Question Number 58639 by rahul 19 last updated on 26/Apr/19
Answered by tanmay last updated on 27/Apr/19
![F(x)=∫_0 ^x tf(t)dt (dF/dx)=∫_0 ^x (∂/(∂x ))(tf(t)dt +xf(x)(dx/dx)−0×f(0)×(d0/dx) (dF/dx)=xf(x) F(x^2 )=x^4 +x^5 =x^4 (1+x)=(x^2 )^2 (1+(√x^2 ) ) F(x)=x^2 (1+(√x))=x^2 +x^(5/2) (dF/dx)=2x+(5/2)x^(3/2) 2x+(5/2)x^(3/2) =xf(x) f(x)=2+(5/2)(√x) a) f(4)=2+(5/2)×(√4) =7 b)f(x)=2+(5/2)(√x) continous function [x>0] c)f(x)=2+(5/2)(√x) (df/dx)=(5/2)×(1/(2(√x)))=(5/(4(√x))) when x>0 (df/dx)>0 so increasing gunction](Q58640.png)
$${F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {tf}\left({t}\right){dt} \\ $$$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{{x}} \frac{\partial}{\partial{x}\:}\left({tf}\left({t}\right){dt}\:+{xf}\left({x}\right)\frac{{dx}}{{dx}}−\mathrm{0}×{f}\left(\mathrm{0}\right)×\frac{{d}\mathrm{0}}{{dx}}\right. \\ $$$$\frac{{dF}}{{dx}}={xf}\left({x}\right) \\ $$$${F}\left({x}^{\mathrm{2}} \right)={x}^{\mathrm{4}} +{x}^{\mathrm{5}} ={x}^{\mathrm{4}} \left(\mathrm{1}+{x}\right)=\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} }\:\right) \\ $$$${F}\left({x}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+\sqrt{{x}}\right)={x}^{\mathrm{2}} +{x}^{\frac{\mathrm{5}}{\mathrm{2}}} \\ $$$$\frac{{dF}}{{dx}}=\mathrm{2}{x}+\frac{\mathrm{5}}{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{2}{x}+\frac{\mathrm{5}}{\mathrm{2}}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} ={xf}\left({x}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\: \\ $$$$\left.{a}\right)\:{f}\left(\mathrm{4}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}×\sqrt{\mathrm{4}}\:=\mathrm{7} \\ $$$$\left.{b}\right){f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\:{continous}\:{function}\:\left[{x}>\mathrm{0}\right] \\ $$$$\left.{c}\right){f}\left({x}\right)=\mathrm{2}+\frac{\mathrm{5}}{\mathrm{2}}\sqrt{{x}}\: \\ $$$$\frac{{df}}{{dx}}=\frac{\mathrm{5}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{5}}{\mathrm{4}\sqrt{{x}}}\:{when}\:{x}>\mathrm{0}\:\:\frac{{df}}{{dx}}>\mathrm{0}\:{so}\:{increasing}\:{gunction} \\ $$
Commented by rahul 19 last updated on 27/Apr/19
$${thanks}\:{sir}! \\ $$