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Question Number 38453    Answers: 0   Comments: 4

find f(x)=∫_0 ^∞ ((1−cos(xt))/t) e^(−xt) dt with x>0 1) find asimple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(πt))/t) e^(−t) dt 3)calculate ∫_0 ^∞ ((1−cos(3t))/t) e^(−2t) dt

$${find}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}\:{e}^{−{xt}} {dt}\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{asimple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{t}\right)}{{t}}\:{e}^{−\mathrm{2}{t}} {dt} \\ $$

Question Number 39416    Answers: 1   Comments: 0

Given that f(x) is a cubic function and f(x) = x^3 − (x^2 /4) + 5x − 7 a) find one factor of f(x) b) find (d^2 y/dx^2 ) for f(x) c) hence Evaluate y = ∫_0 ^∞ f(x).

$${Given}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{cubic}\:{function} \\ $$$${and}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\mathrm{5}{x}\:−\:\mathrm{7} \\ $$$$\left.{a}\right)\:{find}\:{one}\:{factor}\:{of}\:{f}\left({x}\right) \\ $$$$\left.{b}\right)\:{find}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{for}\:{f}\left({x}\right) \\ $$$$\left.{c}\right)\:{hence}\:{Evaluate}\:\:{y}\:=\:\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right). \\ $$

Question Number 38451    Answers: 1   Comments: 1

new attempt to solve qu. 37630 ∫(dx/((√x)+(√(x+1))+(√(x+2))))= [t=x+1 → dx=dt] =∫(dt/((√(t−1))+(√t)+(√(t+1))))= [((to omit the roots)),(((√a)+(√b)+(√c) must be multiplied with)),(((−(√a)−(√b)+(√c))(−(√a)+(√b)−(√c))((√a)−(√b)−(√c)))),(((1/((√a)+(√b)+(√c)))=((a^(3/2) +b^(3/2) +c^(3/2) +2(√(abc))−((a+b)(√c)+(a+c)(√b)+(b+c)(√a)))/(a^2 +b^2 +c^2 −2(ab+ac+bc))))) ] =∫((t(√(t−1))+t(√t)+t(√(t+1))+2(√(t−1))−2(√(t+1))−2(√((t−1)t(t+1))))/(3t^2 −4))dt= =∫((t(√(t−1)))/(3t^2 −4))dt+∫((t(√t))/(3t^2 −4))dt+∫((t(√(t+1)))/(3t^2 −4))dt+2∫((√(t−1))/(3t^2 −4))dt−2∫((√(t+1))/(3t^2 −4))−2∫((√((t−1)t(t+1)))/(3t^2 −4))dt I think I can solve them all except the last one so please somebody try ∫((√((t−1)t(t+1)))/(3t^2 −4))dt=? I will do the others tomorrow

