Category: Integration

Question Number 99278 by Rio Michael last updated on 19/Jun/20 $$x0246810$$$$f\left(x\right)2.43.64.96.98.011.9$$$$\mathrm{Given}\mathrm{the}\mathrm{curve}y=f\left(x\right),\mathrm{with}\mathrm{corresponding}\mathrm{values}\mathrm{of}$$$$f\left(x\right)\mathrm{at}\mathrm{certain}x\mathrm{values}.\mathrm{The}\mathrm{curve}y=f\left(x\right)\mathrm{is}\mathrm{rotated}$$$$\mathrm{at}\mathrm{an}\mathrm{angle}\mathrm{of}2\pi \mathrm{about}\mathrm{the}\mathrm{x}-\mathrm{axis}.\mathrm{Find}$$$$\left(1\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{surface}\mathrm{generated}\mathrm{using}{\mathrm{simpson}}^{\prime }\mathrm{s}\mathrm{rule}.$$$$\left(2\right)\mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{surface}\mathrm{generated}\mathrm{using}{\mathrm{simpson}}^{\prime }\mathrm{s}\mathrm{rule}.$$…

Question Number 164809 by mathlove last updated on 22/Jan/22 $$1)\int \frac{\sqrt[3]{x}}{\sqrt{x}+\sqrt[4]{x}}=?$$$$2)\int \frac{x}{{(x^2+2x+2)}^2}=?$$ Answered by Ar Brandon last updated on 22/Jan/22 $$\mathrm{Ostrogradsky}…

Question Number 33736 by prof Abdo imad last updated on 23/Apr/18 $$findthevalueof{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{{(1+x+x^2)}^2}dx$$ Commented by prof Abdo imad last…

Question Number 99261 by Ar Brandon last updated on 19/Jun/20 $$\mathrm{Given}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{2}\int \frac{\mathrm{x}+1}{\mathrm{x}-1}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$$$$\mathrm{Show}\mathrm{that}\mathrm{F}\mathrm{is}\mathrm{defined},\mathrm{continuous},\mathrm{and}\mathrm{derivable}.$$$$\mathrm{And}\mathrm{find}\mathrm{its}\mathrm{derivative}$$ Answered by abdomathmax last updated on 19/Jun/20 $$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}\:\int^{\mathrm{x}}…

Question Number 33703 by math khazana by abdo last updated on 22/Apr/18 $$give{\int }_0^{\mathrm{\infty }}\frac{x{e}^{-x}}{1-e^{-2x}}sin\left(\pi x\right)dxatformofserie.$$ Terms of Service Privacy Policy Contact:…