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Question Number 36217 Answers: 1 Comments: 0

Suppose a_1 ,...,a_n ,are non−negative reals such that S= a_1 +...+a_n < proof that 1 + S≤ (1 + a_1 )._(...) .(1+ a_n ) ≤ (1/(1−s))

$Suppose a_{1}, . . ., a_{n}, are non - negative$ $reals such that S = a_{1} + . . . + a_{n} <$ $p r o o f t h a t$ $1 + S ⩽ (1 + a_{1}) ._{. . .} . (1 + a_{n}) ⩽ \frac{1}{1 - s}$

Question Number 36207 Answers: 0 Comments: 4

x^4 +10x^3 +6x−1=0 for those who want an exact solution... (x−a−bi)(x−a+bi)(x−c−d)(x−c+d)=0 x^4 −2(a+c)x^3 +(a^2 +4ac+b^2 +c^2 −d^2 )x^2 − −2(a(c^2 −d^2 )+c(a^2 +b^2 ))x+(a^2 +b^2 )(c^2 −d^2 )=0 1. −2(a+c)=10 2. a^2 +4ac+b^2 +c^2 −d^2 =0 3. −2(a(c^2 −d^2 )+c(a^2 +b^2 ))=6 4. (a^2 +b^2 )(c^2 −d^2 )=−1 1. a=−c−5e 2. b=(√(−a^2 +4ac−c^2 +d^2 ))=(√(2c^2 +10c+d^2 −25)) 3. d=(√(ac+c^2 +((b^2 c+3)/a)))=(√(−((2c^3 +15c^2 +3)/(2c+5)))) 2. b=(√((2c^3 +15c^2 −128)/(2c+5))) 4. 4c^4 +40c^3 +75c^2 −125c+((689)/4)−((17161)/(4(2c+5)^2 ))=0 64(c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1)=0 c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1=0 c=u−((15)/6) u^6 −((75)/4)u^4 +((1939)/(16))u^2 −((17161)/(64))=0 u=(√v) v^3 −((75)/4)v^2 +((1939)/(16))v−((17161)/(64))=0 v=w+((75)/(12)) w^3 +4w+1=0 w=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =((−(1/2)+((√(849))/(18))))^(1/3) +((−(1/2)−((√(849))/(18))))^(1/3) and now we have to go all the way back... not very friendly... as I mentioned before, it′s almost (or maybe absolutely) impossible to find a nicer exact form of w, so let me use the approximation w≈−.246266 v=((25)/4)+w≈6.00373 u=((√(25+4w))/2)≈2.45025 c=−(5/2)+((√(25+4w))/2)≈−.0497482 d=(√(((25)/2)−w−((131)/(2(√(25+4w))))))≈.787215i a=−(5/2)−((√(25+4w))/2)≈−4.95025 b=(√(−((25)/2)+w+((131)/(2(√(25+4w))))))≈5.11001i x_1 =a+bi≈−10.0603 x_2 =a−bi≈.159762 x_3 =c+d≈−.0497482+.787215i x_4 =c−d≈−.0497482−.787215i

