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Question Number 112962 by mohammad17 last updated on 10/Sep/20

Commented by john santu last updated on 10/Sep/20

$$ithink\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{t}^8{e}^{-4t}dt$$

Commented by Dwaipayan Shikari last updated on 10/Sep/20

$${\int }_0^{\mathrm{\infty }}{t}^8.e^{-4t}dt={\int }_0^{\mathrm{\infty }}\frac{{\left(\frac{u}{4}\right)}^8}{4}.e^{-u}dt4t=u,4=\frac{du}{dt}$$$$\frac{1}{4^9}\mathrm{\Gamma }(1+8)=\frac{8!}{4^9}$$

Commented by Aziztisffola last updated on 10/Sep/20

$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{mean}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^8{\mathrm{e}}^{-4\mathrm{t}}\mathrm{dt}$$$$\mathrm{In}\mathrm{this}\mathrm{case};\mathrm{here}\mathrm{is}\mathrm{the}\mathrm{solution}:$$$$\mathrm{let}\mathrm{f}\left(\mathrm{t}\right)={\mathrm{t}}^8\Rightarrow \mathcal{L}\left\{\mathrm{f}\left(\mathrm{t}\right)\right\}=\mathrm{F}\left(\mathrm{s}\right)=\frac{8!}{{\mathrm{s}}^9}$$$$\mathcal{L}\left\{{\mathrm{t}}^8{\mathrm{e}}^{-4\mathrm{t}}\right\}=\mathcal{L}\left\{\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-4\mathrm{t}}\right\}=\mathrm{F}(\mathrm{s}+4)$$$$=\frac{8!}{{(\mathrm{s}+4)}^9}$$$$\mathrm{then}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{st}}{\mathrm{t}}^8{\mathrm{e}}^{-4\mathrm{t}}\mathrm{dt}=\frac{8!}{{(\mathrm{s}+4)}^9}$$$$\mathrm{let}\mathrm{s}=0\Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{t}}^8{\mathrm{e}}^{-4\mathrm{t}}\mathrm{dt}=\frac{8!}{4^9}$$$$$$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$x^8{\int }_0^{\mathrm{\infty }}{e}^{-4t}dt$$$$\frac{1}{-4}{x}^8{\left[e^{-4t}\right]}_0^{\mathrm{\infty }}=\frac{x^8}{4}$$

Commented by mohammad17 last updated on 10/Sep/20

$$siriwantbylaplaseandgamafunction$$

Answered by Aziztisffola last updated on 10/Sep/20

$$\mathrm{Q}1)F\left(s\right)=\mathcal{L}\left(x^8{e}^{-4t}\right)=x^8\mathcal{L}\left(e^{-4t}\right)=x^8\times \frac{1}{s+4}$$$$F\left(s\right)=\frac{x^8}{s+4}$$$$lets=0\Rightarrow {\int }_0^{\mathrm{\infty }}{x}^8{e}^{-4t}dt=F\left(0\right)=\frac{x^8}{4}$$

Answered by mathmax by abdo last updated on 11/Sep/20

$$\mathrm{if}\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^8{\mathrm{e}}^{-4\mathrm{t}}\mathrm{dt}\mathrm{we}\mathrm{do}\mathrm{tbe}\mathrm{cha}7\mathrm{gement}4\mathrm{t}=\mathrm{u}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{u}}{4}\right)}^8{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{4}=\frac{1}{4^9}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^8{\mathrm{e}}^{-\mathrm{u}}[\mathrm{du}=\frac{1}{4^9}\times \mathrm{\Gamma }\left(9\right)$$$$=\frac{8!}{4^9}$$

Answered by mathmax by abdo last updated on 11/Sep/20

$${\mathrm{q}}_2)\mathrm{A}={\int }_0^{\mathrm{\infty }}{3}^{-{4\mathrm{z}}^2}\mathrm{dz}\left(\mathrm{i}\mathrm{suppose}\mathrm{z}\mathrm{reel}\right)$$$$\mathrm{A}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{4\mathrm{z}}^2\ln \left(3\right)}\mathrm{dz}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(2\sqrt{\ln 3}\mathrm{z}\right)}^2}\mathrm{dz}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}$$$$2\sqrt{\ln 3}\mathrm{z}=\mathrm{u}\Rightarrow \mathrm{A}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\frac{\mathrm{du}}{2\sqrt{\ln 3}}=\frac{1}{2\sqrt{\ln 3}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}$$$$=\frac{1}{2\sqrt{\ln 3}}\times \frac{\sqrt{\pi }}{2}=\frac{\sqrt{\pi }}{4\sqrt{\ln 3}}$$

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