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Question Number 46423 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((arctan(x^2 ))/(1+x^2 ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 46422 Answers: 0 Comments: 0

let f_n (x)=((√(1+x^2 ))−x)^(n ) with n integr natural 1) find ∫ f_n (x)dx 2) calculate f_n ^(−1) (x) 3) find lim_(x→+∞) f_n (x) 4) calculate f_n ^′ (x) and f_n ^′ (0) 5) give the equation of tangent at point E(0,f(0))

$${let}\:{f}_{{n}} \left({x}\right)=\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)^{{n}\:} \:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:\int\:\:{f}_{{n}} \left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}_{{n}} ^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{lim}_{{x}\rightarrow+\infty} \:{f}_{{n}} \left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:{f}_{{n}} ^{'} \left({x}\right)\:\:{and}\:\:{f}_{{n}} ^{'} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{5}\right)\:{give}\:{the}\:{equation}\:{of}\:{tangent}\:{at}\:{point}\:{E}\left(\mathrm{0},{f}\left(\mathrm{0}\right)\right) \\ $$

Question Number 46421 Answers: 0 Comments: 3

let f(x)=(x+(1/x))^n −(x−(1/x))^n with n integr natural and x from R (x≠0) 1) simplify f(x) 2) calculate lim_(x→+∞) f(x) 3) calculate ∫_1 ^3 f(x)dx

$${let}\:{f}\left({x}\right)=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{{n}} −\left({x}−\frac{\mathrm{1}}{{x}}\right)^{{n}} \:{with}\:{n}\:{integr}\:{natural}\:{and}\:{x}\:{from}\:{R}\:\left({x}\neq\mathrm{0}\right) \\ $$$$\left.\mathrm{1}\right)\:{simplify}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}\left({x}\right){dx} \\ $$

Question Number 46420 Answers: 0 Comments: 0

let p(x)=(1+ix)^5 −1 with i^2 =−1 1) solve inside C[x] the equation p(x)=0 2)factorize inside C[x] the polynom p(x)

$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{\mathrm{5}} −\mathrm{1}\:\:{with}\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\left[{x}\right]\:{the}\:{equation}\:{p}\left({x}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right){factorize}\:{inside}\:{C}\left[{x}\right]\:{the}\:{polynom}\:{p}\left({x}\right) \\ $$$$ \\ $$$$ \\ $$

Question Number 46419 Answers: 0 Comments: 4

let f(x)= (√(x+1−(√(x−1)))) 1) find D_f 2) determine f^(−1) (x) 3) find ∫ f(x)dx 4) dtetrmine ∫ f^(−1) (x) 5) let g(x)= (ch(x))^2 calculate fog(x) and (fog)^′ (x) .

$${let}\:{f}\left({x}\right)=\:\sqrt{{x}+\mathrm{1}−\sqrt{{x}−\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:\int\:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{dtetrmine}\:\int\:\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{let}\:{g}\left({x}\right)=\:\left({ch}\left({x}\right)\right)^{\mathrm{2}} \:\:\:{calculate}\:{fog}\left({x}\right)\:{and}\:\left({fog}\right)^{'} \left({x}\right)\:. \\ $$

Question Number 46418 Answers: 0 Comments: 0

find ∫∫_([0,1]×[0,1]) ((√(x+(√y)))−(√(y+(√x))))dxdy

$${find}\:\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right]} \left(\sqrt{{x}+\sqrt{{y}}}−\sqrt{{y}+\sqrt{{x}}}\right){dxdy} \\ $$

Question Number 46411 Answers: 1 Comments: 0

Question Number 46410 Answers: 0 Comments: 0

Question Number 46414 Answers: 2 Comments: 1

Question Number 46402 Answers: 1 Comments: 1

show that cos 7(1/2) =(√2)+(√3)+(√4)+(√6)

$${show}\:{that}\:\mathrm{cos}\:\mathrm{7}\frac{\mathrm{1}}{\mathrm{2}}\:=\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}+\sqrt{\mathrm{6}} \\ $$

Question Number 46401 Answers: 0 Comments: 1

Find the sum of the series: (1/(1 + (√x))) , (1/(1 − x)) , (1/(1 − (√x))) , ...

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\frac{\mathrm{1}}{\mathrm{1}\:+\:\sqrt{\mathrm{x}}}\:,\:\:\frac{\mathrm{1}}{\mathrm{1}\:−\:\mathrm{x}}\:,\:\:\frac{\mathrm{1}}{\mathrm{1}\:−\:\sqrt{\mathrm{x}}}\:\:,\:\:...\: \\ $$

Question Number 46383 Answers: 1 Comments: 3

Question Number 46382 Answers: 1 Comments: 0

Question Number 46378 Answers: 1 Comments: 0

Factorise : 3x^4 +6x^3 +8x^2 −2x−3=0.

$${Factorise}\:: \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}. \\ $$

Question Number 46374 Answers: 1 Comments: 1

At each of three corners of a square of side 2cm is placed a point charge of magnitude 3μC. What will be the magnitude and direction of the resultant force on a point charge −1μC if it were placed (a) at the centre of the square? (b) at the vacant corner of the sqre?

$${At}\:{each}\:{of}\:{three}\:{corners}\:{of}\:{a}\:{square}\: \\ $$$${of}\:{side}\:\mathrm{2}{cm}\:{is}\:{placed}\:{a}\:{point}\:{charge}\: \\ $$$${of}\:{magnitude}\:\mathrm{3}\mu{C}. \\ $$$${What}\:{will}\:{be}\:{the}\:{magnitude}\:{and}\: \\ $$$${direction}\:{of}\:{the}\:{resultant}\:{force}\:{on}\:{a} \\ $$$${point}\:{charge}\:−\mathrm{1}\mu{C}\:{if}\:{it}\:{were}\:{placed} \\ $$$$\:\:\:\left({a}\right)\:\:{at}\:{the}\:{centre}\:{of}\:{the}\:{square}? \\ $$$$\:\:\:\left({b}\right)\:\:{at}\:{the}\:{vacant}\:{corner}\:{of}\:{the}\:{sqre}? \\ $$

Question Number 46369 Answers: 2 Comments: 1

∫ (dx/(x(√(x^2 + 2x − 1)))) = tan^(−1) (((x − 1)/(√(x^2 + 2x − 1)))) + C I′m confused in choosing the right substitution so that we can get the result above. Please help...

$$\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}\:−\:\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{1}}}\right)\:+\:{C} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{confused}\:\mathrm{in}\:\mathrm{choosing}\:\mathrm{the}\:\mathrm{right}\:\mathrm{substitution} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result}\:\mathrm{above}. \\ $$$$\mathrm{Please}\:\mathrm{help}... \\ $$

Question Number 46395 Answers: 1 Comments: 0