Category: Integration

Question Number 119681 by TANMAY PANACEA last updated on 26/Oct/20 $${\int }_0^{\pi }\sqrt{\frac{1+cos2x}{2}}dx$$$${\int }_0^{\mathrm{\infty }}\left[{ne}^{-x}\right]dx$$ Answered by bemath last updated…

Question Number 54074 by rahul 19 last updated on 28/Jan/19 $$Evaluate:$$$$1){\int }_0^1\frac{dx}{\sqrt{1+x}+\sqrt{1-x}+2}$$$$$$$$2){\int }_0^2\frac{ln(1+2x)}{1+x^2}$$$$\left.\mathrm{3}\right) \…

Question Number 54073 by cesar.marval.larez@gmail.com last updated on 28/Jan/19 Answered by estudiante last updated on 28/Jan/19 $$VemosqesunaintegralimpropiadetipoI:$$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\int_{{a}} ^{{R}} {x}^{{n}} {dx}\:=\:\underset{{R}\rightarrow\infty} {\mathrm{lim}}\:\mid\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\mid_{{a}}…

Question Number 119570 by mnjuly1970 last updated on 25/Oct/20 $$\dots \underset{\mathrm{♣}}{\stackrel{\mathrm{♣}}{\mathrm{♣}}}nicecalculus\underset{\mathrm{♣}}{\stackrel{\mathrm{♣}}{\mathrm{♣}}}\dots$$$$provethat::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-2x}ln\left(\frac{1+e^{-x}}{1-e^{-x}}\right)=1$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\bigstar\:\mathscr{M}.\mathscr{N}.\mathrm{1970}\bigstar……

Question Number 54011 by maxmathsup by imad last updated on 27/Jan/19 $$calculatef\left(a\right)=\int \frac{dx}{\sqrt{1+ax}-\sqrt{1-ax}}witha>0.$$$$2)calculate{U}_n={\int }_{-\frac{1}{na}}^{\frac{1}{na}}\frac{dx}{\sqrt{1+ax}-\sqrt{1-ax}}withnfromNandn>1$$$$find{lim}_{n\to +\mathrm{\infty }}{U}_{n}andstudytheconvergenceof\mathrm{\Sigma }{U}_n$$ Commented by…