Question Number 163439 by MathsFan last updated on 07/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163428 by SANOGO last updated on 06/Jan/22 $${jusgifier}\:{la}\:{convergence}\:{de}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dx} \\ $$ Commented by Ar Brandon last updated on 06/Jan/22…
Question Number 32357 by .none. last updated on 23/Mar/18 $${x}^{\mathrm{2}} +{ax}−\mathrm{24}=\mathrm{0} \\ $$$${root}\:{is}\:{integer} \\ $$$${a}\:\:\:{range} \\ $$$$ \\ $$$${i}\:{cant}\:{speak}\:{english}\:{well}.\:{sorry} \\ $$ Answered by mrW2 last…
Question Number 97891 by pranesh last updated on 10/Jun/20 Commented by mr W last updated on 10/Jun/20 $${y}={e}^{{x}} −\mathrm{1} \\ $$ Commented by bemath last…
Question Number 163402 by Ahmed777hamouda last updated on 06/Jan/22 $$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\boldsymbol{{x}}−\mathrm{1}\right)^{\mathrm{10}} \left(\boldsymbol{{x}}−\mathrm{3}\right)^{\mathrm{3}} \boldsymbol{{dx}} \\ $$ Answered by Ar Brandon last updated on 06/Jan/22 $${I}=\int_{\mathrm{0}}…
Question Number 163393 by nurtani last updated on 06/Jan/22 Answered by Ar Brandon last updated on 06/Jan/22 $$\left[\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} −\frac{\mathrm{1}}{\pi}\mathrm{cos}\pi{x}\right]_{\mathrm{1}/\mathrm{2}} ^{\mathrm{1}} \\ $$ Terms of Service…
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Question Number 163368 by nurtani last updated on 06/Jan/22 Answered by Ar Brandon last updated on 06/Jan/22 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\mathrm{4}}{\pi}\mathrm{sin}^{\mathrm{2007}} {x}}{\mathrm{sin}^{\mathrm{2007}} {x}+\mathrm{cos}^{\mathrm{2007}} {x}}{dx}…\left(\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}}…
Question Number 97827 by amrabdelsabour last updated on 10/Jun/20 Commented by Rio Michael last updated on 10/Jun/20 $$\mathrm{Any}\:\mathrm{ideas}\:\mathrm{what}\:\mathrm{this}\:\mathrm{means}? \\ $$ Terms of Service Privacy Policy…
Question Number 163346 by MathsFan last updated on 06/Jan/22 $${how}\:{do}\:{i}\:{calculate}\:{for}\:\frac{\mathrm{1}}{\mathrm{2}}! \\ $$ Commented by Ar Brandon last updated on 06/Jan/22 $$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)={x}\left({x}−\mathrm{1}\right)\Gamma\left({x}−\mathrm{1}\right)…\Gamma\left(\mathrm{1}\right)={x}! \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\:\:,\:\:\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}…