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IntegrationQuestion and Answers: Page 199

Question Number 82232    Answers: 1   Comments: 1

Question Number 82223    Answers: 0   Comments: 1

given f(x) = f(x+(π/6)) , ∀x∈ R if ∫_0 ^(π/6) f(x) dx = T then ∫_π ^((7π)/3) f(x+π) dx = ?

$${given}\:{f}\left({x}\right)\:=\:{f}\left({x}+\frac{\pi}{\mathrm{6}}\right)\:,\:\forall{x}\in\:\mathbb{R} \\ $$$${if}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{6}}} {\int}}\:{f}\left({x}\right)\:{dx}\:=\:{T}\: \\ $$$${then}\:\underset{\pi} {\overset{\frac{\mathrm{7}\pi}{\mathrm{3}}} {\int}}\:{f}\left({x}+\pi\right)\:{dx}\:=\:? \\ $$

Question Number 82185    Answers: 1   Comments: 2

∫ (dx/(sec x + csc x)) = ?

$$\int\:\frac{{dx}}{\mathrm{sec}\:{x}\:+\:{csc}\:{x}}\:=\:?\: \\ $$

Question Number 82174    Answers: 1   Comments: 1

∫_0 ^π x ln(sin x) dx = ?

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:{x}\:{ln}\left(\mathrm{sin}\:{x}\right)\:{dx}\:=\:?\: \\ $$

Question Number 82139    Answers: 1   Comments: 0

∫ ((√(x^4 +x^(−4) +2))/x^3 ) dx

$$\int\:\:\frac{\sqrt{{x}^{\mathrm{4}} +{x}^{−\mathrm{4}} +\mathrm{2}}}{{x}^{\mathrm{3}} }\:{dx}\: \\ $$

Question Number 82022    Answers: 0   Comments: 0

Question Number 82020    Answers: 0   Comments: 2

Question Number 81996    Answers: 0   Comments: 0

calculate I_n =∫∫_([(1/n),n[) e^(−x^2 −3y^2 ) dxdy and find lim_(n→+∞) I_n conclude that ∫_0 ^∞ e^(−x^2 ) dx=((√π)/2)

$${calculate}\:{I}_{{n}} =\int\int_{\left[\frac{\mathrm{1}}{{n}},{n}\left[\right.\right.} \:\:{e}^{−{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} } {dxdy} \\ $$$${and}\:{find}\:{lim}_{{n}\rightarrow+\infty} \:\:{I}_{{n}} \\ $$$${conclude}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

Question Number 81994    Answers: 0   Comments: 0

calculate ∫∫_W (x+y)e^(x−y) dxdy with W is the triangle limited by o,A(1,0)and B(0,1)

$${calculate}\:\int\int_{{W}} \left({x}+{y}\right){e}^{{x}−{y}} {dxdy} \\ $$$${with}\:{W}\:{is}\:{the}\:{triangle}\:{limited}\:{by} \\ $$$${o},{A}\left(\mathrm{1},\mathrm{0}\right){and}\:{B}\left(\mathrm{0},\mathrm{1}\right) \\ $$

Question Number 81993    Answers: 0   Comments: 0

calculate ∫∫_D ln(1+x+y)dxdy with D is the triangle limited by points 0,A(1,0) and B(0,1)

$${calculate}\:\int\int_{{D}} {ln}\left(\mathrm{1}+{x}+{y}\right){dxdy} \\ $$$${with}\:{D}\:{is}\:{the}\:{triangle}\:{limited}\:{by} \\ $$$${points}\:\mathrm{0},{A}\left(\mathrm{1},\mathrm{0}\right)\:{and}\:{B}\left(\mathrm{0},\mathrm{1}\right) \\ $$

