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Author: Tinku Tara

find-the-laplace-transform-of-0-te-2t-sintdt-

Question Number 219519 by OmoloyeMichael last updated on 27/Apr/25 $$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{laplace}}\:\boldsymbol{{transform}}\:\boldsymbol{{of}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \boldsymbol{{te}}^{−\mathrm{2}\boldsymbol{{t}}} \boldsymbol{{sintdt}} \\ $$ Answered by SdC355 last updated on 27/Apr/25 $$\mathrm{First}\:\mathrm{idea}..\mathrm{Let}'\mathrm{s}\:\mathrm{define}\:{F}\left({s}\right)\:\mathrm{as}\:…

Find-0-1-x-x-2-dx-

Question Number 219515 by hardmath last updated on 27/Apr/25 $$\mathrm{Find}:\:\:\:\boldsymbol{\Omega}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:?\: \\ $$ Answered by SdC355 last updated on 27/Apr/25 $$\mathrm{can}'\mathrm{t}\:\mathrm{Find}\:\mathrm{primitive}\:\mathrm{function}\:\int\:\centerdot\: \\…

Question-219503

Question Number 219503 by Rojarani last updated on 27/Apr/25 Answered by Frix last updated on 27/Apr/25 $$\sqrt{\mathrm{4}{x}}+\sqrt{\mathrm{4}{x}−\mathrm{4}}+\sqrt{\mathrm{4}{x}−\mathrm{8}}=\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{5}}+\sqrt{{x}+\mathrm{9}} \\ $$$$\sqrt{\mathrm{4}{x}}=\sqrt{{x}+\mathrm{9}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\sqrt{\mathrm{4}{x}−\mathrm{4}}=\sqrt{{x}+\mathrm{5}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$$$\sqrt{\mathrm{4}{x}−\mathrm{8}}=\sqrt{{x}+\mathrm{1}}\:\Rightarrow\:{x}=\mathrm{3} \\ $$…

p-t-1-ipi-i-i-e-st-ln-s-0-s-ds-q-t-1-ipi-i-i-pi-2s-L-0-s-pi-2s-iY-0-is-e-st-ds-g-s-0-J-t-J-st-dt-h-s-0-

Question Number 219496 by SdC355 last updated on 27/Apr/25 $${p}\left({t}\right)=−\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\:\frac{{e}^{{st}} \left(\mathrm{ln}\left({s}\right)+\boldsymbol{\gamma}_{\mathrm{0}} \right)}{{s}}\:\mathrm{d}{s} \\ $$$${q}\left({t}\right)=\frac{\mathrm{1}}{\boldsymbol{{i}}\pi}\int_{−\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} ^{\:\:\infty\boldsymbol{{i}}+\boldsymbol{\gamma}} \:\left\{−\frac{\pi}{\mathrm{2}{s}}\boldsymbol{\mathrm{L}}_{\mathrm{0}} \left({s}\right)+\frac{\pi}{\mathrm{2}{s}}\boldsymbol{{i}}{Y}_{\mathrm{0}} \left(−\boldsymbol{{i}}{s}\right)\right\}{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{g}\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu}…