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Question-200862

Question Number 200862 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $$\Phi \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{10}} }{dx}=_{{x}={t}^{\frac{\mathrm{1}}{\mathrm{10}}} } \frac{\mathrm{1}}{\mathrm{10}}\int_{\mathrm{0}}…

Question-200863

Question Number 200863 by sonukgindia last updated on 25/Nov/23 Answered by Mathspace last updated on 25/Nov/23 $${x}^{\mathrm{2}\pi} ={t}\:\Rightarrow{x}={t}^{\frac{\mathrm{1}}{\mathrm{2}\pi}} \:{and} \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }{dx}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}}…

lim-x-xE-x-3-x-2-sin-x-

Question Number 200864 by Rydel last updated on 25/Nov/23 $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{xE}\left({x}\right)+\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{sin}\:{x}}} \\ $$ Answered by Mathspace last updated on 25/Nov/23 $$={lim}_{{x}\rightarrow+\infty} \frac{{xE}\left({x}\right)+\mathrm{3}}{{x}\sqrt{\mathrm{1}+\frac{{sinx}}{{x}^{\mathrm{2}} }}} \\…

Question-200856

Question Number 200856 by sonukgindia last updated on 24/Nov/23 Commented by mr W last updated on 24/Nov/23 $${R}=\mathrm{1}\:{cm}\:={bigger}\:{radius} \\ $$$${r}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{cm}\:={smaller}\:{radius} \\ $$$${yellow}\:{area}={R}×{r}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{cm}^{\mathrm{2}} \\ $$ Terms…

Question-200778

Question Number 200778 by sonukgindia last updated on 23/Nov/23 Answered by MM42 last updated on 23/Nov/23 $${if}\:\:{n}=\mathrm{1}\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.={tan}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{2}}\:…