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Question Number 40097 by maxmathsup by imad last updated on 15/Jul/18
$${let}\:{f}\left({x}\right)={ln}\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}} \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:{D}_{{f}} \:\:\:\:{and}\:{find}\:{the}\:{assymptotes}\:{to}\:{C}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({x}\right)\:{and}\:{give}\:{the}\:{variation}\:{of}\:{f} \\ $$$$\left.\mathrm{3}\right)\:{give}\:{the}\:{graph}\:{of}\:{f} \\ $$$$\left.\mathrm{4}\right)\:{give}\:{the}\:{equation}\:{of}\:{tangent}\:{to}\:{C}_{{f}\:} \:\:\:{at}\:{point}\:\:{E}\left(\frac{\mathrm{1}}{\mathrm{2}},{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$\left.\mathrm{5}\right)\:{calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:. \\ $$
Commented by math khazana by abdo last updated on 03/Aug/18
![1) D_f =]−2,2[ lim_(x→−2^+ ) f(x)=−∞ lim_(x→2^− ) f(x)=+∞ so x=−2 and x=2 are assymptotes to C_f 2) we have f(x)=(1/2){ln(2+x)−ln(2−x)}⇒ f^′ (x)= (1/(2(2+x))) +(1/(2(2−x))) =(1/2){(1/(2+x)) +(1/(2−x))} =(1/2) (4/(4−x^2 )) =(2/(4−x^2 )) > 0 because −2<x<2 so f is increasing on]−2,2[ 4) equ.of tangent is y=f^′ ((1/2))(x−(1/2))+f((1/2)) f^′ ((1/2))= (2/(4−(1/4))) =((2×4)/(15)) =(8/(15)) f((1/2)) =(1/2)ln(((2+(1/2))/(2−(1/2)))) =(1/2)ln((5/3)) ⇒ y =(8/(15))(x−(1/2)) +(1/2)ln((5/3)) y=(8/(15))x −(4/(15)) +(1/2)ln((5/3)).](Q41211.png)
$$\left.\mathrm{1}\left.\right)\:{D}_{{f}} =\right]−\mathrm{2},\mathrm{2}\left[\right. \\ $$$${lim}_{{x}\rightarrow−\mathrm{2}^{+} } \:\:\:{f}\left({x}\right)=−\infty \\ $$$${lim}_{{x}\rightarrow\mathrm{2}^{−} } \:\:{f}\left({x}\right)=+\infty\:\:{so}\:\:{x}=−\mathrm{2}\:{and}\:{x}=\mathrm{2}\:{are} \\ $$$${assymptotes}\:{to}\:{C}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{2}+{x}\right)−{ln}\left(\mathrm{2}−{x}\right)\right\}\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}+{x}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}−{x}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{2}+{x}}\:+\frac{\mathrm{1}}{\mathrm{2}−{x}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{4}−{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{4}−{x}^{\mathrm{2}} }\:>\:\mathrm{0}\:{because}\:\:−\mathrm{2}<{x}<\mathrm{2}\:{so}\:{f} \\ $$$$\left.{is}\:{increasing}\:{on}\right]−\mathrm{2},\mathrm{2}\left[\right. \\ $$$$\left.\mathrm{4}\right)\:{equ}.{of}\:{tangent}\:{is}\:{y}={f}^{'} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${f}^{'} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\:\frac{\mathrm{2}}{\mathrm{4}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\frac{\mathrm{2}×\mathrm{4}}{\mathrm{15}}\:=\frac{\mathrm{8}}{\mathrm{15}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)\:\Rightarrow \\ $$$${y}\:=\frac{\mathrm{8}}{\mathrm{15}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right) \\ $$$${y}=\frac{\mathrm{8}}{\mathrm{15}}{x}\:\:−\frac{\mathrm{4}}{\mathrm{15}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{5}}{\mathrm{3}}\right). \\ $$
Commented by math khazana by abdo last updated on 03/Aug/18
![5) let I = ∫_0 ^1 ln((√((2+x)/(2−x))))dx I = (1/2)∫_0 ^1 ln(2+x)dx −(1/2) ∫_0 ^1 ln(2−x)dx but ∫_0 ^1 ln(2+x)dx =_(2+x=t) ∫_2 ^3 ln(t)dt=[tln(t)−t]_2 ^3 =3ln(3)−3−2ln(2)+2 =3ln(3)−2ln(2)−1 ∫_0 ^1 ln(2−x)dx =_(2−x=t) −∫_2 ^1 ln(t)dt =∫_1 ^2 ln(t)dt =[tln(t)−t]_1 ^2 =2ln(2)−2 +1 =2ln(2)−1 ⇒ ∫_0 ^1 f(x)dx=(3/2)ln(3)−ln(2)−(1/2) −ln(2) +(1/2) =(3/2)ln(3)−2ln(2).](Q41213.png)
$$\left.\mathrm{5}\right)\:{let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\frac{\mathrm{2}+{x}}{\mathrm{2}−{x}}}\right){dx} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{x}\right){dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}−{x}\right){dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{x}\right){dx}\:=_{\mathrm{2}+{x}={t}} \:\:\int_{\mathrm{2}} ^{\mathrm{3}} {ln}\left({t}\right){dt}=\left[{tln}\left({t}\right)−{t}\right]_{\mathrm{2}} ^{\mathrm{3}} \\ $$$$=\mathrm{3}{ln}\left(\mathrm{3}\right)−\mathrm{3}−\mathrm{2}{ln}\left(\mathrm{2}\right)+\mathrm{2}\:=\mathrm{3}{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}−{x}\right){dx}\:=_{\mathrm{2}−{x}={t}} \:\:−\int_{\mathrm{2}} ^{\mathrm{1}} {ln}\left({t}\right){dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} {ln}\left({t}\right){dt}\:=\left[{tln}\left({t}\right)−{t}\right]_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}\:+\mathrm{1} \\ $$$$=\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)−{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:−{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right). \\ $$