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Author: Tinku Tara

x-2-3x-cos-x-dx-

Question Number 135259 by leena12345 last updated on 11/Mar/21 $$\int\left({x}^{\mathrm{2}} +\mathrm{3}{x}\right)\mathrm{cos}\:\left({x}\right){dx} \\ $$ Answered by Ar Brandon last updated on 11/Mar/21 $$\mathcal{I}=\int\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}\right)\mathrm{cosxdx} \\ $$$$\:\:\:=\left(\mathrm{x}^{\mathrm{2}}…

Question-135253

Question Number 135253 by otchereabdullai@gmail.com last updated on 11/Mar/21 Commented by mr W last updated on 11/Mar/21 $$\angle{YXZ}=\mathrm{90}°\:\Rightarrow{YX}=\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }=\mathrm{4} \\ $$$${area}\:{of}\:\Delta{XYZ}=\frac{{YX}×{XZ}}{\mathrm{2}}=\frac{{YZ}×{XN}}{\mathrm{2}} \\ $$$$\Rightarrow{XN}=\frac{\mathrm{4}×\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{12}}{\mathrm{5}} \\…

advanced-calculus-first-prove-that-1-0-1-ln-1-x-ln-1-x-x-dx-5-8-3-then-conclude-that-2-0-1-ln-2-1-

Question Number 135252 by mnjuly1970 last updated on 11/Mar/21 $$\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:\:{calculus}…. \\ $$$$\:\:\:\:{first}\:{prove}\:{that}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}=\frac{−\mathrm{5}}{\mathrm{8}}\:\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:{then}\:{conclude}\:{that}:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{{x}}{dx}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}…

Question-135254

Question Number 135254 by 0731619177 last updated on 11/Mar/21 Answered by Dwaipayan Shikari last updated on 11/Mar/21 $${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{−{ax}^{\mathrm{2}} } −{e}^{−{bx}^{\mathrm{2}} } }{{x}}\right)^{\mathrm{2}} {dx}…

Question-69710

Question Number 69710 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by mathmax by abdo last updated on 27/Sep/19 $${we}\:{have}\:{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{5}\:={x}^{\mathrm{2}} −{x}\:−\mathrm{5}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)−\mathrm{5}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}}…

Question-69709

Question Number 69709 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by kaivan.ahmadi last updated on 26/Sep/19 $${lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}{x}}{\mathrm{3}^{{x}} {ln}\mathrm{3}}={lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}}{\mathrm{3}^{{x}} \left({ln}\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ Commented…