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AllQuestion and Answers: Page 62

Question Number 218673    Answers: 3   Comments: 1

Question Number 218672    Answers: 2   Comments: 0

Question Number 218662    Answers: 4   Comments: 0

Prove: ∫^∞ _0 ((sin(x))/x) dx = (π/2)

$$\: \\ $$$$\:\:\:\:{Prove}:\:\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{sin}\left({x}\right)}{{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 218658    Answers: 2   Comments: 0

Question Number 218769    Answers: 1   Comments: 0

Fourier Series f(θ)=e^(izsin(θ))

$$\mathrm{Fourier}\:\mathrm{Series}\:{f}\left(\theta\right)={e}^{\boldsymbol{{i}}{z}\mathrm{sin}\left(\theta\right)} \\ $$

Question Number 218656    Answers: 2   Comments: 0

resolve the equation with unknow p P is polynom 1) P(X^2 )=(X^2 +1)P(X) 2) P 0P =P

$${resolve}\:{the}\:{equation}\:{with}\:{unknow}\:{p} \\ $$$${P}\:\:{is}\:{polynom}\: \\ $$$$\left.\mathrm{1}\right)\:{P}\left({X}^{\mathrm{2}} \right)=\left({X}^{\mathrm{2}} +\mathrm{1}\right){P}\left({X}\right) \\ $$$$\left.\mathrm{2}\right)\:{P}\:\mathrm{0}{P}\:={P} \\ $$

Question Number 218652    Answers: 0   Comments: 4

Angle of incidence is 40° if the direction of the incidence ray is constant and the mirror is rotated through 20° the angle between the new reflected ray and new normal is

Angle of incidence is 40° if the direction of the incidence ray is constant and the mirror is rotated through 20° the angle between the new reflected ray and new normal is

Question Number 218651    Answers: 1   Comments: 0

Question Number 218650    Answers: 1   Comments: 0

Question Number 218649    Answers: 2   Comments: 0

Question Number 218648    Answers: 1   Comments: 0

Question Number 218646    Answers: 1   Comments: 1

Question Number 218645    Answers: 4   Comments: 0

Question Number 218644    Answers: 2   Comments: 0

Question Number 218636    Answers: 1   Comments: 0

Question Number 218629    Answers: 1   Comments: 0

Question Number 218626    Answers: 0   Comments: 0

Prove that for all real numbers a and b with a<b, the following inequality holds; (∫_a ^b 1 dx)^3 ≤ (b−a)(∫_a ^b (x−a+1)^2 dx)(∫_(a ) ^b (1/((a−x+1)^3 ))dx)

$$ \\ $$$$\:{Prove}\:{that}\:{for}\:{all}\:{real}\:{numbers}\:{a}\:{and}\:{b} \\ $$$${with}\:{a}<{b},\:{the}\:{following}\:{inequality}\:{holds}; \\ $$$$\left(\int_{{a}} ^{{b}} \mathrm{1}\:{dx}\right)^{\mathrm{3}} \leqslant\:\left({b}−{a}\right)\left(\int_{{a}} ^{{b}} \left({x}−{a}+\mathrm{1}\right)^{\mathrm{2}} {dx}\right)\left(\int_{{a}\:} ^{{b}} \frac{\mathrm{1}}{\left({a}−{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}\right) \\ $$$$ \\ $$

Question Number 218624    Answers: 1   Comments: 0

Prove; ∫^e _(1/e) (t^2 /e^t^(2 ) ) dt ≤ 1− (1/e^(2 ) ) e−the base of natural logarithm

$$ \\ $$$$\:\:\:\:{Prove};\:\underset{\frac{\mathrm{1}}{{e}}} {\int}^{{e}} \:\frac{{t}^{\mathrm{2}} }{{e}^{{t}^{\mathrm{2}\:} } }\:{dt}\:\leqslant\:\mathrm{1}−\:\frac{\mathrm{1}}{{e}^{\mathrm{2}\:\:} } \\ $$$$\:\:\:{e}−{the}\:{base}\:{of}\:{natural}\:{logarithm} \\ $$$$ \\ $$

Question Number 219017    Answers: 0   Comments: 0

Question Number 218609    Answers: 1   Comments: 0

Can y=x be expressed as them su of two periodic functions?

$$\mathrm{Can}\:{y}={x}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{them} \\ $$$$\mathrm{su}\:\mathrm{of}\:\mathrm{two}\:\mathrm{periodic}\:\mathrm{functions}? \\ $$

Question Number 218607    Answers: 3   Comments: 0

Question Number 218606    Answers: 3   Comments: 0

Question Number 218605    Answers: 1   Comments: 0

Question Number 218603    Answers: 0   Comments: 3

Question Number 218599    Answers: 0   Comments: 2

∫_0 ^∞ x^(s−1) Π_(n=1 ) ^∞ (1−e^(−nx) )^(−24) dx

$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \boldsymbol{{x}}^{\boldsymbol{{s}}−\mathrm{1}} \:\underset{\boldsymbol{{n}}=\mathrm{1}\:} {\overset{\infty} {\prod}}\left(\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{nx}}} \right)^{−\mathrm{24}} \:\boldsymbol{{dx}} \\ $$$$ \\ $$

Question Number 218598    Answers: 0   Comments: 0

∫^∞ _0 (x/(sinh(x)))ln((x/(e^x −1)))dx

$$ \\ $$$$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{\boldsymbol{{x}}}{\boldsymbol{{sinh}}\left(\boldsymbol{{x}}\right)}\boldsymbol{{ln}}\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}\right)\boldsymbol{{dx}}\: \\ $$$$ \\ $$

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