$$\mathrm{new}\:\mathrm{attempt}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{qu}.\:\mathrm{37630} \\ $$$$ \\ $$$$\int\frac{{dx}}{\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\mathrm{1}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\int\frac{{dt}}{\sqrt{{t}−\mathrm{1}}+\sqrt{{t}}+\sqrt{{t}+\mathrm{1}}}= \\ $$$$ \\ $$$$\:\:\:\:\:\begin{bmatrix}{\mathrm{to}\:\mathrm{omit}\:\mathrm{the}\:\mathrm{roots}}\\{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\:\mathrm{must}\:\mathrm{be}\:\mathrm{multiplied}\:\mathrm{with}}\\{\left(−\sqrt{{a}}−\sqrt{{b}}+\sqrt{{c}}\right)\left(−\sqrt{{a}}+\sqrt{{b}}−\sqrt{{c}}\right)\left(\sqrt{{a}}−\sqrt{{b}}−\sqrt{{c}}\right)}\\{\frac{\mathrm{1}}{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}=\frac{{a}^{\mathrm{3}/\mathrm{2}} +{b}^{\mathrm{3}/\mathrm{2}} +{c}^{\mathrm{3}/\mathrm{2}} +\mathrm{2}\sqrt{{abc}}−\left(\left({a}+{b}\right)\sqrt{{c}}+\left({a}+{c}\right)\sqrt{{b}}+\left({b}+{c}\right)\sqrt{{a}}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)}}\end{bmatrix} \\ $$$$ \\ $$$$=\int\frac{{t}\sqrt{{t}−\mathrm{1}}+{t}\sqrt{{t}}+{t}\sqrt{{t}+\mathrm{1}}+\mathrm{2}\sqrt{{t}−\mathrm{1}}−\mathrm{2}\sqrt{{t}+\mathrm{1}}−\mathrm{2}\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}= \\ $$$$=\int\frac{{t}\sqrt{{t}−\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\int\frac{{t}\sqrt{{t}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\int\frac{{t}\sqrt{{t}+\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\mathrm{2}\int\frac{\sqrt{{t}−\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}−\mathrm{2}\int\frac{\sqrt{{t}+\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}−\mathrm{2}\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{them}\:\mathrm{all}\:\mathrm{except}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\ $$$$\mathrm{so}\:\mathrm{please}\:\mathrm{somebody}\:\mathrm{try} \\ $$$$\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}=? \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{do}\:\mathrm{the}\:\mathrm{others}\:\mathrm{tomorrow} \\ $$

Question Number 38460    Answers: 0   Comments: 1

let ∣x∣>1 find the value of F(x)=∫_0 ^∞ ln(1+xt^2 )dt 2)calculate ∫_0 ^∞ ln(1+3t^2 )dt .

$${let}\:\mid{x}\mid>\mathrm{1}\:{find}\:{the}\:{value}\:{of} \\ $$$${F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} \right){dt}\:. \\ $$

Question Number 38459    Answers: 0   Comments: 7

x^2 (x−b^2 )+a^2 b^2 (x−a^2 )=0 Solve for x.

$${x}^{\mathrm{2}} \left({x}−{b}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({x}−{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${Solve}\:{for}\:{x}. \\ $$

Question Number 38458    Answers: 0   Comments: 3

let ∣x∣<1 calculate F(x)=∫_0 ^1 ln(1+xt^2 )dt 2) find the value of ∫_0 ^1 ln(1 +(1/2)t^2 )dt 3)find the value of A(θ) =∫_0 ^1 ln(1+sinθ t^2 )dt .

$${let}\:\mid{x}\mid<\mathrm{1}\:\:{calculate}\:\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{sin}\theta\:{t}^{\mathrm{2}} \right){dt}\:. \\ $$

Question Number 38457    Answers: 0   Comments: 0

f is a C^2 function prove that 1)L(f^′ )=x L(f)−f(0^+ ) 2)L(f^(′′) )=x^2 L(f) −xf(0^+ )−f^′ (0^+ ) L means Laplace transform.

$${f}\:{is}\:{a}\:{C}^{\mathrm{2}} {function}\:\:{prove}\:{that}\: \\ $$$$\left.\mathrm{1}\right){L}\left({f}^{'} \right)={x}\:{L}\left({f}\right)−{f}\left(\mathrm{0}^{+} \right) \\ $$$$\left.\mathrm{2}\right){L}\left({f}^{''} \right)={x}^{\mathrm{2}} {L}\left({f}\right)\:−{xf}\left(\mathrm{0}^{+} \right)−{f}^{'} \left(\mathrm{0}^{+} \right) \\ $$$${L}\:{means}\:{Laplace}\:{transform}. \\ $$

Question Number 38456    Answers: 0   Comments: 0

calculate Σ_(n=1) ^∞ (−1)^n ((cos(nx))/n^2 ) and Σ_(n=1) ^∞ (−1)^n ((sin(nx))/n^2 )

$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }\:\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} } \\ $$

Question Number 38455    Answers: 0   Comments: 0

solve the d.e y^′ −xe^(−2x) y =cos(3x)

$${solve}\:{the}\:{d}.{e}\:\:{y}^{'} \:−{xe}^{−\mathrm{2}{x}} {y}\:={cos}\left(\mathrm{3}{x}\right) \\ $$

Question Number 38454    Answers: 0   Comments: 3

let f(x)=∫_0 ^∞ ((1−cos(xt^2 ))/t^2 ) e^(−xt^2 ) dt with x>0 1) find a simple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(2t^2 ))/t^2 ) e^(−3t^2 ) dt .

