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Evaluate : 1) ∫_(−2) ^( 2) ∫_(−(√(4−x^2 ))) ^( (√(4−x^2 ))) (3−x)dydx . (after changing the integral to polar form). 2) ∫_0 ^4 ∫_0 ^(4−x) ∫_0 ^( 4−(y^2 /4)) dzdydx .

$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\int_{−\mathrm{2}} ^{\:\mathrm{2}} \int_{−\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} ^{\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \:\left(\mathrm{3}−{x}\right){dydx}\:. \\ $$$$\left({after}\:{changing}\:{the}\:{integral}\:{to}\:{polar}\:{form}\right). \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{4}} \int_{\mathrm{0}} ^{\mathrm{4}−{x}} \int_{\mathrm{0}} ^{\:\mathrm{4}−\frac{{y}^{\mathrm{2}} }{\mathrm{4}}} \:{dzdydx}\:. \\ $$

Question Number 73429 Answers: 1 Comments: 3

Solve : ∫(([cos^(−1) x(√(1−x^2 ))]^(−1) )/(log_e [2+((sin(2x(√(1−x^2 ))))/π)]))dx Evaluate ∫_(−π/2) ^( π/2) sin^2 xcos^2 x(cosx+sinx)dx

$$\:\:\:{Solve}\::\:\int\frac{\left[{cos}^{−\mathrm{1}} {x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right]^{−\mathrm{1}} }{{log}_{{e}} \left[\mathrm{2}+\frac{{sin}\left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{\pi}\right]}{dx} \\ $$$$\:\:{Evaluate}\:\:\int_{−\pi/\mathrm{2}} ^{\:\pi/\mathrm{2}} {sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\left({cosx}+{sinx}\right){dx} \\ $$

Question Number 73411 Answers: 2 Comments: 2

calculate 1)cos(1+i) , sin(1+3i) 2) arctan(i), arctan(2i) , arctan(1+i) ,arctan(1−i) , arctan(1+2i). 3) have us conj(arctanz)=arctan(z^− )?

$${calculate}\:\: \\ $$$$\left.\mathrm{1}\right){cos}\left(\mathrm{1}+{i}\right)\:,\:{sin}\left(\mathrm{1}+\mathrm{3}{i}\right) \\ $$$$\left.\mathrm{2}\right)\:{arctan}\left({i}\right),\:{arctan}\left(\mathrm{2}{i}\right)\:,\:{arctan}\left(\mathrm{1}+{i}\right)\:,{arctan}\left(\mathrm{1}−{i}\right)\:, \\ $$$${arctan}\left(\mathrm{1}+\mathrm{2}{i}\right). \\ $$$$\left.\mathrm{3}\right)\:{have}\:{us}\:\:{conj}\left({arctanz}\right)={arctan}\left(\overset{−} {{z}}\right)? \\ $$

Question Number 73406 Answers: 0 Comments: 0

Use the Sandwich( Pinchin or Squeez ) theorem to prove that Lim_(x→a) (√x) = (√a)

$${Use}\:{the}\:{Sandwich}\left(\:{Pinchin}\:{or}\:{Squeez}\:\right)\:{theorem}\:{to}\:{prove} \\ $$$${that}\: \\ $$$$\:\underset{{x}\rightarrow{a}} {\mathrm{Lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$

Question Number 73405 Answers: 0 Comments: 0

can someone please prove the Chinese Remainder theorem, for modula arithmetic?

$${can}\:{someone}\:{please}\:{prove}\:{the}\: \\ $$$${Chinese}\:{Remainder}\:{theorem},\:{for}\: \\ $$$${modula}\:{arithmetic}? \\ $$

Question Number 73399 Answers: 3 Comments: 1

{ ((x^2 +y^2 =65)),(((x−1)(y−1)=17)) :} please help me to solve it...

$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{65}}\\{\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)=\mathrm{17}}\end{cases} \\ $$$$ \\ $$$${please}\:{help}\:{me}\:{to}\:{solve}\:{it}... \\ $$

Question Number 73397 Answers: 1 Comments: 1

find f(x)=∫_0 ^1 e^(−t) ln(1−xt^2 )dt with ∣x∣<1 2)calculate ∫_0 ^1 e^(−t) ln(1−(t^2 /2))dt

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

Question Number 73396 Answers: 1 Comments: 1

find ∫_0 ^1 e^(−t^2 ) ln(1−t)dt

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}−{t}\right){dt} \\ $$

Question Number 73378 Answers: 0 Comments: 3

Hello ,i shar withe you nice problem show that ∀k∈N^∗ ∃n∈N such that k≤Σ_(j=1) ^n (1/j)<k+1 have a very Nice day

$${Hello}\:,{i}\:{shar}\:{withe}\:{you}\:{nice}\:{problem}\: \\ $$$${show}\:{that}\:\forall{k}\in\mathbb{N}^{\ast} \:\exists{n}\in\mathbb{N}\:{such}\:{that} \\ $$$${k}\leqslant\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{j}}<{k}+\mathrm{1} \\ $$$${have}\:{a}\:{very}\:{Nice}\:{day} \\ $$$$ \\ $$

Question Number 73358 Answers: 0 Comments: 1

1/4x2−1/2x−13=0

$$\mathrm{1}/\mathrm{4}{x}\mathrm{2}−\mathrm{1}/\mathrm{2}{x}−\mathrm{13}=\mathrm{0} \\ $$

Question Number 73356 Answers: 0 Comments: 2

Question Number 73347 Answers: 0 Comments: 0

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