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Author: Tinku Tara

if-w-f-u-and-v-where-f-uu-f-vv-0-and-u-x-2-y-2-2-and-v-xy-show-that-w-xx-w-yy-0-pleas-sir-help-me-

Question Number 72883 by mhmd last updated on 04/Nov/19 $${if}\:{w}={f}\left({u}\:{and}\:{v}\right)\:{where}\:{f}_{{uu}} +{f}_{{vv}} =\mathrm{0}\:{and}\:{u}=\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)/\mathrm{2}\:{and}\:{v}={xy}\:{show}\:{that}\:{w}_{{xx}} +{w}_{{yy}} =\mathrm{0}\:? \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$ Answered by mind is power…

A-B-R-f-1-0-0-1-f-x-2-dx-A-and-0-1-xf-x-dx-B-what-is-the-integral-value-of-0-1-xf-x-f-x-1-dx-by-using-trrms-of-A-and-B-

Question Number 138415 by tugu last updated on 13/Apr/21 $${A},{B}\:\in{R},\:\:{f}\left(\mathrm{1}\right)=\mathrm{0}\:,\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\:={A}\:{and}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xf}\left({x}\right){dx}={B}\: \\ $$$${what}\:{is}\:{the}\:{integral}\:{value}\:{of}\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{xf}\left({x}\right)\left({f}\:'\left({x}\right)−\mathrm{1}\right){dx}\:{by}\:{using}\:{trrms}\:{of}\:{A}\:{and}\:{B}\:?\: \\ $$ Answered by Ar Brandon…

if-the-F-x-1-x-1-x-2t-F-t-dt-what-the-F-1-value-using-the-Leibnitz-formula-

Question Number 138407 by tugu last updated on 13/Apr/21 $${if}\:{the}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}}\underset{\mathrm{1}} {\overset{{x}} {\int}}\left(\mathrm{2}{t}−{F}\:'\left({t}\right)\right){dt}\:\:\Rightarrow\:{what}\:{the}\:{F}\:'\left(\mathrm{1}\right)\:{value}\:{using}\:{the}\:{Leibnitz}\:{formula}. \\ $$ Answered by ajfour last updated on 13/Apr/21 $${F}\:'\left({x}\right)=−\frac{{F}\left({x}\right)}{{x}}+\frac{\mathrm{1}}{{x}}\left[\mathrm{2}{x}−{F}\:'\left({x}\right)\right] \\ $$$$\left({x}+\mathrm{1}\right){F}\:'\left({x}\right)+{F}\left({x}\right)=\mathrm{2}{x} \\…

Question-7331

Question Number 7331 by rohit meena last updated on 23/Aug/16 Commented by sandy_suhendra last updated on 24/Aug/16 $${for}\:{the}\:{simetric}\:{root},\:{like}\:{y}_{\mathrm{1}} =\mathrm{2}\alpha\:{and}\:{y}_{\mathrm{2}} =\mathrm{2}\beta,\:{we}\:{can}\:{use}\:{the}\:{subtitute}\:{method} \\ $$$${y}=\mathrm{2}{x}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}{y}\:{substitute}\:{to}\:{ax}^{\mathrm{2}} +\:{bx}\:+{c}\:=\:\mathrm{0} \\ $$$${a}\left(\frac{\mathrm{1}}{\mathrm{2}}{y}\right)^{\mathrm{2}}…