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Question Number 219461 Answers: 0 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: 0 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: 0

∫_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: 0 Comments: 1

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: 0 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: 0 Comments: 2

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: 0 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: 0

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

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

Question Number 219440 Answers: 0 Comments: 0

I=∫tan(((cos(n))/(n(1−cos(n) + cos^2 (n))))) dn

$$ \\ $$$$\:\:\:{I}=\int\mathrm{tan}\left(\frac{\mathrm{cos}\left({n}\right)}{{n}\left(\mathrm{1}−\mathrm{cos}\left({n}\right)\:+\:\mathrm{cos}^{\mathrm{2}} \left({n}\right)\right)}\right)\:{dn} \\ $$$$ \\ $$

Question Number 219434 Answers: 0 Comments: 0

Question Number 219428 Answers: 1 Comments: 1

a , b, ∈ R ∫_( −∞) ^( ∞) (((e^(iax) −1)(e^(ibx) −1))/x^2 ) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\:,\:{b},\:\in\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:−\infty} ^{\:\infty} \frac{\left({e}^{{iax}} −\mathrm{1}\right)\left({e}^{{ibx}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} }\:{dx} \\ $$$$ \\ $$

Question Number 219427 Answers: 0 Comments: 4

Question Number 219425 Answers: 1 Comments: 1

Question Number 219423 Answers: 1 Comments: 0

Question Number 219414 Answers: 0 Comments: 0

given the recursive {a_n } define by setting a_(1 ) ∈ (0,1) , a_(n+1) = a_n (1−a_n ) , n≥1 prove that (1) lim_(n→∞) na_n = 1 (2) b_n = n(1−na_n ) is a incresing sequence and diverge to ∞ (3) lim_(n→∞) ((n(1−na_n ))/(ln(n))) = 1

$$\:\:\:\mathrm{given}\:\mathrm{the}\:\mathrm{recursive}\:\left\{\mathrm{a}_{\mathrm{n}} \right\}\:\mathrm{define}\:\mathrm{by}\:\mathrm{setting} \\ $$$$\:\:\mathrm{a}_{\mathrm{1}\:} \:\in\:\left(\mathrm{0},\mathrm{1}\right)\:\:\:,\:\:\:\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} \left(\mathrm{1}−\mathrm{a}_{\mathrm{n}} \right)\:\:\:,\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\:\mathrm{prove}\:\mathrm{that}\:\:\left(\mathrm{1}\right)\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{na}_{\mathrm{n}} =\:\mathrm{1} \\ $$$$\:\:\left(\mathrm{2}\right)\:\:\mathrm{b}_{\mathrm{n}} \:=\:\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{incresing}\:\mathrm{sequence} \\ $$$$\:\:\:\mathrm{and}\:\mathrm{diverge}\:\mathrm{to}\:\infty \\ $$$$\:\:\:\left(\mathrm{3}\right)\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{n}\left(\mathrm{1}−\mathrm{na}_{\mathrm{n}} \right)}{\mathrm{ln}\left(\mathrm{n}\right)}\:=\:\mathrm{1} \\ $$

Question Number 219408 Answers: 2 Comments: 0

a_1 =(1/2). a_n =(2/(n+3))+(1−(1/(n+3)))^2 a_(n−1) . Find the maximum of a_n .

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}.\:{a}_{{n}} =\frac{\mathrm{2}}{{n}+\mathrm{3}}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{3}}\right)^{\mathrm{2}} {a}_{{n}−\mathrm{1}} . \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{a}_{{n}} . \\ $$

Question Number 219404 Answers: 1 Comments: 2

given g(x)= ((x−2023)/(x−1)) find (gogogogogog)(2024)

$$\:\:\mathrm{given}\:\mathrm{g}\left(\mathrm{x}\right)=\:\frac{\mathrm{x}−\mathrm{2023}}{\mathrm{x}−\mathrm{1}} \\ $$$$\:\:\mathrm{find}\:\left(\mathrm{gogogogogog}\right)\left(\mathrm{2024}\right) \\ $$

Question Number 219403 Answers: 1 Comments: 0

Question Number 219388 Answers: 1 Comments: 0

∫_0 ^( ∞) ((sin^2 (u))/u^2 ) du=I I(t)=∫_0 ^( ∞) ((sin^2 (u))/u^2 )e^(−ut) du=∫_0 ^( ∞) (1/u)∙((sin^2 (u))/u)e^(−ut) du= ∫_( t) ^( ∞) L_u {((sin^2 (p))/p)} du=∫_( t) ^( ∞) ∫_0 ^( ∞) ((sin^2 (p))/p)e^(−up) dpdu ∫_( t) ^( ∞) ∫_( w) ^( ∞) L_s {sin^2 (r)}ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) ∫_0 ^( ∞) sin^2 (r)e^(−sr) dr ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) (2/(s(s^2 +4))) ds dw=∫_( t) ^( ∞) [(1/2)ln(s)−(1/4)ln(s^2 +4)]_(s=w) ^(s=∞) dw ∫_( t) ^( ∞) (1/4)ln((w^2 /(w^2 +4)))dw=[(1/4)w∙ln(1+((2/w))^2 )+tan^(−1) ((1/2)w)]_(w=t) ^(w=∞) (π/2)−(1/4)t∙ln(1+((2/t))^2 )−tan^(−1) ((1/2)t) lim_(t→0) {tan^(−1) ((2/t))−(1/4)t∙ln(1+((2/t))^2 )}=(π/2)

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }{e}^{−{ut}} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{{u}}\centerdot\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}}{e}^{−{ut}} \mathrm{d}{u}= \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\mathcal{L}_{{u}} \left\{\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}\right\}\:\mathrm{d}{u}=\int_{\:{t}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}{e}^{−{up}} \mathrm{d}{p}\mathrm{d}{u} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\int_{\:{w}} ^{\:\infty} \:\mathcal{L}_{{s}} \left\{\mathrm{sin}^{\mathrm{2}} \left({r}\right)\right\}\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \left({r}\right){e}^{−{sr}} \mathrm{d}{r}\:\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \:\:\frac{\mathrm{2}}{{s}\left({s}^{\mathrm{2}} +\mathrm{4}\right)}\:\mathrm{d}{s}\:\mathrm{d}{w}=\int_{\:{t}} ^{\:\infty} \:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({s}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left({s}^{\mathrm{2}} +\mathrm{4}\right)\right]_{{s}={w}} ^{{s}=\infty} \mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} +\mathrm{4}}\right)\mathrm{d}{w}=\left[\frac{\mathrm{1}}{\mathrm{4}}{w}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{w}}\right)^{\mathrm{2}} \right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{w}\right)\right]_{{w}={t}} ^{{w}=\infty} \\ $$$$\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{t}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)\right\}=\frac{\pi}{\mathrm{2}} \\ $$

Question Number 219384 Answers: 1 Comments: 0

Question Number 219377 Answers: 1 Comments: 0

If ((fog)^(−1) of)(x)= 3x−8 find g(5).

$$\:\mathrm{If}\:\left(\left(\mathrm{fog}\right)^{−\mathrm{1}} \mathrm{of}\right)\left(\mathrm{x}\right)=\:\mathrm{3x}−\mathrm{8} \\ $$$$\:\mathrm{find}\:\mathrm{g}\left(\mathrm{5}\right). \\ $$

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