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Question Number 38466    Answers: 0   Comments: 1

let a from R find F_a (t)= ∫_(−∞) ^(+∞) ((cos(tx))/(a^2 +x^2 ))dx 2) calculate F_2 (3) and F_3 (2)

$${let}\:{a}\:{from}\:{R}\:\:{find}\:{F}_{{a}} \left({t}\right)=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{F}_{\mathrm{2}} \left(\mathrm{3}\right)\:\:{and}\:{F}_{\mathrm{3}} \left(\mathrm{2}\right) \\ $$

Question Number 38465    Answers: 0   Comments: 2

find f(x)= ∫_0 ^1 ln(1+xt^3 )dt with ∣x∣<1 . 2) calculate ∫_0 ^1 ln(1+4t^3 )dt and ∫_0 ^1 ln(2+t^3 )dt.

$${find}\:\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{3}} \right){dt}\:{with}\:\mid{x}\mid<\mathrm{1}\:. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\mathrm{4}{t}^{\mathrm{3}} \right){dt}\:\:\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{t}^{\mathrm{3}} \right){dt}. \\ $$

Question Number 38464    Answers: 0   Comments: 1

let f(x)= ∫_0 ^1 ((ln(1−x^2 t^2 ))/t^2 )dt with ∣x∣<1 find f(x) at a simple form .

$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${find}\:\:{f}\left({x}\right)\:{at}\:{a}\:{simple}\:{form}\:. \\ $$

Question Number 38463    Answers: 0   Comments: 1

calculate I = ∫_0 ^1 ((ln (1−(t^2 /4)))/t^2 )dt

$${calculate}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\:\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\right)}{{t}^{\mathrm{2}} }{dt}\: \\ $$

Question Number 38462    Answers: 0   Comments: 0

find ∫_1 ^(+∞) arctan(x −(1/x))dx

$${find}\:\:\int_{\mathrm{1}} ^{+\infty} {arctan}\left({x}\:−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$

Question Number 38461    Answers: 0   Comments: 0

find f(x) = ∫_0 ^∞ arctan(1+e^(−xt) )dt with x>0 2) find ∫_0 ^∞ arctan(1+e^(−2t) )dt.

$${find}\:\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:{arctan}\left(\mathrm{1}+{e}^{−{xt}} \right){dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:{arctan}\left(\mathrm{1}+{e}^{−\mathrm{2}{t}} \right){dt}. \\ $$

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)=?

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

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