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({xt}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−{xt}^{\mathrm{2}} } {dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\:{e}^{−\mathrm{3}{t}^{\mathrm{2}} } {dt}\:. \\ $$

Question Number 38405    Answers: 1   Comments: 2

If f(3)=3;f(1)=2 ⇒∫_1 ^3 f(x)f^′ (x)dx=.....

$$\mathrm{If}\:\mathrm{f}\left(\mathrm{3}\right)=\mathrm{3};\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2}\: \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\mathrm{3}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{f}^{'} \left(\mathrm{x}\right)\mathrm{dx}=..... \\ $$

Question Number 38397    Answers: 2   Comments: 2

evaluate ∫secxdx

$${evaluate}\:\:\int{secxdx} \\ $$

Question Number 38395    Answers: 1   Comments: 1

Find roots of the equation z^2 +2(1+i)z +2=0 leaving your answer in a+ib

$${Find}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${z}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{1}+{i}\right){z}\:+\mathrm{2}=\mathrm{0} \\ $$$${leaving}\:{your}\:{answer}\:{in}\:{a}+{ib} \\ $$

Question Number 38391    Answers: 1   Comments: 0

a+2b+3c=12 2ab+3ac+6bc=48 a+b+c=...

$${a}+\mathrm{2}{b}+\mathrm{3}{c}=\mathrm{12} \\ $$$$\mathrm{2}{ab}+\mathrm{3}{ac}+\mathrm{6}{bc}=\mathrm{48} \\ $$$${a}+{b}+{c}=... \\ $$

Question Number 38420    Answers: 0   Comments: 0

In the figure below,a particle A of mass 2kg is lying on a rough wooden block.The particle A is connected by a light inextensible horizontal string passing over a smooth light fixed pulley at the edge of the block,to a particle B of mass 3kg which hangs freely. The coefficent of friction between the particle A and the surface of the block is μ. Given that the string is taut and the system is released from rest such that the particle move with an acceleration of 4ms^(−1) . Find a) the tension b) the value of μ.

$${In}\:{the}\:{figure}\:{below},{a}\:{particle}\:{A}\:{of} \\ $$$${mass}\:\mathrm{2}{kg}\:{is}\:{lying}\:{on}\:{a}\:{rough}\:{wooden} \\ $$$${block}.{The}\:{particle}\:{A}\:{is}\:{connected}\:{by} \\ $$$${a}\:{light}\:{inextensible}\:{horizontal}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{light}\:{fixed} \\ $$$${pulley}\:{at}\:{the}\:{edge}\:{of}\:{the}\:{block},{to}\:{a} \\ $$$${particle}\:{B}\:{of}\:{mass}\:\mathrm{3}{kg}\:{which}\:{hangs} \\ $$$${freely}.\:{The}\:{coefficent}\:{of}\:{friction} \\ $$$${between}\:{the}\:{particle}\:{A}\:{and}\:{the}\:{surface} \\ $$$${of}\:{the}\:{block}\:{is}\:\mu. \\ $$$${Given}\:{that}\:{the}\:{string}\:{is}\:{taut}\:{and}\:{the}\:{system}\:{is}\:{released}\:{from}\:{rest}\:\:{such}\:{that}\:{the}\:{particle}\:{move}\:{with}\:{an} \\ $$$${acceleration}\:{of}\:\mathrm{4}{ms}^{−\mathrm{1}} .\:{Find} \\ $$$$\left.{a}\right)\:{the}\:{tension} \\ $$$$\left.{b}\right)\:{the}\:{value}\:{of}\:\mu. \\ $$

