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Category: Integration

lets-f-0-1-R-given-by-f-x-x-x-Q-1-x-x-Q-is-f-continuos-at-x-1-2-is-0-1-fdx-riemann-integrable-is-0-1-fdx-lebesgue-integable-

Question Number 1124 by 123456 last updated on 17/Jun/15 $$\mathrm{lets}\:{f}:\left[\mathrm{0},\mathrm{1}\right]\rightarrow\mathbb{R}\:\mathrm{given}\:\mathrm{by} \\ $$$${f}\left({x}\right)=\begin{cases}{{x}}&{{x}\in\mathbb{Q}}\\{\mathrm{1}−{x}}&{{x}\notin\mathbb{Q}}\end{cases} \\ $$$$\mathrm{is}\:{f}\:\mathrm{continuos}\:\mathrm{at}\:{x}=\mathrm{1}/\mathrm{2}? \\ $$$$\mathrm{is}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{fdx}\:\mathrm{riemann}\:\mathrm{integrable}? \\ $$$$\mathrm{is}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{fdx}\:\mathrm{lebesgue}\:\mathrm{integable}? \\ $$ Commented…

I-0-pi-x-pi-x-sin-x-dx-

Question Number 1085 by 123456 last updated on 10/Jun/15 $$\mathrm{I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}}{dx} \\ $$ Commented by 123456 last updated on 10/Jun/15 $${f}\left({x}\right)=\frac{{x}\left(\pi−{x}\right)}{\mathrm{sin}\:{x}} \\ $$$${f}\left(\mathrm{0}^{+} \right)\overset{?}…

sinx-1-sinx-sin2x-dx-

Question Number 66589 by Tanmay chaudhury last updated on 17/Aug/19 $$\int\frac{{sinx}}{\mathrm{1}+{sinx}+{sin}\mathrm{2}{x}}{dx} \\ $$ Commented by MJS last updated on 17/Aug/19 $$\mathrm{Weierstrass}\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right]\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{4}\int\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}}…

1-ln-x-dx-

Question Number 132110 by Raxreedoroid last updated on 11/Feb/21 $$\int\frac{\mathrm{1}}{\mathrm{ln}\:{x}}{dx}=? \\ $$ Answered by Olaf last updated on 11/Feb/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{ln}{x}} \\ $$$$\mathrm{Let}\:{x}\:=\:{e}^{{u}} \\ $$$$\mathrm{F}\left({e}^{{u}} \right)\:=\:\int\frac{{d}\left({e}^{{u}}…

evaluate-0-2-x-2-dx-

Question Number 66561 by Rio Michael last updated on 17/Aug/19 $${evaluate}\: \\ $$$$\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \mid\:{x}+\:\mathrm{2}\mid\:{dx}. \\ $$ Commented by kaivan.ahmadi last updated on 17/Aug/19 $$\mathrm{0}<{x}<\mathrm{2}\Rightarrow{x}+\mathrm{2}>\mathrm{0}\Rightarrow\mid{x}+\mathrm{2}\mid={x}+\mathrm{2}…