$x^{4} + 10 x^{3} + 6 x - 1 = 0$ $for those who want an exact solution . . .$ $(x - a - b i) (x - a + b i) (x - c - d) (x - c + d) = 0$ $x^{4} - 2 (a + c) x^{3} + (a^{2} + 4 a c + b^{2} + c^{2} - d^{2}) x^{2} -$ $- 2 (a (c^{2} - d^{2}) + c (a^{2} + b^{2})) x + (a^{2} + b^{2}) (c^{2} - d^{2}) = 0$ $1 . - 2 (a + c) = 10$ $2 . a^{2} + 4 a c + b^{2} + c^{2} - d^{2} = 0$ $3 . - 2 (a (c^{2} - d^{2}) + c (a^{2} + b^{2})) = 6$ $4 . (a^{2} + b^{2}) (c^{2} - d^{2}) = - 1$ $1 . a = - c - 5 e$ $2 . b = \sqrt{- a^{2} + 4 a c - c^{2} + d^{2}} = \sqrt{2 c^{2} + 10 c + d^{2} - 25}$ $3 . d = \sqrt{a c + c^{2} + \frac{b^{2} c + 3}{a}} = \sqrt{- \frac{2 c^{3} + 15 c^{2} + 3}{2 c + 5}}$ $2 . b = \sqrt{\frac{2 c^{3} + 15 c^{2} - 128}{2 c + 5}}$ $4 .$ $4 c^{4} + 40 c^{3} + 75 c^{2} - 125 c + \frac{689}{4} - \frac{17161}{4 {(2 c + 5)}^{2}} = 0$ $64 (c^{6} + 15 c^{5} + 75 c^{4} + 125 c^{3} + 4 c^{2} + 20 c + 1) = 0$ $c^{6} + 15 c^{5} + 75 c^{4} + 125 c^{3} + 4 c^{2} + 20 c + 1 = 0$ $c = u - \frac{15}{6}$ $u^{6} - \frac{75}{4} u^{4} + \frac{1939}{16} u^{2} - \frac{17161}{64} = 0$ $u = \sqrt{v}$ $v^{3} - \frac{75}{4} v^{2} + \frac{1939}{16} v - \frac{17161}{64} = 0$ $v = w + \frac{75}{12}$ $w^{3} + 4 w + 1 = 0$ $w = \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{p^{3}}{27} + \frac{q^{2}}{4}}} =$ $= \sqrt[3]{- \frac{1}{2} + \frac{\sqrt{849}}{18}} + \sqrt[3]{- \frac{1}{2} - \frac{\sqrt{849}}{18}}$ $and now we have to go all the way back . . .$ $not very friendly . . .$ $as I mentioned before, {it}^{'} s almost (or maybe$ $absolutely) impossible to find a nicer exact$ $form of w, so let me use the approximation$ $w \approx - . 246266$ $v = \frac{25}{4} + w \approx 6.00373$ $u = \frac{\sqrt{25 + 4 w}}{2} \approx 2.45025$ $c = - \frac{5}{2} + \frac{\sqrt{25 + 4 w}}{2} \approx - . 0497482$ $d = \sqrt{\frac{25}{2} - w - \frac{131}{2 \sqrt{25 + 4 w}}} \approx . 787215 i$ $a = - \frac{5}{2} - \frac{\sqrt{25 + 4 w}}{2} \approx - 4.95025$ $b = \sqrt{- \frac{25}{2} + w + \frac{131}{2 \sqrt{25 + 4 w}}} \approx 5 . 11001 i$ $x_{1} = a + b i \approx - 10.0603$ $x_{2} = a - b i \approx . 159762$ $x_{3} = c + d \approx - . 0497482 + . 787215 i$ $x_{4} = c - d \approx - . 0497482 - . 787215 i$

Question Number 36205 Answers: 0 Comments: 0

calculate ∫_0 ^∞ ((x^2 −1)/((x^2 +1)^2 )) x^(1/3) dx

$c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} - 1}{{(x^{2} + 1)}^{2}} x^{\frac{1}{3}} d x$

Question Number 36204 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sin(x))/x)dx

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{x^{2} - 1}{x^{2} + 1} \frac{s i n (x)}{x} d x$

Question Number 36203 Answers: 0 Comments: 1

let f(t) = ∫_0 ^∞ ((cos(tx))/((2+x^2 )^2 ))dx 1) find a simple form of f(t) 2) calculate ∫_0 ^∞ ((cos(3x))/((2+x^2 )^2 ))dx

$l e t f (t) = \int_{0}^{\infty} \frac{c o s (t x)}{{(2 + x^{2})}^{2}} d x$ $1) f i n d a s i m p l e f o r m o f f (t)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (3 x)}{{(2 + x^{2})}^{2}} d x$

Question Number 36202 Answers: 0 Comments: 1

calculate ∫_0 ^∞ ((x^2 dx)/((x^2 +1)^3 ))

$c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} d x}{{(x^{2} + 1)}^{3}}$

Question Number 36201 Answers: 0 Comments: 1

calculate ∫_0 ^(π/2) (dθ/(1+2sin^2 θ))

$c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + 2 {s i n}^{2} θ}$

Question Number 36200 Answers: 0 Comments: 4

calculate ∫_0 ^(2π) (dθ/((2+cosθ)^2 ))

$c a l c u l a t e \int_{0}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}$

Question Number 36198 Answers: 0 Comments: 1

let f(z) = ((z^2 +1)/(z^4 −1)) find (a_(k)) the poles of f and calculate Res(f,a_k )

$l e t f (z) = \frac{z^{2} + 1}{z^{4} - 1}$ $f i n d (a_{k)} t h e p o l e s o f f a n d c a l c u l a t e$ $R e s (f, a_{k})$

Question Number 36197 Answers: 0 Comments: 1

find the value of ∫_0 ^(2π) (dx/(cos^2 x +3 sin^2 x))

$f i n d t h e v a l u e o f \int_{0}^{2 π} \frac{d x}{{c o s}^{2} x + 3 {s i n}^{2} x}$

