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IntegrationQuestion and Answers: Page 194

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))

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

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

Question Number 81313 Answers: 1 Comments: 5

∫(((x−2)^3 )/((x+2)^5 )) dx

$$\int\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{5}} }\:\mathrm{dx} \\ $$

Question Number 81290 Answers: 0 Comments: 3

∫_(−1) ^1 arctan(x)arctan((x/(√(1−x^2 ))))arctan(((1+x)/(1−x)))dx

$$\int_{−\mathrm{1}} ^{\mathrm{1}} {arctan}\left({x}\right){arctan}\left(\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right){arctan}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right){dx} \\ $$

Question Number 81279 Answers: 0 Comments: 2

∫ ((√(2x+1))/(3x)) dx = ?

$$\int\:\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}\:{dx}\:=\:? \\ $$