Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 74017 by mathmax by abdo last updated on 17/Nov/19

$$letf\left(x\right)={\int }_x^{x^2+3}{e}^{-xt}ln(1+e^{-xt})dtwithx>0$$ $$1)calculatef\left(x\right)$$ $$2)find{lim}_{x\to +\mathrm{\infty }}f\left(x\right).$$

Commented bymathmax by abdo last updated on 17/Nov/19

$$1)calculate{f}^{{}^{\prime }}\left(x\right)$$

Commented bymind is power last updated on 17/Nov/19

$$\mathrm{u}=\mathrm{xt}\Rightarrow \mathrm{du}=\mathrm{xdt}$$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}{\int }_{{\mathrm{x}}^2}^{{\mathrm{x}}^3+3\mathrm{x}}{\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})\mathrm{du}$$ $$\int {\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})\mathrm{du}$$ $$=-{\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})-\int \frac{{\mathrm{e}}^{-2\mathrm{u}}}{1+{\mathrm{e}}^{-\mathrm{u}}}\mathrm{du}$$ $$=-{\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})-\int \frac{({\mathrm{e}}^{-\mathrm{u}}+1)({\mathrm{e}}^{-\mathrm{u}}-1)}{(1+{\mathrm{e}}^{-\mathrm{u}})}-\int \frac{{\mathrm{e}}^{\mathrm{u}}}{{\mathrm{e}}^{\mathrm{u}}+1}\mathrm{du}$$ $$=-{\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})+\int (-{\mathrm{e}}^{-\mathrm{u}}+1)\mathrm{du}-\ln ({\mathrm{e}}^{\mathrm{u}}+1)$$ $$=-{\mathrm{e}}^{-\mathrm{u}}\ln (1+{\mathrm{e}}^{-\mathrm{u}})+{\mathrm{e}}^{-\mathrm{u}}+\mathrm{u}-\ln ({\mathrm{e}}^{\mathrm{u}}+1)+\mathrm{c}$$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}\{[-{\mathrm{e}}^{-({\mathrm{x}}^3+3\mathrm{x})}\ln (1+{\mathrm{e}}^{-{\mathrm{x}}^3-3\mathrm{x}})+{\mathrm{e}}^{-({\mathrm{x}}^3+3\mathrm{x})}+({\mathrm{x}}^3+3\mathrm{x})-\ln ({\mathrm{e}}^{{\mathrm{x}}^3+3\mathrm{x}}+1)]$$ $$+{\mathrm{e}}^{-{\mathrm{x}}^2}\ln (1+{\mathrm{e}}^{-{\mathrm{x}}^2})-{\mathrm{e}}^{-{\mathrm{x}}^2}-{\mathrm{x}}^2+\ln ({\mathrm{e}}^{{\mathrm{x}}^2}+1))]$$ $$\lim \mathrm{f}\left(\mathrm{x}\right)=0$$

Commented bymathmax by abdo last updated on 21/Nov/19

$$1)wehavef\left(x\right)={\int }_x^{x^2+3}{e}^{-xt}ln(1+e^{-xt})dt(x>0)changement$$ $$xt=ugivef\left(x\right)=\frac{1}{x}{\int }_{x^2}^{x^3+3x}{e}^{-u}ln(1+e^{-u})du\Rightarrow$$ $$xf\left(x\right)={\int }_{x^2}^{x^3+3x}{e}^{-u}ln(1+e^{-u})dy{=}_{bypsrts}{[-e^{-u}ln(1+e^{-u})]}_{x^2}^{x^3+3x}$$ $$-{\int }_{x^2}^{x^3+3x}(-e^{-u})\times \frac{-e^{-u}}{1+e^{-u}}du=e^{-x^2}ln(1+e^{-x^2})$$ $$-e^{-(x^3+3x)}ln(1+e^{-(x^3+3x)})+{\int }_{x^2}^{x^3+3x}\frac{e^{-2u}}{1+e^{-u}}duwehave$$ $${\int }_{x^2}^{x^3+3x}\frac{e^{-2u}}{1+e^{-u}}du={\int }_{x^2}^{x^3+3x}\frac{1}{e^{2u}+e^u}du{=}_{e^u=z}{\int }_{e^{x^2}}^{e^{x^3+3x}}\frac{1}{z^2+z}\frac{dz}{z}$$ $$={\int }_{e^{x^2}}^{e^{x^3+3x}}\frac{dz}{z^2(z+1)}decompositionofF\left(z\right)=\frac{1}{z^2(z+1)}$$ $$F\left(z\right)=\frac{a}{z}+\frac{b}{z^2}+\frac{c}{z+1}$$ $$b=1,c=1\Rightarrow F\left(z\right)=\frac{a}{z}+\frac{1}{z^2}+\frac{1}{z+1}$$ $${lim}_{z\to +\mathrm{\infty }}zF\left(z\right)=0=a+c\Rightarrow a=-c=-1\Rightarrow F\left(z\right)=-\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z+1}$$ $${\int }_{e^{x^2}}^{e^{x^3+3x}}\frac{dz}{z^2(z+1)}={[ln\mid \frac{z+1}{z}\mid ]}_{e^{x^2}}^{e^{x^3+3x}}+{[-\frac{1}{z}]}_{e^{x^2}}^{e^{x^3+3x}}$$ $$=ln\left(\frac{1+e^{x^3+3x}}{e^{x^3+3x}}\right)-ln\left(\frac{1+e^{x^2}}{e^{x^2}}\right)+(e^{-x^2}-e^{-(x^3+3x)})\Rightarrow$$ $$f\left(x\right)=\frac{1}{x}\{e^{-x^2}ln(1+e^{-x^2})-e^{-(x^3+3x)}ln(1+e^{-(x^3+3x)})$$ $$ln\left(\frac{1+e^{x^3+3x}}{e^{x^3+3x}}\right)-ln\left(\frac{1+e^{x^2}}{e^{x^2}}\right)+e^{-x^2}-e^{-(x^3+3x)}\}$$ $$2){lim}_{x\to +\mathrm{\infty }}f\left(x\right)=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com