Question Number 67628 by mhmd last updated on 29/Aug/19 $$\int{x}^{{n}} {lnx}/{n}^{{x}\:} \:{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 2092 by Filup last updated on 02/Nov/15 $$\mathrm{If}\:\boldsymbol{{A}}=\langle{p}_{{x}} ,\:{p}_{{y}} ,\:{p}_{{z}} \rangle\:\mathrm{is}\:\mathrm{a}\:\mathrm{position}\:\mathrm{vector}\: \\ $$$$\mathrm{in}\:\mathrm{standard}\:\mathrm{position},\:\mathrm{vector}\:\boldsymbol{{r}}=\langle{r}_{{x}} ,\:{r}_{{y}} ,\:{r}_{{z}} \rangle \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{3}\:\mathrm{dimensional}\:\mathrm{circle} \\ $$$$\mathrm{with}\:\mathrm{focal}\:\mathrm{point}\:\mathrm{at}\:\boldsymbol{{A}},\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{vector}\:\mathrm{equation}\:\boldsymbol{{r}}\left(\theta\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is}\:\mathrm{the} \\…
Question Number 133161 by abdullahquwatan last updated on 19/Feb/21 $$\mathrm{give}\:\mathrm{me}\:\mathrm{problems}\:\mathrm{algebra}\:\mathrm{limit}\:\mathrm{function}\:\mathrm{hard} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 2089 by Yozzi last updated on 01/Nov/15 $${If}\:{x}\in\left[\mathrm{0},\mathrm{0}.\mathrm{5}\pi\right],\:{show}\:{that}\:{sinx}\geqslant{x}−\frac{\mathrm{1}}{\mathrm{6}}{x}^{\mathrm{3}} . \\ $$$${Hence}\:{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{3000}}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}/\mathrm{10}} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}−{x}+{sinx}\right)^{\mathrm{2}} }{dx}\leqslant\frac{\mathrm{2}}{\mathrm{5999}}. \\ $$ Commented by prakash jain…
Question Number 2088 by Yozzi last updated on 01/Nov/15 $${Show}\:{that}\:\forall{n}\in\mathbb{Z}^{+} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{4}} {tan}^{\mathrm{2}{n}+\mathrm{1}} \theta{d}\theta=\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{\mathrm{2}{m}}\right). \\ $$$${Deduce}\:{that}\:\underset{\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}−\mathrm{1}} }{\mathrm{2}{m}}=\mathrm{0}.\mathrm{5}{ln}\mathrm{2}. \\…
Question Number 67623 by mr W last updated on 29/Aug/19 Commented by mr W last updated on 29/Aug/19 $${find}\:{the}\:{shaded}\:{area}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67618 by mhmd last updated on 29/Aug/19 $${find}\:{the}\:{area}\:{abovnded}\:{r}={cos}\mathrm{2}\theta \\ $$ Answered by mr W last updated on 29/Aug/19 $${A}=\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{r}^{\mathrm{2}} {d}\theta}{\mathrm{2}} \\…
Question Number 67617 by aliesam last updated on 29/Aug/19 Commented by kaivan.ahmadi last updated on 29/Aug/19 $$\int\frac{\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2}} −\mathrm{2}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{6}} }+{x}}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }}{dx}=\int\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}}}−\mathrm{2}\sqrt[{\mathrm{4}}]{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{{x}^{\mathrm{2}} }}\right){dx}= \\ $$$$\int\left({x}^{\frac{−\mathrm{1}}{\mathrm{4}}} −\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{4}}}…
Question Number 67615 by TawaTawa last updated on 29/Aug/19 Commented by TawaTawa last updated on 29/Aug/19 $$\mathrm{i}\:\mathrm{did}.\:\:\:\:\:\:\mathrm{A}\:\mathrm{of}\:\mathrm{Big}\:\mathrm{semi}\:\mathrm{circle}\:=\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\pi.\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\mathrm{2}\pi \\ $$$$\mathrm{A}\:\mathrm{of}\:\mathrm{second}\:\mathrm{big}\:\mathrm{semicircle}\:\:=\:\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:\mathrm{to}\:\mathrm{do}\:\mathrm{again} \\…
Question Number 2075 by 123456 last updated on 01/Nov/15 $${f}^{−\mathrm{1}} \left({x}\right)={f}'\left({x}\right) \\ $$$${f}\left({x}\right)=?? \\ $$ Commented by Yozzi last updated on 01/Nov/15 $$\mathrm{1}={f}\left({x}\right){f}^{'} \left({x}\right) \\…