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Question-135402

Question Number 135402 by mohammad17 last updated on 12/Mar/21 Answered by Ñï= last updated on 13/Mar/21 $${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}\left(\mathrm{4}{xe}^{{x}} \mathrm{co}{s}\:{x}+{xe}^{−{x}} +{x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{1}>>\frac{\mathrm{1}}{{D}^{\mathrm{2}} −\mathrm{2}{D}+\mathrm{2}}\mathrm{4}{xe}^{{x}}…

Question-135395

Question Number 135395 by 0731619177 last updated on 12/Mar/21 Answered by Dwaipayan Shikari last updated on 12/Mar/21 $${I}\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {sin}\left({log}\left(\frac{\mathrm{1}}{{x}}\right)\right)\frac{{x}^{{b}} −{x}^{{a}} }{{log}\left({x}\right)}{dx} \\ $$$${I}'\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}}…

Question-4317

Question Number 4317 by shiv009 last updated on 10/Jan/16 Commented by prakash jain last updated on 10/Jan/16 $$\mathrm{Vertices}\:\mathrm{of}\:\mathrm{triangle}\:\left(\mathrm{1},\mathrm{3}\right),\:\left(\mathrm{4},−\mathrm{1}\right)\: \\ $$$$\mathrm{area}\:\mathrm{5}.\:\mathrm{Find}\:\mathrm{position}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \mathrm{vertex}. \\ $$$$\left(\mathrm{0},\:\mathrm{2tan}^{−\mathrm{1}} \frac{\mathrm{5}}{\mathrm{4}}\right) \\…

1-1-2-1-x-2-1-2-3-4-1-x-4-1-2-3-4-5-6-1-x-6-

Question Number 135385 by rs4089 last updated on 12/Mar/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\mathrm{1}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{3}}{\mathrm{4}}\frac{\mathrm{5}}{\mathrm{6}}\:\frac{\mathrm{1}}{{x}^{\mathrm{6}} }+….=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com