Question Number 81921    Answers: 1   Comments: 0

Question Number 81889    Answers: 1   Comments: 0

Question Number 81888    Answers: 1   Comments: 1

Question Number 81801    Answers: 1   Comments: 1

Question Number 81739    Answers: 2   Comments: 1

∫(dx/(cos^3 x−sin^3 x))

$$\int\frac{{dx}}{{cos}^{\mathrm{3}} {x}−{sin}^{\mathrm{3}} {x}} \\ $$

Question Number 81719    Answers: 0   Comments: 1

1) find ∫ (dx/((x+2)^5 (x−3)^9 )) 2) calculate ∫_4 ^(+∞) (dx/((x+2)^5 (x−3)^9 ))

$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{5}} \left({x}−\mathrm{3}\right)^{\mathrm{9}} } \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{4}} ^{+\infty} \:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{5}} \left({x}−\mathrm{3}\right)^{\mathrm{9}} } \\ $$

Question Number 81636    Answers: 0   Comments: 4

∫ ((x(tan^(−1) (x))^2 )/((1+x^2 )^(3/2) )) dx =

$$\int\:\frac{{x}\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx}\:=\: \\ $$

Question Number 81591    Answers: 0   Comments: 6

if g(−2)=−5 and g′(x)= (x^2 /(cos^2 (x)+1)) find g(4)

$$\mathrm{if}\:\mathrm{g}\left(−\mathrm{2}\right)=−\mathrm{5}\:\mathrm{and}\: \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{1}} \\ $$$$\mathrm{find}\:\mathrm{g}\left(\mathrm{4}\right)\: \\ $$

Question Number 81549    Answers: 0   Comments: 5

∫_0 ^4 ⌊x⌋^2 dx = ∫_0 ^4 ⌊x^2 ⌋dx=

$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\lfloor\mathrm{x}\rfloor^{\mathrm{2}} \:\mathrm{dx}\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\lfloor\mathrm{x}^{\mathrm{2}} \rfloor\mathrm{dx}= \\ $$

Question Number 81482    Answers: 1   Comments: 1

Evaluate ∫_(−∞) ^∞ (dx/(x^2 +4x+13)).

$${Evaluate}\:\:\int_{−\infty} ^{\infty} \frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{13}}. \\ $$

Question Number 81433    Answers: 1   Comments: 0

calculate ∫_2 ^(+∞) ((2x+3)/((x−1)^3 (x^2 +x+1)^2 ))dx

$${calculate}\:\int_{\mathrm{2}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$

Question Number 81432    Answers: 1   Comments: 1

find ∫ (dx/((x+1)^3 (x^2 +3)^2 ))

$${find}\:\:\int\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$

Question Number 81427    Answers: 1   Comments: 4

1) calculate ∫_0 ^∞ cos(x^3 )dx 2)find the value of ∫_0 ^∞ cos(x^n )dx with n≥2

$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{3}} \right){dx} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{{n}} \right){dx}\:{with}\:{n}\geqslant\mathrm{2} \\ $$

Question Number 81400    Answers: 1   Comments: 3

Hello Nice day im thinking of this one a close forme? ∫(√(1+x^p ))dx p∈R_+ , x∈[0,1[

$${Hello}\:\:{Nice}\:{day}\:{im}\:{thinking}\:{of}\:{this}\:{one}\:\:{a}\:{close}\:{forme}? \\ $$$$\int\sqrt{\mathrm{1}+{x}^{{p}} }{dx} \\ $$$${p}\in\mathbb{R}_{+} , \\ $$$${x}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$

