Question Number 55638 by gunawan last updated on 01/Mar/19 $$\mathrm{For}\:\mathrm{all}\:{n}\:\in\:\mathbb{N} \\ $$$${f}_{{n}} \left({x}\right)=\begin{cases}{\frac{{nx}}{\mathrm{2}{n}−\mathrm{1}},\:\:\:\:\:{x}\:\in\:\left[\mathrm{0},\:\frac{\mathrm{2}{n}−\mathrm{1}}{{n}}\right]}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:,\:\:\:\:\:\:{x}\:\in\left[\frac{\mathrm{2}{n}−\mathrm{1}}{{n}},\:\mathrm{2}\right]}\end{cases} \\ $$$$\mathrm{then}\:\mathrm{for}\:{n}\rightarrow\infty \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} {f}_{{n}} \left({x}\right)\:{dx}\:\mathrm{convergences}\:\mathrm{to}.. \\ $$$$ \\ $$ Answered…
Question Number 121164 by mnjuly1970 last updated on 05/Nov/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{ADVANCED}\:\:\mathrm{CALCULUS}… \\ $$$$\:\:\:\:\mathrm{If}\:\:\:\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx}\:=\lambda\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}\: \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:{find}\:\:{the}\:\:{value}\:{of}\:''\lambda''\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{m}.\mathrm{n}.\mathrm{july}.\mathrm{1970}… \\…
Question Number 55615 by Abdo msup. last updated on 28/Feb/19 $${let}\:{F}\left(\alpha\right)=\int_{\alpha} ^{\mathrm{1}+\alpha^{\mathrm{2}} } \:\:\frac{{sin}\left(\alpha{x}\right)}{\mathrm{1}+\alpha{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\frac{{dF}}{{d}\alpha}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:{lim}_{\alpha\rightarrow\mathrm{0}} \:\:{F}\left(\alpha\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 121117 by benjo_mathlover last updated on 05/Nov/20 $$\:\mathrm{J}=\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{x}+\mathrm{1}}}\:\mathrm{dx}\:? \\ $$ Answered by liberty last updated on 05/Nov/20 $$\Rightarrow\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}=\left(\mathrm{2x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)\: \\…
Question Number 121119 by benjo_mathlover last updated on 05/Nov/20 $$\mathrm{M}=\:\int\underset{−\mathrm{15}} {\overset{−\mathrm{8}} {\:}}\left(\:\frac{\mathrm{dx}}{\mathrm{x}\sqrt{\mathrm{1}−\mathrm{x}}}\right)\:?\: \\ $$ Answered by liberty last updated on 05/Nov/20 $$\mathrm{M}\:=\underset{−\mathrm{15}} {\overset{−\mathrm{8}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}^{−\mathrm{1}}…
Question Number 55577 by rahul 19 last updated on 27/Feb/19 Commented by rahul 19 last updated on 27/Feb/19 $${Prove}\:{that}: \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 186637 by Mingma last updated on 07/Feb/23 Answered by mr W last updated on 07/Feb/23 $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{2}−\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\mathrm{2}−\mathrm{cos}\:\mathrm{2}{x}}…
Question Number 55571 by maxmathsup by imad last updated on 26/Feb/19 $${let}\:{u}_{{n}} =\:\int_{\frac{\pi}{{n}+\mathrm{1}}} ^{\frac{\pi}{{n}}} \sqrt{{tan}\left({x}\right)}{dx}\:\:{with}\:{n}\geqslant\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{U}_{{n}} \:{interms}\:{of}\:{n}\:\:\:{and}\:{calculate}\:{lim}_{{n}\rightarrow+\infty\:\:} \:{U}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nature}\:{of}\:{the}\:{serie}\:\sum_{{n}\geqslant\mathrm{3}} \:{U}_{{n}} \\ $$ Commented…
Question Number 121102 by bramlexs22 last updated on 05/Nov/20 $$\:\int\:\frac{\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)}\:\mathrm{dx}\:? \\ $$ Answered by liberty last updated on 05/Nov/20 $$\Omega=\int\frac{\left(\mathrm{x}−\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}}…
Question Number 55560 by Tawa1 last updated on 26/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 27/Feb/19 $${x}={tana}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{6}} {a}×{sec}^{\mathrm{2}} {ada}}{\left({tan}^{\mathrm{4}} {a}+{sec}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{5}}{\mathrm{2}}}…