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solve ∫∫_G (7x−y)dxdy, where G is given by y=0 x+2y=3, x=y^2 i want to know if the integral below is a correct representation of the integral above. (∫_0 ^(3/2) ∫_0 ^(9/4) (7x−y)dxdy).

$${solve}\:\int\int_{{G}} \left(\mathrm{7}{x}−{y}\right){dxdy},\:{where}\:{G}\:{is}\:{given}\:{by}\:{y}=\mathrm{0} \\ $$$${x}+\mathrm{2}{y}=\mathrm{3},\:{x}={y}^{\mathrm{2}} \\ $$$$ \\ $$$${i}\:{want}\:{to}\:{know}\:{if}\:{the}\:{integral}\:{below}\:{is}\:{a}\:{correct} \\ $$$${representation}\:{of}\:{the}\:{integral}\:{above}. \\ $$$$\:\left(\underset{\mathrm{0}} {\overset{\frac{\mathrm{3}}{\mathrm{2}}} {\int}}\underset{\mathrm{0}} {\overset{\frac{\mathrm{9}}{\mathrm{4}}} {\int}}\left(\mathrm{7}{x}−{y}\right){dxdy}\right). \\ $$

Question Number 130053 Answers: 4 Comments: 0

Show that ∫_0 ^( ∞) ((x^2 ln(x))/(x^4 +x^2 +1))dx = (π^2 /(12))

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$

Question Number 130029 Answers: 1 Comments: 2

1) decompose F(x)=(1/((x^2 −4)^3 (x^2 +1)^2 )) 2) find ∫_3 ^∞ (dx/((x^2 −4)^3 (x^2 +1)^2 ))

$$\left.\mathrm{1}\right)\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\int_{\mathrm{3}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Question Number 130011 Answers: 2 Comments: 0

... advanced calculus... prove that:: Φ=∫_( R) e^((−e^x +2x)) x^2 dx=(1−γ)^2 +((π^2 −6)/6)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\Phi=\underset{\:\:\:\:\:\mathbb{R}} {\int}{e}^{\left(−{e}^{{x}} +\mathrm{2}{x}\right)} {x}^{\mathrm{2}} {dx}=\left(\mathrm{1}−\gamma\right)^{\mathrm{2}} +\frac{\pi^{\mathrm{2}} −\mathrm{6}}{\mathrm{6}} \\ $$$$ \\ $$

Question Number 129996 Answers: 0 Comments: 0

∫((x^2 )^(1/5) /( ((1+(√x^5 )))^(1/3) ))dx

$$\int\frac{\sqrt[{\mathrm{5}}]{{x}^{\mathrm{2}} }}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+\sqrt{{x}^{\mathrm{5}} }}}{dx} \\ $$

Question Number 129991 Answers: 0 Comments: 0

∫ (e^x )(((x^6 +x^5 +5x^4 )/((1+x)^6 )))dx = ...

$$\:\int\:\left(\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \right)\left(\frac{\boldsymbol{\mathrm{x}}^{\mathrm{6}} +\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{4}} }{\left(\mathrm{1}+\boldsymbol{\mathrm{x}}\right)^{\mathrm{6}} }\right)\boldsymbol{\mathrm{dx}}\:=\:... \\ $$

Question Number 129989 Answers: 1 Comments: 0

Find the area bounded xy^2 = 4a^2 (2a−x) and its asymptotes.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{xy}^{\mathrm{2}} \:=\:\mathrm{4a}^{\mathrm{2}} \left(\mathrm{2a}−\mathrm{x}\right) \\ $$$$\mathrm{and}\:\mathrm{its}\:\mathrm{asymptotes}. \\ $$

Question Number 129978 Answers: 1 Comments: 0

is this true for n∈N^★ ? someone please prove or falsify! ∫_0 ^∞ e^(−x^(2n) ) dx=Γ ((2n+1)/(2n))

$$\mathrm{is}\:\mathrm{this}\:\mathrm{true}\:\mathrm{for}\:{n}\in\mathbb{N}^{\bigstar} ?\:\mathrm{someone}\:\mathrm{please}\:\mathrm{prove} \\ $$$$\mathrm{or}\:\mathrm{falsify}! \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{x}^{\mathrm{2}{n}} } {dx}=\Gamma\:\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}} \\ $$

Question Number 129975 Answers: 2 Comments: 0

Question Number 129972 Answers: 1 Comments: 0

....nice calculus... evaluation: Ω=∫_0 ^( ∞) t^2 e^(−t) ln(t)dt=?? solution: f(s)=∫_0 ^( ∞) t^(2+s) e^(−t) dt Ω=f ′(0)=... f(s)=Γ(3+s) f ′(s)=Γ′(3+s)=ψ(3+s)Γ(3+s) f ′(0)=ψ(3)Γ(3)=2((3/2) −γ) =3−2γ ∴ Ω=∫_0 ^( ∞) t^2 e^(−t) ln(t)=3−2γ ...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}... \\ $$$$\:\:{evaluation}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}} {e}^{−{t}} {ln}\left({t}\right){dt}=?? \\ $$$$\:\:{solution}: \\ $$$$\:\:\:{f}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}+{s}} {e}^{−{t}} {dt} \\ $$$$\:\:\:\Omega={f}\:'\left(\mathrm{0}\right)=... \\ $$$$\:\:\:{f}\left({s}\right)=\Gamma\left(\mathrm{3}+{s}\right) \\ $$$$\:\:\:\:{f}\:'\left({s}\right)=\Gamma'\left(\mathrm{3}+{s}\right)=\psi\left(\mathrm{3}+{s}\right)\Gamma\left(\mathrm{3}+{s}\right) \\ $$$${f}\:'\left(\mathrm{0}\right)=\psi\left(\mathrm{3}\right)\Gamma\left(\mathrm{3}\right)=\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\:−\gamma\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}−\mathrm{2}\gamma \\ $$$$\:\:\:\therefore\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {t}^{\mathrm{2}} {e}^{−{t}} {ln}\left({t}\right)=\mathrm{3}−\mathrm{2}\gamma\:... \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$

Question Number 129946 Answers: 1 Comments: 0

∫ ((tan ϕ+3)/(sin ϕ)) dϕ ?

$$\:\int\:\frac{\mathrm{tan}\:\varphi+\mathrm{3}}{\mathrm{sin}\:\varphi}\:\mathrm{d}\varphi\:? \\ $$

Question Number 129926 Answers: 1 Comments: 0

...nice calculus... evaluate: Σ_(k=0) ^∞ Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((1/((k+m+n)!)))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:{calculus}... \\ $$$$\:\:{evaluate}: \\ $$$$\:\:\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\left({k}+{m}+{n}\right)!}\right)=? \\ $$$$ \\ $$

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