Question Number 212899 by vasil92 last updated on 26/Oct/24 Answered by MrGaster last updated on 02/Nov/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{dx}=\mathrm{1} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[{x}−\frac{{x}^{{n}−\mathrm{1}} }{\left({n}+\mathrm{1}\right)\centerdot^{\mathrm{2}}…
Question Number 212635 by Nadirhashim last updated on 19/Oct/24 $$\:\:\boldsymbol{{m}}\leqslant\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{1}\:}{\:\sqrt{\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{7}\:}}\:.\boldsymbol{{dx}}\leqslant\boldsymbol{{k}}\:\boldsymbol{{find}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{constant}} \\ $$$$\:\:\boldsymbol{{m}}\:\boldsymbol{{and}}\:\boldsymbol{{k}} \\ $$ Commented by Ghisom last updated on…
Question Number 212626 by Ghisom last updated on 19/Oct/24 $$\mathrm{let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)}} \\ $$$$\mathrm{let}\:{a},\:{b},\:{c}\:\in\mathbb{R}\:\wedge{a}<{b}<{c} \\ $$$$\Rightarrow\:{D}\left({f}\left({x}\right)\right)=\left({a},\:{b}\right)\cup\left({c},\:\infty\right) \\ $$$$\mathrm{prove}\:\underset{{a}} {\overset{{b}} {\int}}{f}\left({x}\right){dx}=\underset{{c}} {\overset{\infty} {\int}}{f}\left({x}\right){dx} \\ $$ Answered by MrGaster…
Question Number 212553 by MrGaster last updated on 17/Oct/24 $$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt[{{n}}]{\mathrm{2}}−\mathrm{1}}{\:\sqrt[{{n}}]{\mathrm{2}{n}+\mathrm{1}}}\mid\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}{n}}} {e}^{−{y}^{\mathrm{2}} } {dy}+\ldots+\int^{\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}{n}}} {e}^{−{y}^{\mathrm{2}} } {dy}\mid=? \\ $$$$ \\ $$ Terms…
Question Number 212319 by Ghisom last updated on 09/Oct/24 $$\mathrm{find} \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\:{dx} \\ $$ Commented by Spillover last updated on 10/Oct/24 $$\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}\:\:{right}? \\…
Question Number 212164 by mnjuly1970 last updated on 04/Oct/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}}\:−\:\mathrm{ln}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\right)\:=\:? \\ $$$$\: \\ $$$$ \\ $$ Answered by…
Question Number 212139 by mnjuly1970 last updated on 03/Oct/24 $$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{I}_{{n}} \:=\:\int_{−\pi} ^{\:\pi} \frac{\:\mathrm{sin}\left({nx}\:\right)}{\left(\mathrm{1}\:+\:{e}^{{x}} \right)\mathrm{sin}{x}}\:{dx}\:=?\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$ Answered…
Question Number 212099 by vahid last updated on 30/Sep/24 Answered by mehdee7396 last updated on 30/Sep/24 $$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}\:\:\:\:\:;\:{let}\:\:\:{tan}\frac{{x}}{\mathrm{2}}={u} \\ $$$$=\int\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}={ln}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}+{c} \\ $$$$={ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\right)+{c}\:\:\checkmark \\…
Question Number 212053 by universe last updated on 28/Sep/24 $$ \\ $$$$\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\:\boldsymbol{{y}}} \:\boldsymbol{{e}}^{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} } \boldsymbol{{dx}}\right)\boldsymbol{{dy}}\:+\int_{\mathrm{1}} ^{\mathrm{2}} \left(\int_{\mathrm{0}} ^{\:\mathrm{2}−\boldsymbol{{y}}} \:\boldsymbol{{e}}^{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} }…
Question Number 212023 by Spillover last updated on 27/Sep/24 Answered by Frix last updated on 27/Sep/24 $$\int\:\frac{\left({x}+\mathrm{1}\right)\mathrm{tan}\:{x}}{\left(\mathrm{1}+\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left({x}+\mathrm{1}−\frac{\mathrm{1}}{\underset{\left[{t}=\mathrm{tan}\:{x}\right]} {\underbrace{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}}}−\frac{{x}}{\underset{\left[\mathrm{by}\:\mathrm{parts}\right]} {\underbrace{\mathrm{1}+\mathrm{sin}\:\mathrm{2}{x}}}}\right){dx}= \\ $$$$… \\…