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Author: Tinku Tara

why-d-dx-0-y-e-t-dt-e-y-dy-dx-

Question Number 17480 by Arnab Maiti last updated on 06/Jul/17 $$\mathrm{why}\:\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\int_{\mathrm{0}} ^{\:\:\mathrm{y}} \mathrm{e}^{\mathrm{t}} \mathrm{dt}\right)=\mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}} \\ $$ Answered by mrW1 last updated on 06/Jul/17 $$\mathrm{let}\:\mathrm{F}\left(\mathrm{y}\right)=\int_{\mathrm{0}}…

If-f-x-is-a-periodic-function-with-period-time-t-prove-that-a-a-t-f-x-dx-is-a-indipendent-

Question Number 17479 by Arnab Maiti last updated on 06/Jul/17 $$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{periodic}\:\mathrm{function}\:\mathrm{with}\:\mathrm{period} \\ $$$$\mathrm{time}\:\mathrm{t}\:;\:\mathrm{prove}\:\mathrm{that}\int_{\mathrm{a}} ^{\:\mathrm{a}+\mathrm{t}} {f}\left({x}\right)\mathrm{d}{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{indipendent}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

A-0-0-B-2-4-and-C-6-14-if-the-triangle-has-vertices-find-the-lenght-of-the-median-drawn-from-the-vertx-C-

Question Number 148550 by mathdanisur last updated on 29/Jul/21 $$\boldsymbol{{A}}\left(\mathrm{0};\mathrm{0}\right)\:\:;\:\:\boldsymbol{{B}}\left(−\mathrm{2};\mathrm{4}\right)\:\:{and}\:\:\boldsymbol{{C}}\left(−\mathrm{6};\mathrm{14}\right) \\ $$$${if}\:{the}\:{triangle}\:{has}\:{vertices},\:{find}\:{the} \\ $$$${lenght}\:{of}\:{the}\:{median}\:{drawn}\:{from} \\ $$$${the}\:{vertx}\:\boldsymbol{{C}} \\ $$ Commented by MJS_new last updated on 29/Jul/21…

Prove-that-e-e-2-lnx-1-lnx-2-dx-e-6-2e-3-

Question Number 17475 by Arnab Maiti last updated on 06/Jul/17 $$\mathrm{Prove}\:\mathrm{that}\:\int_{\mathrm{e}} ^{\mathrm{e}^{\mathrm{2}} } \frac{\mathrm{ln}{x}}{\left(\mathrm{1}+\mathrm{ln}{x}\right)^{\mathrm{2}} }\mathrm{d}{x}=\frac{\mathrm{e}}{\mathrm{6}}\left(\mathrm{2e}−\mathrm{3}\right) \\ $$ Answered by ajfour last updated on 06/Jul/17 $$\mathrm{lnx}=\mathrm{t}\:\:\:\Rightarrow\:\:\:\frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{dt}\:\:\mathrm{or}\:\:\mathrm{dx}=\mathrm{e}^{\mathrm{t}}…