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hi-guyz-please-i-need-ur-help-I-1-5-xe-x-3-dx-

Question Number 136211 by greg_ed last updated on 19/Mar/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{guyz}}\:!\: \\ $$$$\boldsymbol{\mathrm{please}},\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{need}}\:\boldsymbol{\mathrm{ur}}\:\boldsymbol{\mathrm{help}}\:! \\ $$$$\boldsymbol{\mathrm{I}}=\int_{\mathrm{1}} ^{\:\mathrm{5}} \boldsymbol{{xe}}^{\boldsymbol{{x}}^{\mathrm{3}} } \boldsymbol{{dx}} \\ $$ Answered by Olaf last updated…

2-to-the-power-x-2-to-the-power-x-1-3-lt-0-ease-ans-asap-

Question Number 5130 by Apoorva last updated on 16/Apr/16 $$\mathrm{2}\:{to}\:{the}\:{power}\:{x}+\mathrm{2}\:{to}\:{the}\:{power}\:\left(−{x}+\mathrm{1}\right)−\mathrm{3}<\mathrm{0}.{ease}\:{ans}\:{asap} \\ $$ Commented by prakash jain last updated on 17/Apr/16 $$\mathrm{2}^{\left({x}+\mathrm{2}\right)^{\left(−{x}+\mathrm{1}\right)−\mathrm{3}} } <\mathrm{0} \\ $$$$\mathrm{If}\:{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{2}^{\left({x}+\mathrm{2}\right)^{\left(−{x}+\mathrm{1}\right)−\mathrm{3}}…

a-Prove-that-for-any-real-constant-a-0-e-a-x-2-dx-b-If-a-and-b-are-real-constants-explain-why-we-cannot-split-the-integral-0-e-a-x-2-e-b-x-2-dx-as-the-difference-

Question Number 136197 by Ar Brandon last updated on 19/Mar/21 $$\mathrm{a}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{constant}\:\mathrm{a}\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}=\infty \\ $$$$\mathrm{b}.\:\mathrm{If}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{real}\:\mathrm{constants},\:\mathrm{explain}\:\mathrm{why}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{split}\:\mathrm{the} \\ $$$$\mathrm{integral}\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{e}^{−\frac{\mathrm{a}}{\mathrm{x}^{\mathrm{2}} }} −\mathrm{e}^{−\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }} \right)\mathrm{dx}\:\mathrm{as}\:\mathrm{the}\:\mathrm{difference}\:\int_{\mathrm{0}}…

Question-5125

Question Number 5125 by Rojaye Shegz last updated on 16/Apr/16 Commented by Rojaye Shegz last updated on 16/Apr/16 $$\mathrm{Sorry},\:\mathrm{that}\:\mathrm{was}\:\mathrm{an}\:\mathrm{error}. \\ $$$$\frac{\mathrm{p}}{\mathrm{q}}\:\mathrm{is}\:\mathrm{the}\:\left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{p}+\mathrm{q}−\mathrm{1}\right)\left(\mathrm{p}+\mathrm{q}−\mathrm{2}\right)+\mathrm{p}\right\}\mathrm{th}\:\mathrm{term} \\ $$ Commented by…