({rw}\right)}{\mathrm{2}{w}} \\ $$

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