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Question Number 79932    Answers: 1   Comments: 1

Question Number 79929    Answers: 0   Comments: 0

∫e^(√(sin x)) dx=?

$$\int{e}^{\sqrt{\mathrm{sin}\:{x}}} {dx}=? \\ $$

Question Number 79913    Answers: 0   Comments: 1

Convergence of I=∫_0 ^( ∞) (e^t /(e^(−t) +e^(2t) ∣sint∣))dt

$$\:{Convergence}\:\:{of}\:\:{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{{t}} }{{e}^{−{t}} +{e}^{\mathrm{2}{t}} \mid{sint}\mid}{dt} \\ $$

Question Number 79903    Answers: 1   Comments: 11

Question Number 79883    Answers: 0   Comments: 9

to Sir Jagoll (and of course everybody else) (1) y=((x^2 −x−6)/(x^2 −3x−4))= =(((x−3)(x+2))/((x−4)(x+1))) ⇒ ⇒ { ((zeros at x=−2; x=3)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x−6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y−3))/(y−1))=0 D=((25y^2 −46y+25)/(4(y−1)^2 ))>0∀x∈R ⇒ ⇒ { ((range=R)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=1 ⇒ y=1 y′=−((2(x^2 −2x+7))/((x−4)^2 (x+1)^2 )) no real zeros ⇒ ⇒ no local extremes y′′=((4(x^3 −3x^2 +21x−25))/((x−4)^3 (x+1)^3 ))=0 at x≈1.33131 ⇒ ⇒ turning point (2) y=((x^2 −x−6)/(x^2 −3x+4))= =(((x−3)(x+2))/(x^2 −3x+4)); x^2 −3x+4=0 no real zeros ⇒ ⇒ { ((zeros at x=−2; x=3)),((no vertical asymptote)) :} defined for x∈R range: transforming y=((x^2 −x−6)/(x^2 −3x+4)) to x^2 −((3y−1)/(y−1))x+((2(2y+3))/(y−1))=0 D=−((7y^2 +14y−25)/(4(y−1)^2 ))≥0 for −1−((4(√(14)))/7)≤y≤−1+((4(√(14)))/7) ⇒ ⇒ { ((range=[−1−((4(√(14)))/7); −1+((4(√(14)))/7)])),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=5 ⇒ y=1 y′=−((2(x^2 −10x+11))/((x^2 −3x+4)^2 ))=0 at x=5±(√(14)) y^(′′) =((4(x^3 −15x^2 +33x−13))/((x^2 −3x+4)^3 )) y′′ { ((>0 at x=5−(√(14)) ⇒ local minimum)),((<0 at x=5+(√(14)) ⇒ local maximum)),((=0 at { ((x≈.506699)),((x≈2.06421)),((x≈12.4291)) :} ⇒ 3 turning points )) :} (3) y=((x^2 −x+6)/(x^2 −3x−4))= =((x^2 −x+6)/((x−4)(x+1))) ⇒ ⇒ { ((no real zeros)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x+6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y+3))/(y−1))=0 D=((25y^2 +2y−23)/(4(y−1)^2 ))<0 for −1<y<((23)/(25)) ⇒ ⇒ { ((range=R\]−1; ((23)/(25))[)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=−5 ⇒ y=1 y′=−((2(x^2 +10x−11))/((x−4)^2 (x+1)^2 ))=0 at x=−11; x=1 y′′=((4(x^3 +15x^2 −33x+53))/((x−4)^3 (x+1)^3 )) y′′ { ((>0 at x=−11 ⇒ local minimum)),((<0 at x=1 ⇒ local maximum)),((=0 at x≈−17.1098 ⇒ turning point)) :}

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} −\mathrm{46}{y}+\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\mathbb{R}}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{local}\:\mathrm{extremes} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{25}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{33131}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{turning}\:\mathrm{point} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}};\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{no}\:\mathrm{vertical}\:\mathrm{asymptote}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}+\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=−\frac{\mathrm{7}{y}^{\mathrm{2}} +\mathrm{14}{y}−\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\mathrm{0}\:\mathrm{for}\:−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\leqslant{y}\leqslant−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\left[−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}};\:−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\right]}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{11}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}\pm\sqrt{\mathrm{14}} \\ $$$${y}^{''} =\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{33}{x}−\mathrm{13}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}−\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}+\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:\begin{cases}{{x}\approx.\mathrm{506699}}\\{{x}\approx\mathrm{2}.\mathrm{06421}}\\{{x}\approx\mathrm{12}.\mathrm{4291}}\end{cases}\:\Rightarrow\:\mathrm{3}\:\mathrm{turning}\:\mathrm{points}\:}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{23}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{0}\:\mathrm{for}\:−\mathrm{1}<{y}<\frac{\mathrm{23}}{\mathrm{25}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left.\mathrm{range}=\mathbb{R}\backslash\right]−\mathrm{1};\:\frac{\mathrm{23}}{\mathrm{25}}\left[\right.}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{11}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11};\:{x}=\mathrm{1} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{33}{x}+\mathrm{53}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:{x}\approx−\mathrm{17}.\mathrm{1098}\:\Rightarrow\:\mathrm{turning}\:\mathrm{point}}\end{cases} \\ $$