Question Number 38419    Answers: 1   Comments: 0

Question Number 38384    Answers: 0   Comments: 1

∫x^(−3cosx) dx please is this possible? If possibld then prove it. Thanks in advance.

$$\int{x}^{−\mathrm{3}{cosx}} {dx} \\ $$$$ \\ $$$${please}\:{is}\:{this}\:{possible}? \\ $$$${If}\:{possibld}\:{then}\:{prove}\:{it}. \\ $$$$ \\ $$$${Thanks}\:{in}\:{advance}. \\ $$

Question Number 38376    Answers: 1   Comments: 2

Question Number 38373    Answers: 1   Comments: 7

Question Number 38367    Answers: 2   Comments: 2

If f(x)=x^3 +1 then f^(−1) (x)=?

$${If}\:{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1}\:{then}\:{f}^{−\mathrm{1}} \left({x}\right)=? \\ $$

Question Number 38366    Answers: 1   Comments: 0

At time t,the force acting on a particle P of mass 2kg is (2ti + 4j)N.P is initially at rest at the point with position vector (i + 2j). Find: a) the velocity of P when t = 2. b) the position vector when t = 2.

$${At}\:{time}\:{t},{the}\:{force}\:{acting}\:{on}\:{a}\:{particle} \\ $$$${P}\:{of}\:{mass}\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}\boldsymbol{{ti}}\:+\:\mathrm{4}\boldsymbol{{j}}\right){N}.{P} \\ $$$${is}\:{initially}\:{at}\:{rest}\:{at}\:{the}\:{point}\:{with} \\ $$$${position}\:{vector}\:\left(\boldsymbol{{i}}\:+\:\mathrm{2}\boldsymbol{{j}}\right). \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$$$\left.{b}\right)\:{the}\:{position}\:{vector}\:{when}\:{t}\:=\:\mathrm{2}. \\ $$

Question Number 38362    Answers: 0   Comments: 0

Question Number 38365    Answers: 1   Comments: 0

A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.

$${A}\:{particle}\:{P}\:{moves}\:{on}\:{a}\:{straightline} \\ $$$${from}\:{a}\:{fixed}\:{point}\:{O}\:{and}\:{the}\:{distance} \\ $$$${x}\:{from}\:{O}\:{after}\:{t}\:{seconds}\:{is}\:{given}\:{as} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}\:}\:{t}^{\mathrm{4}} \:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{t}^{\mathrm{2}} \:+\:\mathrm{2}{t}. \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{b}\right)\:{the}\:{acceleration}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{c}\right)\:{the}\:{time}\:{at}\:{which}\:{the}\:{speed}\:{P}\: \\ $$$${is}\:{Minimum}. \\ $$

Question Number 38332    Answers: 1   Comments: 0

((3+(√5))/2) −(√((3+(√5))/2)) = ?!

$$\:\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−\sqrt{\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\:=\:?! \\ $$

Question Number 38310    Answers: 1   Comments: 4

let f(x)=∫_0 ^(+∞) ((arctan(xt))/(1+t^2 ))dt with x≥0 1) calculate f^′ (x) then a simple form of f(x) 2) calculate ∫_0 ^(+∞) ((arctant)/(1+t^2 ))dt 3) calculate ∫_0 ^(+∞) ((arctan(2t))/(1+t^2 ))dt

$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctan}\left({xt}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{then}\:{a}\:{simple}\:{form}\:{of}\:\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left(\mathrm{2}{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$

Question Number 38323    Answers: 1   Comments: 1

find Σ_(n=1) ^(+∞) ((4n)/((2n−1)^2 (2n+1)^2 ))

$${find}\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{4}{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

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