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Question Number 88955    Answers: 0   Comments: 10

hello floor function ∫_a ^b ⌊x⌋ dx a,b∈z and b>a =∫_0 ^b ⌊x⌋ dx −∫_0 ^a ⌊x⌋ dx =((b^2 −b)/2)−((a^2 −a)/2) ....(1) now ∫_m ^k ⌊x⌋ dx when m,k∉z when m<a<b<k b=[k] and a=[m] ∫_m ^a ⌊x⌋dx +∫_a ^b ⌊x⌋dx +∫_b ^k ⌊x⌋dx =(a−m)⌊m⌋+((b^2 −b)/2)−((a^2 −a)/2)+(k−b)⌊k⌋ =(⌊m⌋−m)⌊m⌋+((⌊k⌋^2 −⌊k⌋)/2)−((⌊m⌋^2 −⌊m⌋)/2)+(k−⌊k⌋)⌊k⌋ ⌊m⌋^2 −m⌊m⌋ +(1/2)⌊k⌋^2 −(1/2)⌊k⌋−(1/2)⌊m⌋^2 −(1/2)⌊m⌋+k⌊k⌋−⌊k⌋^2 =k⌊k⌋−m⌊m⌋+(1/2)⌊m⌋^2 −(1/2)⌊k⌋^2 +(1/2)⌊m⌋−(1/2)⌊k⌋ ∴∫_m ^k ⌊x⌋dx=k⌊k⌋−m⌊m⌋+(1/2)(⌊m⌋−⌊k⌋)(⌊m⌋+⌊k⌋)+1 example ∫_(−1.5) ^(3.7) ⌊x⌋dx=(3.7)3−((−1.5)(−2))+(1/2)(−2−3)(−2+3+1) =11.1−3−5=3.1

$${hello}\: \\ $$$${floor}\:{function} \\ $$$$\int_{{a}} ^{{b}} \lfloor{x}\rfloor\:\:{dx}\:\:\:\:\:\:\:\:\:\:{a},{b}\in{z}\:\:\:{and}\:{b}>{a} \\ $$$$=\int_{\mathrm{0}} ^{{b}} \lfloor{x}\rfloor\:{dx}\:−\int_{\mathrm{0}} ^{{a}} \lfloor{x}\rfloor\:{dx}\:=\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}\:....\left(\mathrm{1}\right) \\ $$$${now} \\ $$$$\int_{{m}} ^{{k}} \lfloor{x}\rfloor\:{dx}\:\:\:{when}\:{m},{k}\notin{z}\:\:\:\:{when}\:{m}<{a}<{b}<{k} \\ $$$${b}=\left[{k}\right]\:\:\:\:{and}\:{a}=\left[{m}\right] \\ $$$$\int_{{m}} ^{{a}} \lfloor{x}\rfloor{dx}\:+\int_{{a}} ^{{b}} \lfloor{x}\rfloor{dx}\:+\int_{{b}} ^{{k}} \lfloor{x}\rfloor{dx} \\ $$$$=\left({a}−{m}\right)\lfloor{m}\rfloor+\frac{{b}^{\mathrm{2}} −{b}}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} −{a}}{\mathrm{2}}+\left({k}−{b}\right)\lfloor{k}\rfloor \\ $$$$=\left(\lfloor{m}\rfloor−{m}\right)\lfloor{m}\rfloor+\frac{\lfloor{k}\rfloor^{\mathrm{2}} −\lfloor{k}\rfloor}{\mathrm{2}}−\frac{\lfloor{m}\rfloor^{\mathrm{2}} −\lfloor{m}\rfloor}{\mathrm{2}}+\left({k}−\lfloor{k}\rfloor\right)\lfloor{k}\rfloor \\ $$$$\lfloor{m}\rfloor^{\mathrm{2}} −{m}\lfloor{m}\rfloor\:+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor+{k}\lfloor{k}\rfloor−\lfloor{k}\rfloor^{\mathrm{2}} \\ $$$$={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\lfloor{m}\rfloor−\frac{\mathrm{1}}{\mathrm{2}}\lfloor{k}\rfloor \\ $$$$\therefore\int_{{m}} ^{{k}} \lfloor{x}\rfloor{dx}={k}\lfloor{k}\rfloor−{m}\lfloor{m}\rfloor+\frac{\mathrm{1}}{\mathrm{2}}\left(\lfloor{m}\rfloor−\lfloor{k}\rfloor\right)\left(\lfloor{m}\rfloor+\lfloor{k}\rfloor\right)+\mathrm{1} \\ $$$${example} \\ $$$$\int_{−\mathrm{1}.\mathrm{5}} ^{\mathrm{3}.\mathrm{7}} \lfloor{x}\rfloor{dx}=\left(\mathrm{3}.\mathrm{7}\right)\mathrm{3}−\left(\left(−\mathrm{1}.\mathrm{5}\right)\left(−\mathrm{2}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{2}−\mathrm{3}\right)\left(−\mathrm{2}+\mathrm{3}+\mathrm{1}\right) \\ $$$$=\mathrm{11}.\mathrm{1}−\mathrm{3}−\mathrm{5}=\mathrm{3}.\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 88951    Answers: 1   Comments: 0

