Question and Answers Forum

AllQuestion and Answers: Page 1752

Question Number 27691 Answers: 0 Comments: 1

let give A=∫∫_(0≤y≤x≤1) ((dxdxy)/((1+x^2 )(1+y^2 ))) and B= ∫_0 ^(π/4) ((ln(2cos^2 θ))/(2cos(2θ)))dθ calculate A and prove that B=A.

$${let}\:{give}\:\:{A}=\int\int_{\mathrm{0}\leqslant{y}\leqslant{x}\leqslant\mathrm{1}} \:\:\:\:\:\:\frac{{dxdxy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:\:{and} \\ $$$${B}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta\right)}{\mathrm{2}{cos}\left(\mathrm{2}\theta\right)}{d}\theta\:\:{calculate}\:{A}\:{and}\:{prove}\:{that}\:{B}={A}. \\ $$

Question Number 27690 Answers: 0 Comments: 1

find I= ∫∫_D ln(1+x+y)dxdy with D= {(x,y)∈R^2 / x+y≤1 and x≥0 and y≥0 }.

$${find}\:\:\:{I}=\:\:\int\int_{{D}} {ln}\left(\mathrm{1}+{x}+{y}\right){dxdy}\:\:{with} \\ $$$${D}=\:\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} \:\:\:/\:\:{x}+{y}\leqslant\mathrm{1}\:{and}\:{x}\geqslant\mathrm{0}\:{and}\:{y}\geqslant\mathrm{0}\:\right\}. \\ $$

Question Number 27757 Answers: 1 Comments: 0

calculate I= ∫_0 ^(π/2) (dx/(1+cosx)) and J= ∫_0^ ^(π/2) ((cosx)/(1+cosx))dx .

$${calculate}\:\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+{cosx}}\:{and}\:{J}=\:\int_{\mathrm{0}^{} } ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{\mathrm{1}+{cosx}}{dx}\:. \\ $$

Question Number 27688 Answers: 0 Comments: 0

Write the series,indicating the 5th term,the 5th partial sum 0+1+3+...+(((n^2 +n)/2))+...

$${Write}\:{the}\:{series},{indicating}\:{the} \\ $$$$\mathrm{5}{th}\:{term},{the}\:\mathrm{5}{th}\:{partial}\:{sum} \\ $$$$ \\ $$$$\mathrm{0}+\mathrm{1}+\mathrm{3}+...+\left(\frac{{n}^{\mathrm{2}} +{n}}{\mathrm{2}}\right)+... \\ $$

Question Number 28038 Answers: 0 Comments: 0

let give f(x)=(√(x+y)) +1 and D={(x,y)∈R^2 / 0≤x≤1 and −1≤y≤1} find the value of ∫∫ f(x,y)dxdy .

$${let}\:{give}\:{f}\left({x}\right)=\sqrt{{x}+{y}}\:+\mathrm{1}\:\:{and}\:{D}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\right. \\ $$$$\left.{and}\:−\mathrm{1}\leqslant{y}\leqslant\mathrm{1}\right\}\:\:{find}\:{the}\:{value}\:{of}\:\:\int\int\:{f}\left({x},{y}\right){dxdy}\:. \\ $$

Question Number 27685 Answers: 0 Comments: 2

Write the first five series indicating the 5th term,5th partial sum Σ_(n=1) ^∞ t_n , where t_n = { ((1 for n=1)),(((1/2) for n=2)),((1−(1/2)+...+(−1)^(n+1) ((1/n)) for n>2)) :}

$${Write}\:{the}\:{first}\:{five}\:{series}\:{indicating} \\ $$$${the}\:\mathrm{5}{th}\:{term},\mathrm{5}{th}\:{partial}\:{sum} \\ $$$$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{t}_{{n}} ,\:{where} \\ $$$${t}_{{n}} =\begin{cases}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{for}\:{n}=\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{for}\:{n}=\mathrm{2}}\\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+...+\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left(\frac{\mathrm{1}}{{n}}\right)\:\:\:{for}\:\:\:{n}>\mathrm{2}}\end{cases} \\ $$

Question Number 27684 Answers: 0 Comments: 1

1) prove the existence of the integral I=∫_0 ^(π/2) ((ln(1+cosx))/(cosx))dx 2)prove that I= ∫∫_D ((siny)/(1+cosx cosy))dxdy with D=[0,(π/2)]^2 3)find the value of I.

$$\left.\mathrm{1}\right)\:{prove}\:{the}\:{existence}\:{of}\:{the}\:{integral} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}}{dx} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{I}=\:\int\int_{{D}} \:\:\frac{{siny}}{\mathrm{1}+{cosx}\:{cosy}}{dxdy}\:{with}\: \\ $$$${D}=\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{I}. \\ $$

Question Number 27706 Answers: 0 Comments: 0

the number of xε[0,2π]for which∣(√)2sin^(4 ) x+18cos^2 x−(√(2cos^4 x+18sin^2 x))∣=1

$${the}\:{number}\:{of}\:{x}\epsilon\left[\mathrm{0},\mathrm{2}\pi\right]{for}\:{which}\mid\sqrt{}\mathrm{2}{sin}^{\mathrm{4}\:} {x}+\mathrm{18}{cos}^{\mathrm{2}} {x}−\sqrt{\mathrm{2}{cos}^{\mathrm{4}} {x}+\mathrm{18}{sin}^{\mathrm{2}} {x}}\mid=\mathrm{1} \\ $$

Question Number 27677 Answers: 1 Comments: 4