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IntegrationQuestion and Answers: Page 23

Question Number 201184 Answers: 1 Comments: 0

Question Number 201044 Answers: 2 Comments: 0

Ω = ∫_0 ^( 1) ∫_0 ^( 1) (x−y )^2 sin^( 2) ( x+y )dxdy=?

$$ \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}−{y}\:\right)^{\mathrm{2}} {sin}^{\:\mathrm{2}} \:\left(\:{x}+{y}\:\right){dxdy}=? \\ $$

Question Number 201016 Answers: 0 Comments: 0

Question Number 201011 Answers: 0 Comments: 0

Prove that ∫_0 ^∞ ((2arctan((t/x)))/(e^(2𝛑t) −1))dt=In𝚪(x)−xIn(x)+x−(1/2)In(((2𝛑)/x)) Michael faraday

$$\boldsymbol{{Prove}}\:\boldsymbol{{that}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{t}}}{\boldsymbol{{x}}}\right)}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi{t}}} −\mathrm{1}}\boldsymbol{{dt}}=\boldsymbol{{In}\Gamma}\left(\boldsymbol{{x}}\right)−\boldsymbol{{xIn}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{In}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\boldsymbol{{x}}}\right) \\ $$$$\boldsymbol{{Michael}}\:\boldsymbol{{faraday}} \\ $$

Question Number 200933 Answers: 0 Comments: 0

∫coth (ln [(√(tanh (ln ((√(sec^(−1) (x)^(1/4) )))))) ])

$$ \\ $$$$\int\mathrm{coth}\:\left(\mathrm{ln}\:\left[\sqrt{\mathrm{tanh}\:\left(\mathrm{ln}\:\left(\sqrt{\mathrm{sec}^{−\mathrm{1}} \:\:\sqrt[{\mathrm{4}}]{{x}}\:\:}\right)\right)}\:\right]\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 200923 Answers: 0 Comments: 0

Question Number 203714 Answers: 1 Comments: 0

∫2x^2

$$\int\mathrm{2}{x}^{\mathrm{2}} \\ $$

Question Number 200915 Answers: 1 Comments: 0

Question Number 200901 Answers: 0 Comments: 0

Question Number 200930 Answers: 0 Comments: 1

If I_n =∫_0 ^1 (1−x^4 )^n dx and (I_n /I_(n−1) )=((λn)/(λn+1)) then find λ

$${If}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{{n}} {dx}\:\:{and}\:\:\frac{{I}_{{n}} }{{I}_{{n}−\mathrm{1}} }=\frac{\lambda{n}}{\lambda{n}+\mathrm{1}} \\ $$$${then}\:{find}\:\:\lambda \\ $$

Question Number 200844 Answers: 2 Comments: 2

Question Number 200802 Answers: 1 Comments: 0

Question Number 200801 Answers: 1 Comments: 0

Question Number 200748 Answers: 2 Comments: 0

Question Number 200747 Answers: 2 Comments: 0

Question Number 200737 Answers: 0 Comments: 0

Solve: The position vector of a particle at any time t is given by r_− =(acoswt)i+(asinwt)j+bt^2 k (a) show that,although the speed of the particle increases with time,the magnitude of the acceleration is always constant (b) describe the motion of the particle geometrically

$$\boldsymbol{{Solve}}:\:\boldsymbol{{The}}\:\boldsymbol{{position}}\:\boldsymbol{{vector}}\:\boldsymbol{{of}}\:\boldsymbol{{a}}\:\boldsymbol{{particle}}\:\boldsymbol{{at}}\:\boldsymbol{{any}}\:\boldsymbol{{time}}\:\boldsymbol{{t}} \\ $$$$\boldsymbol{{is}}\:\boldsymbol{{given}}\:\boldsymbol{{by}}\:\:\underset{−} {\boldsymbol{{r}}}=\left(\boldsymbol{{acoswt}}\right)\boldsymbol{{i}}+\left(\boldsymbol{{asinwt}}\right)\boldsymbol{{j}}+\boldsymbol{{bt}}^{\mathrm{2}} \boldsymbol{{k}} \\ $$$$\left(\boldsymbol{{a}}\right)\:\boldsymbol{{show}}\:\boldsymbol{{that}},\boldsymbol{{although}}\:\boldsymbol{{the}}\:\boldsymbol{{speed}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{particle}} \\ $$$$\boldsymbol{{increases}}\:\boldsymbol{{with}}\:\boldsymbol{{time}},\boldsymbol{{the}}\:\boldsymbol{{magnitude}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{acceleration}}\:\boldsymbol{{is}}\:\boldsymbol{{always}}\:\boldsymbol{{constant}} \\ $$$$\left(\boldsymbol{{b}}\right)\:\boldsymbol{{describe}}\:\boldsymbol{{the}}\:\boldsymbol{{motion}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{particle}}\:\boldsymbol{{geometrically}} \\ $$

Question Number 200736 Answers: 0 Comments: 0

Solve: A particle moves along the space curve r_− =(t^2 +t)i+(3t−2)j+(2t^3 −4t^2 )k. find (a)velocity (b)speed or magnitude of velocity (c)acceleration (d)magnitude of acceleration at time t=2

$$\boldsymbol{{Solve}}:\:\boldsymbol{{A}}\:\boldsymbol{{particle}}\:\boldsymbol{{moves}}\:\boldsymbol{{along}}\:\boldsymbol{{the}}\:\boldsymbol{{space}} \\ $$$$\boldsymbol{{curve}}\:\underset{−} {\boldsymbol{{r}}}=\left(\boldsymbol{{t}}^{\mathrm{2}} +\boldsymbol{{t}}\right)\boldsymbol{{i}}+\left(\mathrm{3}\boldsymbol{{t}}−\mathrm{2}\right)\boldsymbol{{j}}+\left(\mathrm{2}\boldsymbol{{t}}^{\mathrm{3}} −\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} \right)\boldsymbol{{k}}. \\ $$$$\boldsymbol{{find}} \\ $$$$\left(\boldsymbol{{a}}\right)\boldsymbol{{velocity}} \\ $$$$\left(\boldsymbol{{b}}\right)\boldsymbol{{speed}}\:\boldsymbol{{or}}\:\boldsymbol{{magnitude}}\:\boldsymbol{{of}}\:\boldsymbol{{velocity}} \\ $$$$\left(\boldsymbol{{c}}\right)\boldsymbol{{acceleration}} \\ $$$$\left(\boldsymbol{{d}}\right)\boldsymbol{{magnitude}}\:\boldsymbol{{of}}\:\boldsymbol{{acceleration}}\:\boldsymbol{{at}}\:\boldsymbol{{time}}\:\boldsymbol{{t}}=\mathrm{2} \\ $$

Question Number 200697 Answers: 1 Comments: 0