∫_a ^b ⌈x⌉ dx=? a,b∈R ⌈..⌉ is ceil

$$\int_{{a}} ^{{b}} \lceil{x}\rceil\:{dx}=? \\ $$$${a},{b}\in{R}\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\lceil..\rceil\:{is}\:{ceil}\: \\ $$

Question Number 88943    Answers: 0   Comments: 2

Question Number 88940    Answers: 1   Comments: 0

Question Number 88937    Answers: 0   Comments: 0

Question Number 88936    Answers: 1   Comments: 0

Question Number 88933    Answers: 0   Comments: 5

Question Number 88930    Answers: 0   Comments: 2

find A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx with λ>0 2)find the value of ∫_0 ^∞ ((cos(3x))/((x^2 −x+1)^2 ))dx

$${find}\:{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\lambda{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:{with}\:\lambda>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 88929    Answers: 0   Comments: 1

cakculate ∫_0 ^∞ ((arctan(ch(x)))/(4+x^2 ))dx

$${cakculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ch}\left({x}\right)\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 88928    Answers: 0   Comments: 4

calculate ∫_0 ^∞ (dx/((x^4 +x^2 +3)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{4}} \:\:+{x}^{\mathrm{2}} \:\:+\mathrm{3}\right)^{\mathrm{2}} } \\ $$

Question Number 88927    Answers: 0   Comments: 0

calculate ∫_0 ^∞ ((sin(∣arctanx∣))/(x^2 +1))dx

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

Question Number 88921    Answers: 2   Comments: 0

find x,y x−2y−(√(xy))=0 (√(x−1))−(√(2y−1))=1

$${find}\:{x},{y} \\ $$$${x}−\mathrm{2}{y}−\sqrt{{xy}}=\mathrm{0} \\ $$$$\sqrt{{x}−\mathrm{1}}−\sqrt{\mathrm{2}{y}−\mathrm{1}}=\mathrm{1} \\ $$

Question Number 88918    Answers: 0   Comments: 10

Question Number 88902    Answers: 1   Comments: 4

∫_(1/e) ^e ln∣x∣ dx

$$\int_{\frac{\mathrm{1}}{{e}}} ^{{e}} {ln}\mid{x}\mid\:{dx} \\ $$

Question Number 88899    Answers: 0   Comments: 14

Simplify ((((35+18i(√3)))^(1/3) +((35−18i(√3)))^(1/3) −4)/3)

$${Simplify} \\ $$$$\frac{\sqrt[{\mathrm{3}}]{\mathrm{35}+\mathrm{18}{i}\sqrt{\mathrm{3}}}+\sqrt[{\mathrm{3}}]{\mathrm{35}−\mathrm{18}{i}\sqrt{\mathrm{3}}}−\mathrm{4}}{\mathrm{3}} \\ $$

Question Number 88896    Answers: 1   Comments: 2

find lim_(x→0) ((ln(sin(3x)+cos(3x)))/(ln(sin(x)+cos(x))))

$${find}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{ln}\left({sin}\left(\mathrm{3}{x}\right)+{cos}\left(\mathrm{3}{x}\right)\right)}{{ln}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right)} \\ $$