Question Number 36196 Answers: 0 Comments: 0

let ρ>0 and C the circle x^2 +y^2 =ρ^2 calculate ∫_C ydx +xy dy

$l e t ρ > 0 a n d C t h e c i r c l e x^{2} + y^{2} = ρ^{2}$ $c a l c u l a t e \int_{C} y d x + x y d y$

Question Number 36195 Answers: 0 Comments: 1

let C ={(x,y)∈R^2 / 0≤x≤1 and y=2x^2 } calculate ∫_C x^2 ydx +(x^2 −y^2 )dy

$l e t C = {(x, y) \in R^{2} / 0 ⩽ x ⩽ 1 a n d y = 2 x^{2}}$ $c a l c u l a t e \int_{C} x^{2} y d x + (x^{2} - y^{2}) d y$

Question Number 36194 Answers: 0 Comments: 0

let D ={(x,y,z)∈R^2 / 0<z<1 and x^2 +y^2 <z^2 } calculate ∫∫_D xyzdxdydz

$l e t D = {(x, y, z) \in R^{2} / 0 < z < 1 a n d x^{2} + y^{2} < z^{2}}$ $c a l c u l a t e \int \int_{D} x y z d x d y d z$

Question Number 36193 Answers: 0 Comments: 1

let D ={(x,y)∈ R^2 / x^2 +y^2 −x<0 and x^2 +y^2 −y >0 and y>0} calculate∫∫_D (x+y)^2 dxdy

$l e t D = {(x, y) \in R^{2} / x^{2} + y^{2} - x < 0 a n d$ $x^{2} + y^{2} - y > 0 a n d y > 0}$ $c a l c u l a t e \int \int_{D} {(x + y)}^{2} d x d y$

Question Number 36192 Answers: 0 Comments: 1

let D ={(x,y)∈ R^2 /x^2 +y^2 <1} find the value of ∫∫_D ((dxdy)/(x^2 +y^(2 ) + 2))

$l e t D = {(x, y) \in R^{2} / x^{2} + y^{2} < 1}$ $f i n d t h e v a l u e o f \int \int_{D} \frac{d x d y}{x^{2} + y^{2} + 2}$

Question Number 36191 Answers: 0 Comments: 1

let D = {(x,y)∈R^2 /x>0 ,y>0,x+y<1} 1) calculate ∫∫_D ((xy)/(x^2 +y^2 ))dxdy 2) let a>0 ,b>0 calculate ∫∫_D a^x b^y dxdy

$l e t D = {(x, y) \in R^{2} / x > 0, y > 0, x + y < 1}$ $1) c a l c u l a t e \int \int_{D} \frac{x y}{x^{2} + y^{2}} d x d y$ $2) l e t a > 0, b > 0 c a l c u l a t e \int \int_{D} a^{x} b^{y} d x d y$

Question Number 36190 Answers: 0 Comments: 1

calculate ∫∫_D (x+y)e^(x+y) dxdy with D = {(x,y)∈R^2 / 0<x<2 and 1<y<2 }

$c a l c u l a t e \int \int_{D} (x + y) e^{x + y} d x d y w i t h$ $D = {(x, y) \in R^{2} / 0 < x < 2 a n d 1 < y < 2}$

Question Number 36189 Answers: 0 Comments: 1

let F(x)=∫_0 ^∞ ((e^(−x^2 t) (√t))/(1+t^2 ))dt calculate lim_(x→+∞) F(x) .

$l e t F (x) = \int_{0}^{\infty} \frac{e^{- x^{2} t} \sqrt{t}}{1 + t^{2}} d t$ $c a l c u l a t e {l i m}_{x \to + \infty} F (x) .$

Question Number 36188 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ ((√t)/(1+t^2 ))dt

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t$

Question Number 36187 Answers: 0 Comments: 3

let I_n (x)= ∫_0 ^∞ ((t sin(t))/((t^2 +x^2 )^n ))dt 1) find a relation between I_(n+1) and I_n 2) calculate I_2 (x) and I_3 (x) 3) calculate ∫_0 ^∞ ((tsin(t))/((2+t^2 )^2 ))dt

$l e t I_{n} (x) = \int_{0}^{\infty} \frac{t s i n (t)}{{(t^{2} + x^{2})}^{n}} d t$ $1) f i n d a r e l a t i o n b e t w e e n I_{n + 1} a n d I_{n}$ $2) c a l c u l a t e I_{2} (x) a n d I_{3} (x)$ $3) c a l c u l a t e \int_{0}^{\infty} \frac{t s i n (t)}{{(2 + t^{2})}^{2}} d t$