Question Number 39840 by math khazana by abdo last updated on 12/Jul/18
$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} \:\:{ln}\left(\mathrm{1}+{t}\right)\:{e}^{−{t}} {dt}\: \\ $$
Commented by abdo mathsup 649 cc last updated on 12/Jul/18
![∃?c∈ ]x+1,x^2 +1[ / A(x)=∫_(x+1) ^(x^2 +1) ln(1+t)e^(−t) dt = e^(−c) ∫_(x+1) ^(x^2 +1) ln(1+t)dt but ∫_(x+1) ^(x^2 +1) ln(1+t) dt =_(1+t =u) ∫_(x+2) ^(x^2 +2) ln(u)du =[uln(u)−u]_(x+2) ^(x^2 +2) =(x^2 +2)ln(x^(2 ) +2)−(x^2 +2) (x+2)ln(x+2) +x+2 ⇒ lim_(x→0) ∫_(x+1) ^(x^2 +1) ln(1+t)dt =0 also x→0⇒c→1 ⇒ lim_(x→0) A(x) =e^(−1) .0 =0](https://www.tinkutara.com/question/Q39872.png)
$$\left.\exists?{c}\in\:\right]{x}+\mathrm{1},{x}^{\mathrm{2}} \:+\mathrm{1}\left[\:\:/\right. \\ $$$${A}\left({x}\right)=\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right){e}^{−{t}} \:{dt}\:=\:{e}^{−{c}} \:\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right){dt}\:{but} \\ $$$$\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right)\:{dt}\:=_{\mathrm{1}+{t}\:={u}} \:\:\int_{{x}+\mathrm{2}} ^{{x}^{\mathrm{2}} \:+\mathrm{2}} \:{ln}\left({u}\right){du} \\ $$$$=\left[{uln}\left({u}\right)−{u}\right]_{{x}+\mathrm{2}} ^{{x}^{\mathrm{2}} \:+\mathrm{2}} =\left({x}^{\mathrm{2}} +\mathrm{2}\right){ln}\left({x}^{\mathrm{2}\:} +\mathrm{2}\right)−\left({x}^{\mathrm{2}} \:+\mathrm{2}\right) \\ $$$$\left({x}+\mathrm{2}\right){ln}\left({x}+\mathrm{2}\right)\:+{x}+\mathrm{2}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\int_{{x}+\mathrm{1}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} {ln}\left(\mathrm{1}+{t}\right){dt}\:=\mathrm{0}\:{also}\:{x}\rightarrow\mathrm{0}\Rightarrow{c}\rightarrow\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} {A}\left({x}\right)\:={e}^{−\mathrm{1}} .\mathrm{0}\:=\mathrm{0} \\ $$