Question Number 88892    Answers: 0   Comments: 2

Question Number 88894    Answers: 0   Comments: 1

Question Number 88889    Answers: 0   Comments: 1

Question Number 88886    Answers: 1   Comments: 5

Question Number 88882    Answers: 1   Comments: 1

The coefficient of x^3 in ((√x^5 ) +(3/(√x^3 )))^6 is

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\:{x}^{\mathrm{3}} \:\mathrm{in}\:\left(\sqrt{{x}^{\mathrm{5}} }\:+\frac{\mathrm{3}}{\sqrt{{x}^{\mathrm{3}} }}\right)^{\mathrm{6}} \:\:\mathrm{is} \\ $$

Question Number 88881    Answers: 1   Comments: 0

Question Number 88876    Answers: 0   Comments: 2

Given a 10−digit number X = 1345789026 How many 10−digit number that can be made using every digit from X, with condition: If a number n is located in k^(th) position of X, then the new created number must not contain number n in k^(th) position Example: • Number 1 is located in 1^(st) position of X, hence 1234567890 is not valid, but 2134567890 is valid • Number 5 and 0 are located in 4^(th) and 8^(th) position of X, hence 9435162087 is not valid, but 9431506287 is valid. • 1345026789 is not valid • and so on...

$$\mathrm{Given}\:\mathrm{a}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:{X}\:=\:\mathrm{1345789026} \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{10}−\mathrm{digit}\:\mathrm{number}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{made} \\ $$$$\mathrm{using}\:\mathrm{every}\:\mathrm{digit}\:\mathrm{from}\:{X},\:\mathrm{with}\:\mathrm{condition}: \\ $$$$\mathrm{If}\:\mathrm{a}\:\mathrm{number}\:{n}\:\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{created}\:\mathrm{number}\:\mathrm{must}\:\mathrm{not}\:\mathrm{contain} \\ $$$$\mathrm{number}\:{n}\:\mathrm{in}\:{k}^{{th}} \:\mathrm{position} \\ $$$$ \\ $$$$\mathrm{Example}: \\ $$$$\bullet\:\mathrm{Number}\:\mathrm{1}\:\mathrm{is}\:\mathrm{located}\:\mathrm{in}\:\mathrm{1}^{{st}} \:\mathrm{position}\:\mathrm{of}\:{X},\:\mathrm{hence} \\ $$$$\mathrm{1234567890}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid},\:\mathrm{but}\:\mathrm{2134567890} \\ $$$$\mathrm{is}\:\mathrm{valid} \\ $$$$\bullet\:\mathrm{Number}\:\mathrm{5}\:\mathrm{and}\:\mathrm{0}\:\mathrm{are}\:\mathrm{located}\:\mathrm{in}\:\mathrm{4}^{{th}} \:\mathrm{and}\:\mathrm{8}^{{th}} \:\mathrm{position} \\ $$$$\mathrm{of}\:{X},\:\mathrm{hence}\:\mathrm{9435162087}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid},\:\mathrm{but} \\ $$$$\mathrm{9431506287}\:\mathrm{is}\:\mathrm{valid}. \\ $$$$\bullet\:\mathrm{1345026789}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid} \\ $$$$\bullet\:\mathrm{and}\:\mathrm{so}\:\mathrm{on}... \\ $$

Question Number 88873    Answers: 0   Comments: 1

3+(√(x^2 −5)) > ∣x−1∣

$$\mathrm{3}+\sqrt{{x}^{\mathrm{2}} −\mathrm{5}}\:>\:\mid{x}−\mathrm{1}\mid\: \\ $$

Question Number 88872    Answers: 0   Comments: 1

The system of equations kx+y+z=1, x+ky+z=k and x+y+kz=k^2 have no solution if k equals

$$\mathrm{The}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\:\:{kx}+{y}+{z}=\mathrm{1}, \\ $$$${x}+{ky}+{z}={k}\:\:\:\mathrm{and}\:\:\:{x}+{y}+{kz}={k}^{\mathrm{2}} \:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{if}\:\:\:{k}\:\:\mathrm{equals} \\ $$

Question Number 88871    Answers: 0   Comments: 0

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