Question Number 79879    Answers: 1   Comments: 5

Question Number 79876    Answers: 1   Comments: 1

f is derivable in R. 1) Demonstrate that if f is pair , f ′ is odd(unpair). 1) Demonstrate that if f is unpair , f ′ is pair. Please help me sirs

$${f}\:{is}\:{derivable}\:{in}\:\mathbb{R}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{pair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{odd}\left(\mathrm{unpair}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:{is}\:\mathrm{unpair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{pair}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$

Question Number 79964    Answers: 0   Comments: 3

∫(tan^2 x+tan^4 x)dx=[tan^2 x=t]=∫(t^2 +t^4 )dx=(t^3 /3)+(t^5 /5)+c

$$\int\left(\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{tan}\:^{\mathrm{4}} {x}\right){dx}=\left[\mathrm{tan}\:^{\mathrm{2}} {x}={t}\right]=\int\left({t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dx}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$

Question Number 79947    Answers: 0   Comments: 4

Find ((ln 2)/(2!))+((ln 3)/(3!))+((ln 4)/(4!))+...+((ln n)/(n!))+...=?

$${Find} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}!}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{3}!}+\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{4}!}+...+\frac{\mathrm{ln}\:{n}}{{n}!}+...=? \\ $$

Question Number 79946    Answers: 2   Comments: 0

Find ∫_0 ^( n) [(x)^(1/3) ]dx=? in terms of n. (n∈N)

$${Find}\: \\ $$$$\int_{\mathrm{0}} ^{\:{n}} \left[\sqrt[{\mathrm{3}}]{{x}}\right]{dx}=?\: \\ $$$${in}\:{terms}\:{of}\:{n}.\:\left({n}\in\mathbb{N}\right) \\ $$

Question Number 79943    Answers: 1   Comments: 5

Question Number 79869    Answers: 0   Comments: 1

For witch value of α the integral I=∫_0 ^∞ ((1/(√(1+2x^2 )))−(α/(1+x)))dx converge; and in this case calculate α

$${For}\:\:{witch}\:\:{value}\:\:{of}\:\:\alpha\:\:{the}\:\:{integral} \\ $$$$\:\:{I}=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}−\frac{\alpha}{\mathrm{1}+{x}}\right){dx}\:\:{converge}; \\ $$$$\:\:{and}\:\:{in}\:\:{this}\:\:{case}\:\:{calculate}\:\:\alpha \\ $$

Question Number 79866    Answers: 0   Comments: 1

Chapter (2) Remainder Theorem and Factor Theorem Exercises(2.1) Remainder Throrem 1.Find the remainder when x^(8 ) +2x−5 is divided by (x−1). 2.Find the remainder when 2x^2 13x+10 is divided by (x−3).

$${Chapter}\:\left(\mathrm{2}\right) \\ $$$${Remainder}\:{Theorem} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{and} \\ $$$${Factor}\:{Theorem} \\ $$$${Exercises}\left(\mathrm{2}.\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{Remainder}}\:\mathrm{T}\boldsymbol{\mathrm{hrorem}} \\ $$$$\mathrm{1}.\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\boldsymbol{\mathrm{x}}^{\mathrm{8}\:} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{5} \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{1}\right). \\ $$$$\mathrm{2}.\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{2x}^{\mathrm{2}} \:\mathrm{13x}+\mathrm{10} \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left(\mathrm{x}−\mathrm{3}\right). \\ $$

Question Number 79864    Answers: 0   Comments: 0

Question Number 79861    Answers: 0   Comments: 1

Question Number 79856    Answers: 0   Comments: 2

Question Number 79838    Answers: 1   Comments: 0

∵

$$\because \\ $$

Question Number 79837    Answers: 1   Comments: 10

Question Number 79826    Answers: 0   Comments: 7

Question Number 79825    Answers: 0   Comments: 4

∫_( 0) ^( 1) (√(x^3 + 1)) dx

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\sqrt{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}}\:\:\mathrm{dx} \\ $$

Question Number 79824    Answers: 2   Comments: 7

Question Number 79816    Answers: 1   Comments: 0

hello mister. i need help explaining how determine the range of function of rational functions like (i) f(x)=((ax^2 +bx+c)/(px^2 +qx+r)) (ii) f(x)=((ax^2 +bx+c)/(px+q))

$$\mathrm{hello}\:\mathrm{mister}. \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{explaining}\:\mathrm{how}\:\mathrm{determine} \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{rational}\:\mathrm{functions} \\ $$$$\mathrm{like}\:\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{px}^{\mathrm{2}} +\mathrm{qx}+\mathrm{r}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{px}+\mathrm{q}} \\ $$

Question Number 79814    Answers: 0   Comments: 5

Question Number 79807    Answers: 2   Comments: 2

Question Number 79798    Answers: 0   Comments: 7

Question Number 79794    Answers: 1   Comments: 2

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