Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 28429 by abdo imad last updated on 25/Jan/18

find lim_(x→0)   ∫_(x+1) ^(2x+1)   (t^2 /(ln(1+t)))dt  .

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\int_{{x}+\mathrm{1}} ^{\mathrm{2}{x}+\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{{ln}\left(\mathrm{1}+{t}\right)}{dt}\:\:. \\ $$

Commented by abdo imad last updated on 27/Jan/18

∃ c ∈]x+1,2x+1[/ ∫_(x+1) ^(2x+1) = (1/(ln(1+c))) ∫_(x+1) ^(2x+1)  t^2 dt  = (1/3) (1/(ln(1+c))) ( (2x+1)^3 − (x+1)^3 )  we have x→0 ⇒ c→1  and  lim_(x→0 )  ∫_(x+1) ^(2x+1)  (t^2 /(ln(1+t)))dt =(1/(3ln2))×0=0   (look that the function w(t)= (1/(ln(1+t))) is continue on  [x+1,2x+1] .)

$$\left.\exists\:{c}\:\in\right]{x}+\mathrm{1},\mathrm{2}{x}+\mathrm{1}\left[/\:\int_{{x}+\mathrm{1}} ^{\mathrm{2}{x}+\mathrm{1}} =\:\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{c}\right)}\:\int_{{x}+\mathrm{1}} ^{\mathrm{2}{x}+\mathrm{1}} \:{t}^{\mathrm{2}} {dt}\right. \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{c}\right)}\:\left(\:\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} −\:\left({x}+\mathrm{1}\right)^{\mathrm{3}} \right)\:\:{we}\:{have}\:{x}\rightarrow\mathrm{0}\:\Rightarrow\:{c}\rightarrow\mathrm{1} \\ $$$${and}\:\:{lim}_{{x}\rightarrow\mathrm{0}\:} \:\int_{{x}+\mathrm{1}} ^{\mathrm{2}{x}+\mathrm{1}} \:\frac{{t}^{\mathrm{2}} }{{ln}\left(\mathrm{1}+{t}\right)}{dt}\:=\frac{\mathrm{1}}{\mathrm{3}{ln}\mathrm{2}}×\mathrm{0}=\mathrm{0} \\ $$$$\:\left({look}\:{that}\:{the}\:{function}\:{w}\left({t}\right)=\:\frac{\mathrm{1}}{{ln}\left(\mathrm{1}+{t}\right)}\:{is}\:{continue}\:{on}\right. \\ $$$$\left.\left[{x}+\mathrm{1},\mathrm{2}{x}+\mathrm{1}\right]\:.\right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com