Question Number 81340    Answers: 1   Comments: 4

Quation posted Times ago i can′t find it .after Somme Try i got close Fofme ∫_0 ^π sin(x^2 )dx=(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4),−(π^4 /4))? sin(x)=Σ_(k≥0) (((−1)^k x^(2k+1) )/((2k+1)!)) ⇒sin(x^2 )=Σ_(k≥0) (((−1)^k x^(4k+2) )/((2k+1)!)) ∫_0 ^π sin(x^2 )dx=Σ_(k≥0) (((−1)^k )/((2k+1)!))∫_0 ^π x^(4k+2) dx =Σ_(k≥0) (((−1)^k π^(4k+3) )/((2k+1)!(4k+3))) =(π^3 /3)Σ_(k≥0) ((3(−1)^k π^(4k) )/((2k+1)!(4k+3))) (2k+1)!=k!.Π_(j=1) ^(k+1) (k+j)=k!.2^k .3....(2k+1) =(π^3 /3).Σ_(k≥0) ((3(−1)^k π^(4k) )/(k!.2^k .3....(2k+1)(4k+3))) =(π^3 /3)Σ_(k≥0) ((2^k .3)/(3...(2k+1).(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) ((2^k .3.....(4k−1))/(3.....(2k+1).(7.....(4k−1)(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) (((1/4^k ).(3)...(4k−1))/((1/2^k ).(3)...(2k+1).(1/4^k )(7)...(4k+3))).(−(π^4 /4))^k .(1/(k!)) =(π^3 /3)Σ_(k≥0) ((((3/4))....((3/4)+k−1))/(((3/2))....((3/2)+k−1).((7/4))....((7/4)+k−1))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4);−(π^4 /4))

$${Quation}\:\:{posted}\:{Times}\:{ago} \\ $$$${i}\:{can}'{t}\:{find}\:{it}\:.{after}\:{Somme}\:{Try}\:{i}\:{got}\:{close}\:{Fofme} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}},−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)? \\ $$$${sin}\left({x}\right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\Rightarrow{sin}\left({x}^{\mathrm{2}} \right)=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{4}{k}+\mathrm{2}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\pi} {sin}\left({x}^{\mathrm{2}} \right){dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\pi} {x}^{\mathrm{4}{k}+\mathrm{2}} {dx} \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}+\mathrm{3}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)!={k}!.\underset{{j}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\prod}}\left({k}+{j}\right)={k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{3}\left(−\mathrm{1}\right)^{{k}} \pi^{\mathrm{4}{k}} }{{k}!.\mathrm{2}^{{k}} .\mathrm{3}....\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}}{\mathrm{3}...\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{2}^{{k}} .\mathrm{3}.....\left(\mathrm{4}{k}−\mathrm{1}\right)}{\mathrm{3}.....\left(\mathrm{2}{k}+\mathrm{1}\right).\left(\mathrm{7}.....\left(\mathrm{4}{k}−\mathrm{1}\right)\left(\mathrm{4}{k}+\mathrm{3}\right)\right.}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}.\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\frac{\mathrm{1}}{\mathrm{4}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{4}{k}−\mathrm{1}\right)}{\frac{\mathrm{1}}{\mathrm{2}^{{k}} }.\left(\mathrm{3}\right)...\left(\mathrm{2}{k}+\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{4}^{{k}} }\left(\mathrm{7}\right)...\left(\mathrm{4}{k}+\mathrm{3}\right)}.\left(−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)....\left(\frac{\mathrm{3}}{\mathrm{4}}+{k}−\mathrm{1}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)....\left(\frac{\mathrm{3}}{\mathrm{2}}+{k}−\mathrm{1}\right).\left(\frac{\mathrm{7}}{\mathrm{4}}\right)....\left(\frac{\mathrm{7}}{\mathrm{4}}+{k}−\mathrm{1}\right)}.\left(\frac{−\pi^{\mathrm{4}} }{\mathrm{4}}\right)^{{k}} .\frac{\mathrm{1}}{{k}!} \\ $$$$=\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{7}}{\mathrm{4}};−\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\right) \\ $$$$ \\ $$

Question Number 81336    Answers: 1   Comments: 7

decompose F(x)=(1/((x−1)^3 (x+3)^7 )) and detrmine ∫ F(x)dx 2) calculate ∫_2 ^(+∞) F(x)dx

$${decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{3}\right)^{\mathrm{7}} } \\ $$$${and}\:{detrmine}\:\:\int\:{F}\left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:{F}\left({x}\right){dx} \\ $$

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